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Reduce the L to reduce the resistive loss - the essence of L
is the energy stored in its current carrying, and it is the current that causes I^2 R losses. The energy stored in the C is static. (Yes, there are some losses in polarising the dielectrics but these are small enough to be ignored) "Paul Burridge" wrote in message ... ISTR that one can improve Q in resonant tanks by having a low L-C ratio. Or was it high L-C ratio. I can't remember but need to know. |
Reduce the L to reduce the resistive loss - the essence of L
is the energy stored in its current carrying, and it is the current that causes I^2 R losses. The energy stored in the C is static. (Yes, there are some losses in polarising the dielectrics but these are small enough to be ignored) "Paul Burridge" wrote in message ... ISTR that one can improve Q in resonant tanks by having a low L-C ratio. Or was it high L-C ratio. I can't remember but need to know. |
Tank circuits: achieving maximum Q
Hi guys,
ISTR that one can improve Q in resonant tanks by having a low L-C ratio. Or was it high L-C ratio. I can't remember but need to know. Can any kind soul help me out? Thanks. p. -- The BBC: Licensed at public expense to spread lies. |
Airy,
What you said would be relevant only if you were trying to determine circuit losses due to "unloaded" Q of the components. I believe Paul is trying to determine the 'loaded" Q in order to obtain best selectivity (narrowest bandwidth). Is this true Paul? In order to obtain maximum selectivity, the loaded Q needs to be as high as possible. In the case of a resonant 'tank', the tank reactances are loaded by the external environment. The circuit Q (or 'loaded' Q) in this case is Q=R/X. In order to maximize loaded Q, the X term (reactance) needs to be minimized. This means low L and high C. In any case, the actual circuit losses will be a function of the ratio of unloaded Q (Q of the components) to loaded Q. The higher the unloaded Q of the components, the lower the losses in the circuit. Joe W3JDR "Airy R. Bean" wrote in message ... Reduce the L to reduce the resistive loss - the essence of L is the energy stored in its current carrying, and it is the current that causes I^2 R losses. The energy stored in the C is static. (Yes, there are some losses in polarising the dielectrics but these are small enough to be ignored) "Paul Burridge" wrote in message ... ISTR that one can improve Q in resonant tanks by having a low L-C ratio. Or was it high L-C ratio. I can't remember but need to know. |
Airy,
What you said would be relevant only if you were trying to determine circuit losses due to "unloaded" Q of the components. I believe Paul is trying to determine the 'loaded" Q in order to obtain best selectivity (narrowest bandwidth). Is this true Paul? In order to obtain maximum selectivity, the loaded Q needs to be as high as possible. In the case of a resonant 'tank', the tank reactances are loaded by the external environment. The circuit Q (or 'loaded' Q) in this case is Q=R/X. In order to maximize loaded Q, the X term (reactance) needs to be minimized. This means low L and high C. In any case, the actual circuit losses will be a function of the ratio of unloaded Q (Q of the components) to loaded Q. The higher the unloaded Q of the components, the lower the losses in the circuit. Joe W3JDR "Airy R. Bean" wrote in message ... Reduce the L to reduce the resistive loss - the essence of L is the energy stored in its current carrying, and it is the current that causes I^2 R losses. The energy stored in the C is static. (Yes, there are some losses in polarising the dielectrics but these are small enough to be ignored) "Paul Burridge" wrote in message ... ISTR that one can improve Q in resonant tanks by having a low L-C ratio. Or was it high L-C ratio. I can't remember but need to know. |
On Sun, 21 Mar 2004 16:02:18 GMT, "W3JDR" wrote:
Airy, What you said would be relevant only if you were trying to determine circuit losses due to "unloaded" Q of the components. I believe Paul is trying to determine the 'loaded" Q in order to obtain best selectivity (narrowest bandwidth). Is this true Paul? Some clarification is necessary! The application is the tank in a frequency multiplier. I am seeking to select for the 5th harmonic. Therefore, the tank needs to have as little loss as possible given the fact that the 5th will be way down dB-wise on the fundamental. I can't afford to attenuate it too much as it's already weak to begin with. Ergo, I need the lowest loss components and the best selectivity for the desired 5th harmonic. Thanks, p. -- The BBC: Licensed at public expense to spread lies. |
On Sun, 21 Mar 2004 16:02:18 GMT, "W3JDR" wrote:
Airy, What you said would be relevant only if you were trying to determine circuit losses due to "unloaded" Q of the components. I believe Paul is trying to determine the 'loaded" Q in order to obtain best selectivity (narrowest bandwidth). Is this true Paul? Some clarification is necessary! The application is the tank in a frequency multiplier. I am seeking to select for the 5th harmonic. Therefore, the tank needs to have as little loss as possible given the fact that the 5th will be way down dB-wise on the fundamental. I can't afford to attenuate it too much as it's already weak to begin with. Ergo, I need the lowest loss components and the best selectivity for the desired 5th harmonic. Thanks, p. -- The BBC: Licensed at public expense to spread lies. |
High L/C ratio increases Q.
