|
On Tue, 16 Mar 2004 20:26:15 +0000, Paul Burridge
wrote: If I'm not mistaken, "tuned amplification" IS "filtering". An argument over semantics, then. AFAIC it's not filtering as such. It introduces a high degree of selectivity, certainly. But when someone says "filtering" I assume they're taking about a pi-network or something of that sort, between stages or at the end of a chain of stages. Wow - the strange things you learn on this thread! So how many poles does a circuit need for it to be called a "filter"? Tony (remove the "_" to reply by email) |
On Tue, 16 Mar 2004 20:26:15 +0000, Paul Burridge
wrote: An argument over semantics, then. AFAIC it's not filtering as such. It introduces a high degree of selectivity, certainly. But when someone says "filtering" I assume they're taking about a pi-network or something of that sort, between stages or at the end of a chain of stages. --- Any network which exhibits frequency selectivity is a filter, whether or not you're concerned about whether or not it is or is not. Think about it... from the lowly filter capacitor to the exalted brickwall filter, they're all discrimating against a frequency or a set of frequencies which we have told them we don't want them to let us see. Filters, every one. Just for grins, take a little trip over to a.b.s.e. (same subject heading)and take a look at what John Larkin's series resonant filter feeding a parallel resonant filter strategy looks like as far as allowing you to get a fifth harmonic from a fundamental square wave goes. -- John Fields |
On Tue, 16 Mar 2004 20:26:15 +0000, Paul Burridge
wrote: An argument over semantics, then. AFAIC it's not filtering as such. It introduces a high degree of selectivity, certainly. But when someone says "filtering" I assume they're taking about a pi-network or something of that sort, between stages or at the end of a chain of stages. --- Any network which exhibits frequency selectivity is a filter, whether or not you're concerned about whether or not it is or is not. Think about it... from the lowly filter capacitor to the exalted brickwall filter, they're all discrimating against a frequency or a set of frequencies which we have told them we don't want them to let us see. Filters, every one. Just for grins, take a little trip over to a.b.s.e. (same subject heading)and take a look at what John Larkin's series resonant filter feeding a parallel resonant filter strategy looks like as far as allowing you to get a fifth harmonic from a fundamental square wave goes. -- John Fields |
On Tue, 16 Mar 2004 19:54:27 -0600, John Fields
wrote: On Tue, 16 Mar 2004 20:26:15 +0000, Paul Burridge wrote: An argument over semantics, then. AFAIC it's not filtering as such. It introduces a high degree of selectivity, certainly. But when someone says "filtering" I assume they're taking about a pi-network or something of that sort, between stages or at the end of a chain of stages. --- Any network which exhibits frequency selectivity is a filter, whether or not you're concerned about whether or not it is or is not. Think about it... from the lowly filter capacitor to the exalted brickwall filter, they're all discrimating against a frequency or a set of frequencies which we have told them we don't want them to let us see. Filters, every one. Just for grins, take a little trip over to a.b.s.e. (same subject heading)and take a look at what John Larkin's series resonant filter feeding a parallel resonant filter strategy looks like as far as allowing you to get a fifth harmonic from a fundamental square wave goes. That's just a standard bandpass. What you do is pick a normalized lowpass filter that has the response shape you like, say a Tchebychev (I know... various spellings) and scale it to the impedance Z' and bandwidth W' you want. Then series resonate each L with a C, and parallel resonate each C with an L, both at some desired center frequency. Voila (pardon my French) a bandpass that's 2W' wide. There's no real reason to cascade lossy Q-killing tuned transistor stages when you can put all your Ls and Cs in one place. John |
On Tue, 16 Mar 2004 19:54:27 -0600, John Fields