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High L/C ratio increases Q.
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Oops - L/R ISTR!
"K9SQG" wrote in message ... High L/C ratio increases Q. |
Oops - L/R ISTR!
"K9SQG" wrote in message ... High L/C ratio increases Q. |
Circuit Q = omega*L/R. Reducing L has little or no effect on Q because,
after winding L, you will find R has decreased in about the same proportion. The fewer the number of turns, the shorter the length of wire, and the lower the resistance. The ratio of L to C also has little effect on circuit Q because the intrinsic Q of capacitors is usually an order of magnitude or more greater than Q of L. L and C values of a tuned circuit are usually selected by the reactances required of them at resonance for reasons independent of circuit Q. Eg., the reactance of L and C may be required to be 300 ohms at resonance because other components will have to be connected to them. Usually it is the value of C which controls the value of L. C may have to be trimmer. If it is a fixed value it will have to conform to a preferred series of values and tolerances. If it is too small it will get lost in stray and other circuit capacitances. IMPORTANT - Intrinsic Q of a solenoid is directly proportional only to its physical size. Double all dimensions, including wire diameter, and Q is doubled. Its the the amount of space you have which decides the value of Q. And there's a similar relationship even for magnetic cored components. If you havn't got the room then you will have to put up with a low coil Q. And it's always lower than what you think it is. Its impossible to measure in situ. Spice is of no help. Don't forget that a tuned circuit is never used in isolation. If it is used as a filter in transistor collector circuit then it forms only part of the transistor load. And whatever else is connected will reduce the effective circuit Q. It could be that it doesn't matter what the intrinsic Q of the coil may be provided it is not ridiculously low. Which I suspect to be true in your case. You may be doing your nut about nothing. To put it crudely, it is seldom that coil Q matters. Nearly always whatever you've got is good enough. If in a particular application you might think it does then you are barking up the wrong tree. ---- Reg, G4FGQ |
Circuit Q = omega*L/R. Reducing L has little or no effect on Q because,
after winding L, you will find R has decreased in about the same proportion. The fewer the number of turns, the shorter the length of wire, and the lower the resistance. The ratio of L to C also has little effect on circuit Q because the intrinsic Q of capacitors is usually an order of magnitude or more greater than Q of L. L and C values of a tuned circuit are usually selected by the reactances required of them at resonance for reasons independent of circuit Q. Eg., the reactance of L and C may be required to be 300 ohms at resonance because other components will have to be connected to them. Usually it is the value of C which controls the value of L. C may have to be trimmer. If it is a fixed value it will have to conform to a preferred series of values and tolerances. If it is too small it will get lost in stray and other circuit capacitances. IMPORTANT - Intrinsic Q of a solenoid is directly proportional only to its physical size. Double all dimensions, including wire diameter, and Q is doubled. Its the the amount of space you have which decides the value of Q. And there's a similar relationship even for magnetic cored components. If you havn't got the room then you will have to put up with a low coil Q. And it's always lower than what you think it is. Its impossible to measure in situ. Spice is of no help. Don't forget that a tuned circuit is never used in isolation. If it is used as a filter in transistor collector circuit then it forms only part of the transistor load. And whatever else is connected will reduce the effective circuit Q. It could be that it doesn't matter what the intrinsic Q of the coil may be provided it is not ridiculously low. Which I suspect to be true in your case. You may be doing your nut about nothing. To put it crudely, it is seldom that coil Q matters. Nearly always whatever you've got is good enough. If in a particular application you might think it does then you are barking up the wrong tree. ---- Reg, G4FGQ |
Absolutely not!!!!!!!!!!!!!!!!!!
He needs best possible loaded Q for maximum selectivity! Good unloaded Q will help minimuze circuit losses, but won't help with selectivity. Joe W3JDR "Airy R. Bean" wrote in message ... Then it is unloaded Q that you are interested in. "Paul Burridge" wrote in message ... On Sun, 21 Mar 2004 16:02:18 GMT, "W3JDR" wrote: Airy, What you said would be relevant only if you were trying to determine circuit losses due to "unloaded" Q of the components. I believe Paul is trying to determine the 'loaded" Q in order to obtain best selectivity (narrowest bandwidth). Is this true Paul? Some clarification is necessary! The application is the tank in a frequency multiplier. I am seeking to select for the 5th harmonic. Therefore, the tank needs to have as little loss as possible given the fact that the 5th will be way down dB-wise on the fundamental. I can't afford to attenuate it too much as it's already weak to begin with. Ergo, I need the lowest loss components and the best selectivity for the desired 5th harmonic. |
Absolutely not!!!!!!!!!!!!!!!!!!