wrote: On Tue, 16 Mar 2004 20:26:15 +0000, Paul Burridge wrote: An argument over semantics, then. AFAIC it's not filtering as such. It introduces a high degree of selectivity, certainly. But when someone says "filtering" I assume they're taking about a pi-network or something of that sort, between stages or at the end of a chain of stages. --- Any network which exhibits frequency selectivity is a filter, whether or not you're concerned about whether or not it is or is not. Think about it... from the lowly filter capacitor to the exalted brickwall filter, they're all discrimating against a frequency or a set of frequencies which we have told them we don't want them to let us see. Filters, every one. Just for grins, take a little trip over to a.b.s.e. (same subject heading)and take a look at what John Larkin's series resonant filter feeding a parallel resonant filter strategy looks like as far as allowing you to get a fifth harmonic from a fundamental square wave goes. That's just a standard bandpass. What you do is pick a normalized lowpass filter that has the response shape you like, say a Tchebychev (I know... various spellings) and scale it to the impedance Z' and bandwidth W' you want. Then series resonate each L with a C, and parallel resonate each C with an L, both at some desired center frequency. Voila (pardon my French) a bandpass that's 2W' wide. There's no real reason to cascade lossy Q-killing tuned transistor stages when you can put all your Ls and Cs in one place. John |
For a comparison of rectangular waveform on-times versus spectral
content, the following calculations were done on my WAVESPEC program for a 0.50 to 0.25 times repetition period and with rise and fall times equal to 0.02 times repetition period. If the fundamental energy is the reference, then the harmonics are down from that 0 db by the values shown: /------------ width rel. to rep. period -==--------\ Harm 0.50 0.45 0.40 0.35 0.30 0.25 1 0 0 0 0 0 0 3 -6.4 -7.3 -8.4 -9.6 -11.1 -12.8 5 -28.0 -17.8 -15.1 -15.7 -20.3 * 7 -28.1 -20.1 -26.2 -27.7 -26.6 -29.1 9 -28.4 -24.6 -32.1 -44.2 -43.0 * 11 -28.7 -32.7 -42.2 -25.1 -35.8 -28.7 13 -29.1 * -29.1 * -29.1 * * too far down to matter, not enough there The above will hold true at any fundamental frequency provided the rise and fall times are equal and each equal to 0.02 times the repetition period. Those numbers will change given faster or slower rise/fall times. All db calculated as 20 x Log (voltage). Width is determined at the baseline, not the 50% amplitude point. Len Anderson retired (from regular hours) electronic engineer person |
For a comparison of rectangular waveform on-times versus spectral
content, the following calculations were done on my WAVESPEC program for a 0.50 to 0.25 times repetition period and with rise and fall times equal to 0.02 times repetition period. If the fundamental energy is the reference, then the harmonics are down from that 0 db by the values shown: /------------ width rel. to rep. period -==--------\ Harm 0.50 0.45 0.40 0.35 0.30 0.25 1 0 0 0 0 0 0 3 -6.4 -7.3 -8.4 -9.6 -11.1 -12.8 5 -28.0 -17.8 -15.1 -15.7 -20.3 * 7 -28.1 -20.1 -26.2 -27.7 -26.6 -29.1 9 -28.4 -24.6 -32.1 -44.2 -43.0 * 11 -28.7 -32.7 -42.2 -25.1 -35.8 -28.7 13 -29.1 * -29.1 * -29.1 * * too far down to matter, not enough there The above will hold true at any fundamental frequency provided the rise and fall times are equal and each equal to 0.02 times the repetition period. Those numbers will change given faster or slower rise/fall times. All db calculated as 20 x Log (voltage). Width is determined at the baseline, not the 50% amplitude point. Len Anderson retired (from regular hours) electronic engineer person |
Avery Fineman wrote:
For a comparison of rectangular waveform on-times versus spectral content, the following calculations were done on my WAVESPEC program for a 0.50 to 0.25 times repetition period and with rise and fall times equal to 0.02 times repetition period. If the fundamental energy is the reference, then the harmonics are down from that 0 db by the values shown: /------------ width rel. to rep. period -==--------\ Harm 0.50 0.45 0.40 0.35 0.30 0.25 1 0 0 0 0 0 0 3 -6.4 -7.3 -8.4 -9.6 -11.1 -12.8 5 -28.0 -17.8 -15.1 -15.7 -20.3 * 7 -28.1 -20.1 -26.2 -27.7 -26.6 -29.1 9 -28.4 -24.6 -32.1 -44.2 -43.0 * 11 -28.7 -32.7 -42.2 -25.1 -35.8 -28.7 13 -29.1 * -29.1 * -29.1 * * too far down to matter, not enough there The above will hold true at any fundamental frequency provided the rise and fall times are equal and each equal to 0.02 times the repetition period. Those numbers will change given faster or slower rise/fall times. All db calculated as 20 x Log (voltage). Width is determined at the baseline, not the 50% amplitude point. Len Anderson retired (from regular hours) electronic engineer person I wonder what happens to these numbers as the rise/fall time tends to zero? Peter Lawton |