He needs best possible loaded Q for maximum selectivity! Good unloaded Q will help minimuze circuit losses, but won't help with selectivity. Joe W3JDR "Airy R. Bean" wrote in message ... Then it is unloaded Q that you are interested in. "Paul Burridge" wrote in message ... On Sun, 21 Mar 2004 16:02:18 GMT, "W3JDR" wrote: Airy, What you said would be relevant only if you were trying to determine circuit losses due to "unloaded" Q of the components. I believe Paul is trying to determine the 'loaded" Q in order to obtain best selectivity (narrowest bandwidth). Is this true Paul? Some clarification is necessary! The application is the tank in a frequency multiplier. I am seeking to select for the 5th harmonic. Therefore, the tank needs to have as little loss as possible given the fact that the 5th will be way down dB-wise on the fundamental. I can't afford to attenuate it too much as it's already weak to begin with. Ergo, I need the lowest loss components and the best selectivity for the desired 5th harmonic. |
I don't agree. The selection of a harmonic is a
signal processing function and not one of power transfer. "W3JDR" wrote in message ... Absolutely not!!!!!!!!!!!!!!!!!! He needs best possible loaded Q for maximum selectivity! Good unloaded Q will help minimuze circuit losses, but won't help with selectivity. "Airy R. Bean" wrote in message ... Then it is unloaded Q that you are interested in. "Paul Burridge" wrote in message ... I am seeking to select for the 5th harmonic. |
I don't agree. The selection of a harmonic is a
signal processing function and not one of power transfer. "W3JDR" wrote in message ... Absolutely not!!!!!!!!!!!!!!!!!! He needs best possible loaded Q for maximum selectivity! Good unloaded Q will help minimuze circuit losses, but won't help with selectivity. "Airy R. Bean" wrote in message ... Then it is unloaded Q that you are interested in. "Paul Burridge" wrote in message ... I am seeking to select for the 5th harmonic. |
Paul, in approaching the problem from your viewpoint havn't you set yourself
the task of winding an inductor to have a particular value of Q ? If you intend to use a solenoid then Q can be increased only by increasing its physical size without changing its proportions too much. Utimately you will need to know what is the Q of a particular size coil, number of turns, wire gauge, etc. It will be reduced by its proximity to other components and circuit board by some indeterminate amount. I think you should stop and check whether you have room for the coil in the equipment space available. ;o) Program SOLNOID2 may be of assistance in this onerous task. Download in a few seconds from website below and run immediately. ---- .................................................. .......... Regards from Reg, G4FGQ For Free Radio Design Software go to http://www.btinternet.com/~g4fgq.regp .................................................. .......... |
Paul, in approaching the problem from your viewpoint havn't you set yourself
the task of winding an inductor to have a particular value of Q ? If you intend to use a solenoid then Q can be increased only by increasing its physical size without changing its proportions too much. Utimately you will need to know what is the Q of a particular size coil, number of turns, wire gauge, etc. It will be reduced by its proximity to other components and circuit board by some indeterminate amount. I think you should stop and check whether you have room for the coil in the equipment space available. ;o) Program SOLNOID2 may be of assistance in this onerous task. Download in a few seconds from website below and run immediately. ---- .................................................. .......... Regards from Reg, G4FGQ For Free Radio Design Software go to http://www.btinternet.com/~g4fgq.regp .................................................. .......... |
Loaded Q or unloaded Q? If the load is in parallel with the tank and
a fixed value, decrease the reactance of the tank elements. If the fixed resistive load is in series with the L and C, increase the reactance of the tank elements. Generally you should design the Q to fit the task. (You could expand on that: generally you should design the circuit to fit the task...) Unloaded Q is increased by things like using the right core material and right winding techniques. There's not a simple answer. For air-core coils, the larger the coil the higher the Q possible, up to the point where radiation becomes significant. You might hear that helical resonators are very high Q, but actually the same coil in freespace will be higher Q, so long as it's not so large it radiates a lot. Reg has a program that estimates unloaded Q of air-core RF coils. It's a fairly complex subject...don't expect one answer to fit all situations. Cheers, Tom Paul Burridge wrote in message . .. Hi guys, ISTR that one can improve Q in resonant tanks by having a low L-C ratio. Or was it high L-C ratio. I can't remember but need to know. Can any kind soul help me out? Thanks. p. |
Loaded Q or unloaded Q? If the load is in parallel with the tank and
a fixed value, decrease the reactance of the tank elements. If the fixed resistive load is in series with the L and C, increase the reactance of the tank elements. Generally you should design the Q to fit the task. (You could expand on that: generally you should design the circuit to fit the task...) Unloaded Q is increased by things like using the right core material and right winding techniques. There's not a simple answer. For air-core coils, the larger the coil the higher the Q possible, up to the point where radiation becomes significant. You might hear that helical resonators are very high Q, but actually the same coil in freespace will be higher Q, so long as it's not so large it radiates a lot. Reg has a program that estimates unloaded Q of air-core RF coils. It's a fairly complex subject...don't expect one answer to fit all situations. Cheers, Tom Paul Burridge wrote in message . .. Hi guys, ISTR that one can improve Q in resonant tanks by having a low L-C ratio. Or was it high L-C ratio. I can't remember but need to know. Can any kind soul help me out? Thanks. p. |
"The selection of a harmonic is a signal processing function and not one of power transfer."
Exactly Airy. The output of the multiplier stage, which contains many frequencies, needs to be processed in order to select the desired component and reject all others to the best extent possible. This is an indirect way of saying you need a very selective tuned circuit. The selectivity of the tuned circuit is defined by it's loaded Q; ie, the Q of the reactances in network loaded when by their operating environment. This means that for best selectivity, the reactances of the L and C should be as low as possible, which in turn means that the C should be as large as can be achieved, consistent with realizably small L and component (unloaded) Q. Unfortunately, as the ratio of unloaded component Q to loaded circuit Q gets small, the losses go up. This implies that there's a tradeoff to be made between selectivity and loss. Thus it is in the world of real components - in the world of DSP things can be different. If you want high selectivity, you have to balance it against tolerable circuit losses, at least with real components. Joe W3JDR "Airy R. Bean" wrote in message ... I don't agree. The selection of a harmonic is a signal processing function and not one of power transfer. "W3JDR" wrote in message ... Absolutely not!!!!!!!!!!!!!!!!!! He needs best possible loaded Q for maximum selectivity! Good unloaded Q will help minimuze circuit losses, but won't help with selectivity. "Airy R. Bean" wrote in message ... Then it is unloaded Q that you are interested in. "Paul Burridge" wrote in message ... I am seeking to select for the 5th harmonic. |
"The selection of a harmonic is a signal processing function and not one of power transfer."
Exactly Airy. The output of the multiplier stage, which contains many frequencies, needs to be processed in order to select the desired component and reject all others to the best extent possible. This is an indirect way of saying you need a very selective tuned circuit. The selectivity of the tuned circuit is defined by it's loaded Q; ie, the Q of the reactances in network loaded when by their operating environment. This means that for best selectivity, the reactances of the L and C should be as low as possible, which in turn means that the C should be as large as can be achieved, consistent with realizably small L and component (unloaded) Q. Unfortunately, as the ratio of unloaded component Q to loaded circuit Q gets small, the losses go up. This implies that there's a tradeoff to be made between selectivity and loss. Thus it is in the world of real components - in the world of DSP things can be different. If you want high selectivity, you have to balance it against tolerable circuit losses, at least with real components. Joe W3JDR "Airy R. Bean" wrote in message ... I don't agree. The selection of a harmonic is a signal processing function and not one of power transfer. "W3JDR" wrote in message ... Absolutely not!!!!!!!!!!!!!!!!!! He needs best possible loaded Q for maximum selectivity! Good unloaded Q will help minimuze circuit losses, but won't help with selectivity. "Airy R. Bean" wrote in message ... Then it is unloaded Q that you are interested in. "Paul Burridge" wrote in message ... I am seeking to select for the 5th harmonic. |
On Sun, 21 Mar 2004 18:54:19 +0000 (UTC), "Reg Edwards"
wrote: Paul, in approaching the problem from your viewpoint havn't you set yourself the task of winding an inductor to have a particular value of Q ? If you intend to use a solenoid then Q can be increased only by increasing its physical size without changing its proportions too much. Utimately you will need to know what is the Q of a particular size coil, number of turns, wire gauge, etc. It will be reduced by its proximity to other components and circuit board by some indeterminate amount. I think you should stop and check whether you have room for the coil in the equipment space available. ;o) Program SOLNOID2 may be of assistance in this onerous task. Download in a few seconds from website below and run immediately. Reg, SOLNOID2 has been withdrawn from your site IIRC. I *had* been using it to great effect, but you presumably made some improvements, implemented them, and renamed it SOLNOID3 which is what I now use. Great program! I'm still none the wiser as to whether it's better to have big-L || small C or vice versa, though. :-/ BTW, Reg - can you write a program to work out how I'm going to afford my Council Tax this year? Thanks! :-) -- The BBC: Licensed at public expense to spread lies. |
On Sun, 21 Mar 2004 18:54:19 +0000 (UTC), "Reg Edwards"
wrote: Paul, in approaching the problem from your viewpoint havn't you set yourself the task of winding an inductor to have a particular value of Q ? If you intend to use a solenoid then Q can be increased only by increasing its physical size without changing its proportions too much. Utimately you will need to know what is the Q of a particular size coil, number of turns, wire gauge, etc. It will be reduced by its proximity to other components and circuit board by some indeterminate amount. I think you should stop and check whether you have room for the coil in the equipment space available. ;o) Program SOLNOID2 may be of assistance in this onerous task. Download in a few seconds from website below and run immediately. Reg, SOLNOID2 has been withdrawn from your site IIRC. I *had* been using it to great effect, but you presumably made some improvements, implemented them, and renamed it SOLNOID3 which is what I now use. Great program! I'm still none the wiser as to whether it's better to have big-L || small C or vice versa, though. :-/ BTW, Reg - can you write a program to work out how I'm going to afford my Council Tax this year? Thanks! :-) -- The BBC: Licensed at public expense to spread lies. |