Avery Fineman wrote:
For a comparison of rectangular waveform on-times versus spectral content, the following calculations were done on my WAVESPEC program for a 0.50 to 0.25 times repetition period and with rise and fall times equal to 0.02 times repetition period. If the fundamental energy is the reference, then the harmonics are down from that 0 db by the values shown: /------------ width rel. to rep. period -==--------\ Harm 0.50 0.45 0.40 0.35 0.30 0.25 1 0 0 0 0 0 0 3 -6.4 -7.3 -8.4 -9.6 -11.1 -12.8 5 -28.0 -17.8 -15.1 -15.7 -20.3 * 7 -28.1 -20.1 -26.2 -27.7 -26.6 -29.1 9 -28.4 -24.6 -32.1 -44.2 -43.0 * 11 -28.7 -32.7 -42.2 -25.1 -35.8 -28.7 13 -29.1 * -29.1 * -29.1 * * too far down to matter, not enough there The above will hold true at any fundamental frequency provided the rise and fall times are equal and each equal to 0.02 times the repetition period. Those numbers will change given faster or slower rise/fall times. All db calculated as 20 x Log (voltage). Width is determined at the baseline, not the 50% amplitude point. Len Anderson retired (from regular hours) electronic engineer person I wonder what happens to these numbers as the rise/fall time tends to zero? Peter Lawton |
On Wed, 17 Mar 2004 11:06:42 +1000, Tony wrote:
On Tue, 16 Mar 2004 20:26:15 +0000, Paul Burridge wrote: If I'm not mistaken, "tuned amplification" IS "filtering". An argument over semantics, then. AFAIC it's not filtering as such. It introduces a high degree of selectivity, certainly. But when someone says "filtering" I assume they're taking about a pi-network or something of that sort, between stages or at the end of a chain of stages. Wow - the strange things you learn on this thread! So how many poles does a circuit need for it to be called a "filter"? "Words mean what I choose them to mean! No more; no less." - the Red Queen :-) -- The BBC: Licensed at public expense to spread lies. |
On Wed, 17 Mar 2004 11:06:42 +1000, Tony wrote:
On Tue, 16 Mar 2004 20:26:15 +0000, Paul Burridge wrote: If I'm not mistaken, "tuned amplification" IS "filtering". An argument over semantics, then. AFAIC it's not filtering as such. It introduces a high degree of selectivity, certainly. But when someone says "filtering" I assume they're taking about a pi-network or something of that sort, between stages or at the end of a chain of stages. Wow - the strange things you learn on this thread! So how many poles does a circuit need for it to be called a "filter"? "Words mean what I choose them to mean! No more; no less." - the Red Queen :-) -- The BBC: Licensed at public expense to spread lies. |
|
|
On Tue, 16 Mar 2004 19:43:24 -0800, John Larkin
wrote: On Tue, 16 Mar 2004 19:54:27 -0600, John Fields wrote: Just for grins, take a little trip over to a.b.s.e. (same subject heading)and take a look at what John Larkin's series resonant filter feeding a parallel resonant filter strategy looks like as far as allowing you to get a fifth harmonic from a fundamental square wave goes. Okay, well at least I can see this one! Not sure about the SA trace, though. Came out clearly enough but I'm not sure what you were trying to prove by it. As for the 'scope traces, there doesn't seem to be any phase correlation between the two and you don't indicate at what point the probe was inserted. The square wave r/f slopes look a bit tardy, too. What was the active device you used to generate them? That's just a standard bandpass. What you do is pick a normalized lowpass filter that has the response shape you like, say a Tchebychev (I know... various spellings) and scale it to the impedance Z' and bandwidth W' you want. Then series resonate each L with a C, and parallel resonate each C with an L, both at some desired center frequency. Voila (pardon my French) a bandpass that's 2W' wide. It's basically the same type as Reg's program designed for me. That was built on Sunday, tested and found to be bang on the money and later today I shall try to see if it can be used to 'extract' the elusive 5th. I have to admit I'll be surprised if there's nothing there at all, but we'll have to wait and see. Don't touch that dial! :-) -- The BBC: Licensed at public expense to spread lies. |
On Tue, 16 Mar 2004 19:43:24 -0800, John Larkin