On Sun, 21 Mar 2004 17:44:09 +0000 (UTC), "Reg Edwards"
wrote: To put it crudely, it is seldom that coil Q matters. Nearly always whatever you've got is good enough. If in a particular application you might think it does then you are barking up the wrong tree. Be that as it may, Reg, how satisfied are you that your program accurately calculates Q in the inductors in which it predicts characteristics? For example, it is accepted by all (I hope) that Q(unloaded) is at a maxium when the length of a coil equals its diameter. Your program doesn't seem to reflect this fact. How come? p. -- The BBC: Licensed at public expense to spread lies. |
On Sun, 21 Mar 2004 17:44:09 +0000 (UTC), "Reg Edwards"
wrote: To put it crudely, it is seldom that coil Q matters. Nearly always whatever you've got is good enough. If in a particular application you might think it does then you are barking up the wrong tree. Be that as it may, Reg, how satisfied are you that your program accurately calculates Q in the inductors in which it predicts characteristics? For example, it is accepted by all (I hope) that Q(unloaded) is at a maxium when the length of a coil equals its diameter. Your program doesn't seem to reflect this fact. How come? p. -- The BBC: Licensed at public expense to spread lies. |
Let me clarify. Because Q is largely out of your control, the logical way
of circuit design is FIRST OF ALL to allocate a practical, reasonably attainable value of Q to the inductor, taking the SPACE AVAILABLE into account. Then design the remainder of the circuit around it to meet the required objectives. You really have no alternative. Choosing a starting value for Q without knowing the inductance depends entirely on experience and visual imagination. But you will find program SOLNOID2 very useful in getting you in the right ballpark - only after you decide on the space and clearance available for a coil. I'm still toying with the idea of using an oscillator locked to the 5th harmonic. Q and size of the coil don't matter two hoots. And you have all the output you could possibly want. On second thoughts I would probably have gone about the whole job in an entirely different manner. ;o) ---- Reg, G4FGQ |
Let me clarify. Because Q is largely out of your control, the logical way
of circuit design is FIRST OF ALL to allocate a practical, reasonably attainable value of Q to the inductor, taking the SPACE AVAILABLE into account. Then design the remainder of the circuit around it to meet the required objectives. You really have no alternative. Choosing a starting value for Q without knowing the inductance depends entirely on experience and visual imagination. But you will find program SOLNOID2 very useful in getting you in the right ballpark - only after you decide on the space and clearance available for a coil. I'm still toying with the idea of using an oscillator locked to the 5th harmonic. Q and size of the coil don't matter two hoots. And you have all the output you could possibly want. On second thoughts I would probably have gone about the whole job in an entirely different manner. ;o) ---- Reg, G4FGQ |
In article , Paul Burridge
writes: On Sun, 21 Mar 2004 16:02:18 GMT, "W3JDR" wrote: Airy, What you said would be relevant only if you were trying to determine circuit losses due to "unloaded" Q of the components. I believe Paul is trying to determine the 'loaded" Q in order to obtain best selectivity (narrowest bandwidth). Is this true Paul? Some clarification is necessary! The application is the tank in a frequency multiplier. I am seeking to select for the 5th harmonic. Therefore, the tank needs to have as little loss as possible given the fact that the 5th will be way down dB-wise on the fundamental. I can't afford to attenuate it too much as it's already weak to begin with. Ergo, I need the lowest loss components and the best selectivity for the desired 5th harmonic. Thanks, Design of that is a two-step process. First, you need to establish the impedance (or admittance) of both source and load. For a parallel-resonant circuit selectivity device, they are both in parallel with the unloaded Q of the resonant circuit. For a series-resonant selectivity device, they are in series with it. With vacuum tube and FET circuits, staying in the linear I/O bias region, the first step is easy. Just parallel drain or plate resistance and a gate or grid circuit resistance for a parallel-resonant circuit. With bipolar transistors, the base resistance is quite low compared to tube (valve) and gate inputs, must be impedance-magnitude adjusted such as with tapping down on an inductor. There are several other ways to do impedance-magnitude adjustment; that coil tap is a very common one. Once into a non-linear operation region the overall impedances become dynamic rather than static and depend on drive level and the amount of time an input spends in non-linear region versus the linear region. Using digital logic devices means that the non-linear regions are above saturation and below cutoff but the saturation does not behave the same as with valve grid current run positive on part of the cycle. That is NOT easy to calculate and quite complex for those who take the time to do that. For home workshop design efforts in getting to the task in the most expeditious way, simply Cut And Try. Reg Edwards pointed that out semi-directly. :-) The second step is to select and inductor with, for your needs in being selective to that elusive 5th harmonic, of the highest Q_u (unloaded or "not in-circuit" quality factor) that will fit in the space (physical space) you've alloted. That selection is a compromise in size - cylindrical or "solenoid" cores mean (as Reg