wrote: On Tue, 16 Mar 2004 19:54:27 -0600, John Fields wrote: Just for grins, take a little trip over to a.b.s.e. (same subject heading)and take a look at what John Larkin's series resonant filter feeding a parallel resonant filter strategy looks like as far as allowing you to get a fifth harmonic from a fundamental square wave goes. Okay, well at least I can see this one! Not sure about the SA trace, though. Came out clearly enough but I'm not sure what you were trying to prove by it. As for the 'scope traces, there doesn't seem to be any phase correlation between the two and you don't indicate at what point the probe was inserted. The square wave r/f slopes look a bit tardy, too. What was the active device you used to generate them? That's just a standard bandpass. What you do is pick a normalized lowpass filter that has the response shape you like, say a Tchebychev (I know... various spellings) and scale it to the impedance Z' and bandwidth W' you want. Then series resonate each L with a C, and parallel resonate each C with an L, both at some desired center frequency. Voila (pardon my French) a bandpass that's 2W' wide. It's basically the same type as Reg's program designed for me. That was built on Sunday, tested and found to be bang on the money and later today I shall try to see if it can be used to 'extract' the elusive 5th. I have to admit I'll be surprised if there's nothing there at all, but we'll have to wait and see. Don't touch that dial! :-) -- The BBC: Licensed at public expense to spread lies. |
On Wed, 17 Mar 2004 12:03:04 +0000, Paul Burridge
wrote: Okay, well at least I can see this one! Not sure about the SA trace, though. Came out clearly enough but I'm not sure what you were trying to prove by it. --- It's an FFT of what's coming out of Larkin's suggested two-stage bandpass filter. The first vertical marker (f1) goes through the first peak, 17.19MHz, which proves the fifth is in a 3.44MHz square wave. The second marker goes through 34.38, so so's the tenth. --- As for the 'scope traces, there doesn't seem to be any phase correlation between the two and you don't indicate at what point the probe was inserted. --- Are you kidding? Count the high frequency cycles between the first leading edge and the same point on the second leading edge of the square wave and you'll find there are exactly five. --- The square wave r/f slopes look a bit tardy, too. --- That's because the generator wasn't isolated from the filter and you're seeing the filter's input dragging it around. --- What was the active device you used to generate them? --- Tektronics FG502. Here's the layout: [TEK FG502]-+-[BPF]-+-[TEK 2465A]---[HP5328A] | | | +-[HP54602B] | [TEK DC504] -- John Fields |
On Wed, 17 Mar 2004 12:03:04 +0000, Paul Burridge
wrote: Okay, well at least I can see this one! Not sure about the SA trace, though. Came out clearly enough but I'm not sure what you were trying to prove by it. --- It's an FFT of what's coming out of Larkin's suggested two-stage bandpass filter. The first vertical marker (f1) goes through the first peak, 17.19MHz, which proves the fifth is in a 3.44MHz square wave. The second marker goes through 34.38, so so's the tenth. --- As for the 'scope traces, there doesn't seem to be any phase correlation between the two and you don't indicate at what point the probe was inserted. --- Are you kidding? Count the high frequency cycles between the first leading edge and the same point on the second leading edge of the square wave and you'll find there are exactly five. --- The square wave r/f slopes look a bit tardy, too. --- That's because the generator wasn't isolated from the filter and you're seeing the filter's input dragging it around. --- What was the active device you used to generate them? --- Tektronics FG502. Here's the layout: [TEK FG502]-+-[BPF]-+-[TEK 2465A]---[HP5328A] | | | +-[HP54602B] | [TEK DC504] -- John Fields |
On Wed, 17 Mar 2004 08:58:18 -0600, John Fields
wrote: [TEK FG502]-+-[BPF]-+-[TEK 2465A]---[HP5328A] | | | +-[HP54602B] | [TEK DC504] Should be: +---------|IN1 | | |TEK2465A| [TEK FG502]-+-[BPF]-+-|IN2 OUT2|---[HP5328A] | | | +-[HP54602B] | [TEK DC504] -- John Fields |
On Wed, 17 Mar 2004 08:58:18 -0600, John Fields
wrote: [TEK FG502]-+-[BPF]-+-[TEK 2465A]---[HP5328A] | | | +-[HP54602B] | [TEK DC504] Should be: +---------|IN1 | | |TEK2465A| [TEK FG502]-+-[BPF]-+-|IN2 OUT2|---[HP5328A] | | | +-[HP54602B] | [TEK DC504] -- John Fields |
On Wed, 17 Mar 2004 08:58:18 -0600, John Fields