said) the bigger the better. I'll also add "the bigger the wire diameter, the higher the Q" for the same coil former size. For iron powder toroid forms, the powder mix is important as well as the size as well as the wire size. Just from memory of a few years ago, a Micrometals T37-6 core (the "37" meaning 0.37 or 3/8ths inch, powder mix 6) will yield a Q_u of 80 minimum at 18 MHz using the largest wire that will fit through the center hole. Q_u at 17 MHz will be very close to that. Unloaded Q is a result of many factors and all of those can be modified by things such as the dielectric material of a solenoidal former and the presence of adjacent shielding and even dielectric material. For the easiest application and less time worrying nit- picky details, pick an iron powder core toroidal form...such can be smaller than cylindrical formers allow and are much more forgiving of adjacent/nearby objects. But only if space is at a premium. Small toroidal forms can be difficult to wind for some and multi-turn inductors need lots of wire which can build up in the center hole, precluding use of larger magnet wire diameters. Part of the second step is to combine what you know (or guess) in the first step with a selection of inductance and capacitance for resonance. As others have said, inductive Q_u is the determining factor at HF and capacitive Q_u will be at least 10 times higher, probably in the neighborhood of 500 to 1000 for ceramic or mica capacitors. Do a quick model of the resonant circuit "resistance" (actually the magnitude of impedance) at parallel resonance - parallel the (inductive Q_u times inductive reactance) and the (capacitive Q_u times the same reactance since capacitive reactance is equal to inductive reactance at resonance). Parallel that with the source and load impedance magnitude combined magnitude and you have the total magnitude at resonance. This can be very quick to do with a scientific pocket calculator. To verify the selectivity, run the whole thing again at adjacent harmonics to get the total magnitude of impedance there. Those off-resonance L-C circuit magnitudes can use just the reactance as an approximate step and be very close to those using the unloaded L-C Qs. The ratio of magnitudes on-resonance versus off-resonance will give you a picture of the selectivity possible. If that "doesn't seem to be good," THEN pick a different L:C ratio and do it again...but use what you know about the inductive Q_u at that different inductance. Compare the new on-resonance impedance magnitude to the adjacent off-resonance magnitudes. Is that magnitude ratio worse than before? If so use an opposite L:C ratio. If better, try the same-direction different L:C ratio and compare that. If better, repeat. If worse, hold on the previous L-C combination...you are zeroing-in on what is useful. SELECTIVITY is the thing desired in your application and the relatively-simple calculation of magnitudes and resulting ratios will point in the right direction for something to try in hardware. Selectivity is needed because the lower harmonics have more energy than the 5th. If stumped for a starting L and C value, try the literature on previous multiplier designs as a starting point...then dance through this two-step procedure. In practical hardware, many others besides myself have been led astray by simplistic "L:C ratio Q determination rules" that can be just the reverse. Lots of those old maxims were generated way back in time of large "coils and condensers" one needed both hands to pick up... Len Anderson retired (from regular hours) electronic engineer person |
In article , Paul Burridge
writes: On Sun, 21 Mar 2004 16:02:18 GMT, "W3JDR" wrote: Airy, What you said would be relevant only if you were trying to determine circuit losses due to "unloaded" Q of the components. I believe Paul is trying to determine the 'loaded" Q in order to obtain best selectivity (narrowest bandwidth). Is this true Paul? Some clarification is necessary! The application is the tank in a frequency multiplier. I am seeking to select for the 5th harmonic. Therefore, the tank needs to have as little loss as possible given the fact that the 5th will be way down dB-wise on the fundamental. I can't afford to attenuate it too much as it's already weak to begin with. Ergo, I need the lowest loss components and the best selectivity for the desired 5th harmonic. Thanks, Design of that is a two-step process. First, you need to establish the impedance (or admittance) of both source and load. For a parallel-resonant circuit selectivity device, they are both in parallel with the unloaded Q of the resonant circuit. For a series-resonant selectivity device, they are in series with it. With vacuum tube and FET circuits, staying in the linear I/O bias region, the first step is easy. Just parallel drain or plate resistance and a gate or grid circuit resistance for a parallel-resonant circuit. With bipolar transistors, the base resistance is quite low compared to tube (valve) and gate inputs, must be impedance-magnitude adjusted such as with tapping down on an inductor. There are several other ways to do impedance-magnitude adjustment; that coil tap is a very common one. Once into a non-linear operation region the overall impedances become dynamic rather than static and depend on drive level and the amount of time an input spends in non-linear region versus the linear region. Using digital logic devices means that the non-linear regions are above saturation and below cutoff but the saturation does not behave the same as with valve grid current run positive on part of the cycle. That is NOT easy to calculate and quite complex for those who take the time to do that. For home workshop