wrote: It's an FFT of what's coming out of Larkin's suggested two-stage bandpass filter. The first vertical marker (f1) goes through the first peak, 17.19MHz, which proves the fifth is in a 3.44MHz square wave. The second marker goes through 34.38, so so's the tenth. Curious that there should be a sizeable pass response at the tenth harmonic, isn't it? It doesn't appear to be *that* much down on the intended pass frequency although there appears to be no indexing for the y axis. Are you kidding? Count the high frequency cycles between the first leading edge and the same point on the second leading edge of the square wave and you'll find there are exactly five. Well, I admit I'm a bit of a greenhorn on these things, but to my eyes there appears to be some phase difference. I'll accept your word for it there isn't. What was the active device you used to generate them? --- Tektronics FG502. Here's the layout: [TEK FG502]-+-[BPF]-+-[TEK 2465A]---[HP5328A] | | | +-[HP54602B] | [TEK DC504] Many thanks. I hope it didn't involve you in too much setting-up time to investigate this and post your findings. -- The BBC: Licensed at public expense to spread lies. |
On Wed, 17 Mar 2004 08:58:18 -0600, John Fields
wrote: It's an FFT of what's coming out of Larkin's suggested two-stage bandpass filter. The first vertical marker (f1) goes through the first peak, 17.19MHz, which proves the fifth is in a 3.44MHz square wave. The second marker goes through 34.38, so so's the tenth. Curious that there should be a sizeable pass response at the tenth harmonic, isn't it? It doesn't appear to be *that* much down on the intended pass frequency although there appears to be no indexing for the y axis. Are you kidding? Count the high frequency cycles between the first leading edge and the same point on the second leading edge of the square wave and you'll find there are exactly five. Well, I admit I'm a bit of a greenhorn on these things, but to my eyes there appears to be some phase difference. I'll accept your word for it there isn't. What was the active device you used to generate them? --- Tektronics FG502. Here's the layout: [TEK FG502]-+-[BPF]-+-[TEK 2465A]---[HP5328A] | | | +-[HP54602B] | [TEK DC504] Many thanks. I hope it didn't involve you in too much setting-up time to investigate this and post your findings. -- The BBC: Licensed at public expense to spread lies. |
In article , Peter John Lawton
writes: The above will hold true at any fundamental frequency provided the rise and fall times are equal and each equal to 0.02 times the repetition period. Those numbers will change given faster or slower rise/fall times. All db calculated as 20 x Log (voltage). Width is determined at the baseline, not the 50% amplitude point. Len Anderson retired (from regular hours) electronic engineer person I wonder what happens to these numbers as the rise/fall time tends to zero? The harmonic content will increase...but also show dips depending on the percentage width relative to the period. I could present those (takes only minutes to run the program and transcribe the results) but that is academic only. The rise and fall times will NOT be zero due to the repetition frequency being high (repetition time short). Consider that a 3 MHz waveform has a period of 333 1/3 nSec and that Paul is using a TTL family inverter to make the square wave. Even with a Schmitt trigger inverter the t_r and t_f are going to be finite, possibly 15 nSec with a fast device (and some capacitive loading or semi-resonant whatever to mess with on- and off-times). 15 nSec is 4.5% of the repetition period, quite finite...more than I showed on the small table given previously. I'm sure someone out there wants to argue minutae on numbers but what is being discussed is a squarish waveform with a repetition frequency in the low HF range. Periods are valued in nanoSeconds and the on/off times of squaring devices are ALSO in nanoSeconds. There's just NOT going to be any sort of "zero" on/off times with practical logic devices used by hobbyists. What is not intuitive to me (and to others) is that harmonic energy of a rectangular waveform drops drastically by the 5th harmonic and is certainly lower than "obvious" numbers bandied about. But, also mentioned before by others is that shortening the rect- angular waveshape DOES increase the 5th harmonic, as evident by the approximate 12 db increase at 40 to 35 percent of the repetition period. An equivalent shortening happens in vacuum tube multipliers through biasing (self, fixed, or both) and that can be adjustable along with the drive level. It's not quite the same with bipolars since the overdrive effects are more saturation than in the self- bias conditions of tubes. It's close, though. From all indications of the Fourier series results, there's a definite reason why so few multipliers went beyond tripling. The amount of energy (relative to fundamental and taking into account the finite rise and fall times) of 4th and higher harmonics just isn't as much as intuition would have everyone believe! Len Anderson retired (from regular hours) electronic engineer person |
In article , Peter John Lawton
writes: The above will hold true at any fundamental frequency provided the rise and fall times are equal and each equal to 0.02 times the repetition period. Those numbers will change given faster or slower rise/fall times. All db calculated as 20 x Log (voltage). Width is determined at the baseline, not the 50% amplitude point. Len Anderson retired (from regular hours) electronic engineer person I wonder what happens to these numbers as the rise/fall time tends to zero? The harmonic content will increase...but also show dips depending on the percentage width relative to the period. I could present those (takes only minutes to run the program and transcribe the results) but that is academic only. The rise and fall times will NOT be zero due to the repetition frequency being high (repetition time short). Consider that a 3 MHz waveform has a period of 333 1/3 nSec and that Paul is using a TTL family inverter to make the square wave. Even with a Schmitt trigger inverter the t_r and t_f are going to be finite, possibly 15 nSec with a fast device (and some capacitive loading or semi-resonant whatever to mess with on- and off-times). 15 nSec is 4.5% of the repetition period, quite finite...more than I showed on the small table given previously. I'm sure someone out there wants to argue minutae on numbers but what is being discussed is a squarish waveform with a repetition frequency in the low HF range. Periods are valued in nanoSeconds and the on/off times of squaring devices are ALSO in nanoSeconds. There's just NOT going to be any sort of "zero" on/off times with practical logic devices used by hobbyists. What is not intuitive to me (and to others) is that harmonic energy of a rectangular waveform drops drastically by the 5th harmonic and is certainly lower than "obvious" numbers bandied about. But, also mentioned before by others is that shortening the rect- angular waveshape DOES increase the 5th harmonic, as evident by the approximate 12 db increase at 40 to 35 percent of the repetition period. An equivalent shortening happens in vacuum tube multipliers through biasing (self, fixed, or both) and that can be adjustable along with the drive level. It's not quite the same with bipolars since the overdrive effects are more saturation than in the self- bias conditions of tubes. It's close, though. From all indications of the Fourier series results, there's a definite reason why so few multipliers went beyond tripling. The amount of energy (relative to fundamental and taking into account the finite rise and fall times) of 4th and higher harmonics just isn't as much as intuition would have everyone believe! Len Anderson retired (from regular hours) electronic engineer person |
Paul Burridge wrote in message . ..
.... What leads you to believe I have enough 5th harmonic in *my* particular case? The trace on the web site you provided a link to. The fact that you're using HC logic (which has inherent rise and fall times rather faster than the square wave source I used in my experiment). Your avering that the duty cycle is very nearly 50%. That's not to say you aren't doing something to kill it, but it's NOT difficult to extract it. Note that the subject you put on this thread really nails it: all you need to do is extract (and possibly amplify, depending on the final power level you need) what's already there. Now go do it. But feeding the whole square wave to the amplifier stage is a BAD idea because you can inadventently change the duty cycle (as seen at that amplifier's output) to one where the fifth is nulled. If you only need a few milliwatts, you can get that from the square wave directly, if the source impedance is low enough, simply by using the proper filter. |
Paul Burridge wrote in message . ..