design efforts in getting to the task in the most expeditious way, simply Cut And Try. Reg Edwards pointed that out semi-directly. :-) The second step is to select and inductor with, for your needs in being selective to that elusive 5th harmonic, of the highest Q_u (unloaded or "not in-circuit" quality factor) that will fit in the space (physical space) you've alloted. That selection is a compromise in size - cylindrical or "solenoid" cores mean (as Reg said) the bigger the better. I'll also add "the bigger the wire diameter, the higher the Q" for the same coil former size. For iron powder toroid forms, the powder mix is important as well as the size as well as the wire size. Just from memory of a few years ago, a Micrometals T37-6 core (the "37" meaning 0.37 or 3/8ths inch, powder mix 6) will yield a Q_u of 80 minimum at 18 MHz using the largest wire that will fit through the center hole. Q_u at 17 MHz will be very close to that. Unloaded Q is a result of many factors and all of those can be modified by things such as the dielectric material of a solenoidal former and the presence of adjacent shielding and even dielectric material. For the easiest application and less time worrying nit- picky details, pick an iron powder core toroidal form...such can be smaller than cylindrical formers allow and are much more forgiving of adjacent/nearby objects. But only if space is at a premium. Small toroidal forms can be difficult to wind for some and multi-turn inductors need lots of wire which can build up in the center hole, precluding use of larger magnet wire diameters. Part of the second step is to combine what you know (or guess) in the first step with a selection of inductance and capacitance for resonance. As others have said, inductive Q_u is the determining factor at HF and capacitive Q_u will be at least 10 times higher, probably in the neighborhood of 500 to 1000 for ceramic or mica capacitors. Do a quick model of the resonant circuit "resistance" (actually the magnitude of impedance) at parallel resonance - parallel the (inductive Q_u times inductive reactance) and the (capacitive Q_u times the same reactance since capacitive reactance is equal to inductive reactance at resonance). Parallel that with the source and load impedance magnitude combined magnitude and you have the total magnitude at resonance. This can be very quick to do with a scientific pocket calculator. To verify the selectivity, run the whole thing again at adjacent harmonics to get the total magnitude of impedance there. Those off-resonance L-C circuit magnitudes can use just the reactance as an approximate step and be very close to those using the unloaded L-C Qs. The ratio of magnitudes on-resonance versus off-resonance will give you a picture of the selectivity possible. If that "doesn't seem to be good," THEN pick a different L:C ratio and do it again...but use what you know about the inductive Q_u at that different inductance. Compare the new on-resonance impedance magnitude to the adjacent off-resonance magnitudes. Is that magnitude ratio worse than before? If so use an opposite L:C ratio. If better, try the same-direction different L:C ratio and compare that. If better, repeat. If worse, hold on the previous L-C combination...you are zeroing-in on what is useful. SELECTIVITY is the thing desired in your application and the relatively-simple calculation of magnitudes and resulting ratios will point in the right direction for something to try in hardware. Selectivity is needed because the lower harmonics have more energy than the 5th. If stumped for a starting L and C value, try the literature on previous multiplier designs as a starting point...then dance through this two-step procedure. In practical hardware, many others besides myself have been led astray by simplistic "L:C ratio Q determination rules" that can be just the reverse. Lots of those old maxims were generated way back in time of large "coils and condensers" one needed both hands to pick up... Len Anderson retired (from regular hours) electronic engineer person |
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In article , Paul Burridge
writes: On 22 Mar 2004 01:46:20 GMT, (Avery Fineman) wrote: [snip] Phew! Thanks, Len. There's rather more to these ''simple inductors" than meets the eye, it seems. :( Yes, but it probably took longer to write the procedure out than to actually do it. :-) As a _starting_ procedure, work with magnitudes of impedance, or the square-root of the sums of the squares of resistance and reactance. That's easy enough on a conventional "four-function" pocket calculator that has the added square-root function. For a single resonant circuit you should be interested in the ratio of magnitude at desired frequency to the magnitudes at the undesired frequency. The bigger the ratio, the better selectivity. If you want to investigate toroidal coils (for their smaller size), you can find Q curves at - http://www.amidoncorp.com. Those look exactly like the Q curves in my Micrometals Q Curve subsection of their (surface mail only) catalog. Under one of the links at the top right you will get a rather long html page with over a dozen curves on it representing over a half dozen typical inductors per graph over a wide frequency range. Save that page if you can, it is good for reference later. www.micrometals.com seems to be down on Monday, probably for website revision; haven't heard any rumors that Micrometals went out of business. Micrometals has foreign (to the USA) distributors and sales reps so I was hoping to show some of those in or near the UK. For cylindrical coil shapes, Reg Edwards program is about as good as it gets to save wear and tear on pencil and paper. Len Anderson retired (from regular hours) electronic engineer person. |