.... What leads you to believe I have enough 5th harmonic in *my* particular case? The trace on the web site you provided a link to. The fact that you're using HC logic (which has inherent rise and fall times rather faster than the square wave source I used in my experiment). Your avering that the duty cycle is very nearly 50%. That's not to say you aren't doing something to kill it, but it's NOT difficult to extract it. Note that the subject you put on this thread really nails it: all you need to do is extract (and possibly amplify, depending on the final power level you need) what's already there. Now go do it. But feeding the whole square wave to the amplifier stage is a BAD idea because you can inadventently change the duty cycle (as seen at that amplifier's output) to one where the fifth is nulled. If you only need a few milliwatts, you can get that from the square wave directly, if the source impedance is low enough, simply by using the proper filter. |
|
|
Ignoring most of the previous posts in this thread (sorry - I deleted them) it
occurs to me that maybe the problem may be the relatively low impedance load on the buffer, whose finite output impedance therefore causes some waveform distortion and loss of 5th harmonic. So how about a hi-Z parallel tank coupling circuit; for 17.2MHz (say 4p7 || 18uH) into a load of 100-500 ohms (say a 1k pot + 100n to Gnd, to see the effect of load variations), then into another 74AC buffer? Tony (remove the "_" to reply by email) |
Ignoring most of the previous posts in this thread (sorry - I deleted them) it
occurs to me that maybe the problem may be the relatively low impedance load on the buffer, whose finite output impedance therefore causes some waveform distortion and loss of 5th harmonic. So how about a hi-Z parallel tank coupling circuit; for 17.2MHz (say 4p7 || 18uH) into a load of 100-500 ohms (say a 1k pot + 100n to Gnd, to see the effect of load variations), then into another 74AC buffer? Tony (remove the "_" to reply by email) |
On Thu, 18 Mar 2004 10:42:22 +1000, Tony wrote:
Ignoring most of the previous posts in this thread (sorry - I deleted them) it occurs to me that maybe the problem may be the relatively low impedance load on the buffer, whose finite output impedance therefore causes some waveform distortion and loss of 5th harmonic. So how about a hi-Z parallel tank coupling circuit; for 17.2MHz (say 4p7 || 18uH) into a load of 100-500 ohms (say a 1k pot + 100n to Gnd, to see the effect of load variations), then into another 74AC buffer? Tony (remove the "_" to reply by email) Actually the tank's output will be biased to half-rail, which won't necessarily bias the CMOS output buffer properly. May need to cap-couple to the buffer, with a feedback resistor so it will self-bias (another case of reading the post AFTER hitting "send"). Tony (remove the "_" to reply by email) |
On Thu, 18 Mar 2004 10:42:22 +1000, Tony wrote:
Ignoring most of the previous posts in this thread (sorry - I deleted them) it occurs to me that maybe the problem may be the relatively low impedance load on the buffer, whose finite output impedance therefore causes some waveform distortion and loss of 5th harmonic. So how about a hi-Z parallel tank coupling circuit; for 17.2MHz (say 4p7 || 18uH) into a load of 100-500 ohms (say a 1k pot + 100n to Gnd, to see the effect of load variations), then into another 74AC buffer? Tony (remove the "_" to reply by email) Actually the tank's output will be biased to half-rail, which won't necessarily bias the CMOS output buffer properly. May need to cap-couple to the buffer, with a feedback resistor so it will self-bias (another case of reading the post AFTER hitting "send"). Tony (remove the "_" to reply by email) |
On Wed, 17 Mar 2004 23:43:57 +0000, Paul Burridge
wrote: On 17 Mar 2004 12:31:14 -0800, (Tom Bruhns) wrote: Paul Burridge wrote in message . .. ... What leads you to believe I have enough 5th harmonic in *my* particular case? The trace on the web site you provided a link to. The fact that you're using HC logic (which has inherent rise and fall times rather faster than the square wave source I used in my experiment). Your avering that the duty cycle is very nearly 50%. That's not to say you aren't doing something to kill it, but it's NOT difficult to extract it. Note that the subject you put on this thread really nails it: all you need to do is extract (and possibly amplify, depending on the final power level you need) what's already there. Now go do it. But feeding the whole square wave to the amplifier stage is a BAD idea because you can inadventently change the duty cycle (as seen at that amplifier's output) to one where the fifth is nulled. If you only need a few milliwatts, you can get that from the square wave directly, if the source impedance is low enough, simply by using the proper filter. Yes, but I'd hoped to avoid any intermediate amplification stages. Looks like I'll have to swallow it. --- That doesn't make any sense from the point of view that you've already posted a schematic showing a couple of gain stages. How much 17 MHz. do you really need and what does what you want to feed it into look like? |
Paul Burridge wrote in message . ..
Yes, but I'd hoped to avoid any intermediate amplification stages. Looks like I'll have to swallow it. Exactly how much power do you need? Exactly how "clean" (free from other harmonics) must it be? Don't you have an amplifier in the circuit you're playing with anyway? 100mW should be easy with a single stage following the digital square wave, and a full watt is certainly feasible with the right design. If you were hoping for 100mW of fifth harmonic from a single HC output, you were probably dreaming. |
Paul Burridge wrote in message . ..