In article , Paul Burridge
writes: On 22 Mar 2004 01:46:20 GMT, (Avery Fineman) wrote: [snip] Phew! Thanks, Len. There's rather more to these ''simple inductors" than meets the eye, it seems. :( Yes, but it probably took longer to write the procedure out than to actually do it. :-) As a _starting_ procedure, work with magnitudes of impedance, or the square-root of the sums of the squares of resistance and reactance. That's easy enough on a conventional "four-function" pocket calculator that has the added square-root function. For a single resonant circuit you should be interested in the ratio of magnitude at desired frequency to the magnitudes at the undesired frequency. The bigger the ratio, the better selectivity. If you want to investigate toroidal coils (for their smaller size), you can find Q curves at - http://www.amidoncorp.com. Those look exactly like the Q curves in my Micrometals Q Curve subsection of their (surface mail only) catalog. Under one of the links at the top right you will get a rather long html page with over a dozen curves on it representing over a half dozen typical inductors per graph over a wide frequency range. Save that page if you can, it is good for reference later. www.micrometals.com seems to be down on Monday, probably for website revision; haven't heard any rumors that Micrometals went out of business. Micrometals has foreign (to the USA) distributors and sales reps so I was hoping to show some of those in or near the UK. For cylindrical coil shapes, Reg Edwards program is about as good as it gets to save wear and tear on pencil and paper. Len Anderson retired (from regular hours) electronic engineer person. |
Of course, maybe you don't need such a high Q, Paul. Qu of 30 is
quite reasonable for small SMT RF inductors, at least the type I use. In the following list, "series" means in series from the gate output to the next gate input, in order, and "shunt" means shunt to ground at that point. Anyway, try this (build it or SPICE it or RFSim99 it...add resistors to any simulation to account for the Qu. I'd suggest 3 ohms series and 12k ohms shunt for each 1.8uH.) 47pF series 1.8uH series 470pF shunt 45pF series 1.8uH shunt 3.3pF series 40pF shunt 1.8uH shunt DC blocking cap series high-value DC bias resistors, and the gate input (I assumed to be about 4k net resistance to ground at the gate input, including the bias resistors). It should give you enough voltage gain at 18MHz to drive the second gate at the fifth harmonic, and should attenuate the third at least 50dB if you build it properly, even with low-ish Qu inductors. This is rather a "hack" circuit, but works. The premise is that it's easier to get three inductors all the same value than muck about tuning the inductors. Make the 47pF, 45pF and 40pF caps variable and you can peak up the response at your desired frequency. Your simulation should show a reasonably flat bandpass characteristic, centered at about 18MHz. Paul Burridge wrote in message . .. Hi guys, ISTR that one can improve Q in resonant tanks by having a low L-C ratio. Or was it high L-C ratio. I can't remember but need to know. Can any kind soul help me out? Thanks. p. |
Of course, maybe you don't need such a high Q, Paul. Qu of 30 is
quite reasonable for small SMT RF inductors, at least the type I use. In the following list, "series" means in series from the gate output to the next gate input, in order, and "shunt" means shunt to ground at that point. Anyway, try this (build it or SPICE it or RFSim99 it...add resistors to any simulation to account for the Qu. I'd suggest 3 ohms series and 12k ohms shunt for each 1.8uH.) 47pF series 1.8uH series 470pF shunt 45pF series 1.8uH shunt 3.3pF series 40pF shunt 1.8uH shunt DC blocking cap series high-value DC bias resistors, and the gate input (I assumed to be about 4k net resistance to ground at the gate input, including the bias resistors). It should give you enough voltage gain at 18MHz to drive the second gate at the fifth harmonic, and should attenuate the third at least 50dB if you build it properly, even with low-ish Qu inductors. This is rather a "hack" circuit, but works. The premise is that it's easier to get three inductors all the same value than muck about tuning the inductors. Make the 47pF, 45pF and 40pF caps variable and you can peak up the response at your desired frequency. Your simulation should show a reasonably flat bandpass characteristic, centered at about 18MHz. Paul Burridge wrote in message . .. Hi guys, ISTR that one can improve Q in resonant tanks by having a low L-C ratio. Or was it high L-C ratio. I can't remember but need to know. Can any kind soul help me out? Thanks. p. |
In article , Paul Burridge
writes On 22 Mar 2004 01:46:20 GMT, (Avery Fineman) wrote: [snip] Phew! Thanks, Len. There's rather more to these ''simple inductors" than meets the eye, it seems. :( I note that the "Q" of capacitors when specified is much higher than the specified value for Ls. Unloaded Q will give maximum selectivity but no power transfer. Optimum power transfer will be at q/2 which will only increase bandwidth x 2, (i think but am not) -- ddwyer |
In article , Paul Burridge
writes On 22 Mar 2004 01:46:20 GMT, (Avery Fineman) wrote: [snip] Phew! Thanks, Len. There's rather more to these ''simple inductors" than meets the eye, it seems. :( I note that the "Q" of capacitors when specified is much higher than the specified value for Ls. Unloaded Q will give maximum selectivity but no power transfer. Optimum power transfer will be at q/2 which will only increase bandwidth x 2, (i think but am not) -- ddwyer |
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