Yes, but I'd hoped to avoid any intermediate amplification stages. Looks like I'll have to swallow it. Exactly how much power do you need? Exactly how "clean" (free from other harmonics) must it be? Don't you have an amplifier in the circuit you're playing with anyway? 100mW should be easy with a single stage following the digital square wave, and a full watt is certainly feasible with the right design. If you were hoping for 100mW of fifth harmonic from a single HC output, you were probably dreaming. |
|
|
On Thu, 18 Mar 2004 11:33:17 +1000, Tony wrote:
On Thu, 18 Mar 2004 10:42:22 +1000, Tony wrote: Ignoring most of the previous posts in this thread (sorry - I deleted them) it occurs to me that maybe the problem may be the relatively low impedance load on the buffer, whose finite output impedance therefore causes some waveform distortion and loss of 5th harmonic. So how about a hi-Z parallel tank coupling circuit; for 17.2MHz (say 4p7 || 18uH) into a load of 100-500 ohms (say a 1k pot + 100n to Gnd, to see the effect of load variations), then into another 74AC buffer? Tony (remove the "_" to reply by email) Actually the tank's output will be biased to half-rail, which won't necessarily bias the CMOS output buffer properly. May need to cap-couple to the buffer, with a feedback resistor so it will self-bias (another case of reading the post AFTER hitting "send"). There's a lot of that goes on here. :-) I'll look into the idea, thanks. -- The BBC: Licensed at public expense to spread lies. |
On Thu, 18 Mar 2004 11:33:17 +1000, Tony wrote:
On Thu, 18 Mar 2004 10:42:22 +1000, Tony wrote: Ignoring most of the previous posts in this thread (sorry - I deleted them) it occurs to me that maybe the problem may be the relatively low impedance load on the buffer, whose finite output impedance therefore causes some waveform distortion and loss of 5th harmonic. So how about a hi-Z parallel tank coupling circuit; for 17.2MHz (say 4p7 || 18uH) into a load of 100-500 ohms (say a 1k pot + 100n to Gnd, to see the effect of load variations), then into another 74AC buffer? Tony (remove the "_" to reply by email) Actually the tank's output will be biased to half-rail, which won't necessarily bias the CMOS output buffer properly. May need to cap-couple to the buffer, with a feedback resistor so it will self-bias (another case of reading the post AFTER hitting "send"). There's a lot of that goes on here. :-) I'll look into the idea, thanks. -- The BBC: Licensed at public expense to spread lies. |
Paul Burridge wrote in message . ..
Exactly how much power do you need? Only enough to feed another inverter gate. Egad, Paul! You've been wasting this much net bandwidth just to drive another HC gate?? All you need is a filter/matching circuit that steps up the voltage. This is DOG SIMPLE! See below. Exactly how "clean" (free from other harmonics) must it be? Preferably filthy. It's another multiplier (this time only 3X, thank God!) Then you need a clean enough input that you'll get the desired output purity. "Filthy" is likely NOT the right answer and will just get you into further trouble. But fortunately, "clean" is simple, and "really clean" isn't at all difficult. Try this: square wave output -- I don't recall your exact freq; I used 3.7MHz -- from HC gate, feeds 4.58pF capacitor (make at least that one tuneable). Other end of cap feeds 20uH inductor, Qu=200. Other end of that inductor connects to next gate input, and net 18.6pF of capacitance to ground: say 15pF cap plus 3.6pF of gate input capacitance. For DC bias, gate input to ground = 22kohms; gate input to Vcc = 47kohms. That keeps the gate in a valid logic state when there's no excitation. Assuming the gate's RF input resistance at 18MHz is at least 2.5kohms, you should get a voltage gain at the fifth harmonic of about 15dB, which will be ample to drive the gate input. The available current from the filter is low enough that the gate's input protection diodes should clamp things nicely at the rails. Be sure to use a gate that has input protection, or else add low-capacitance, fast diodes externally. Gain at the third and seventh is down 20dB or so from that. If it needs to be cleaner than that, you can add a second resonator. The gate biasing suggested may result in an output duty cycle significantly different from 50%. If you will always have 3.7MHz drive, you can bias the input more in the center of its range, or even rearrange the circuit a bit and use a feedback resistor from output to input to set the DC bias. The gate's input impedance is then much lower, but you don't need much voltage to drive it. Don't use that trick with a Schmitt trigger input, though. 69 turns of 36AWG (0.125mm) wire, spaced 2 wire diameters c-c, on an 0.375" former, should give you about 20uH at Qu=200 and first parallel SRF about 50MHz, but you should be able to make it more compact using something like a T-50-2 powdered iron core. |
| All times are GMT +1. The time now is 09:49 AM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
RadioBanter.com