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dBm and Voltage
Hi.
I'm doing some RF experimentation and I need to know the "relation" between dBm specificatons and voltage level for a signal. I have a RF mixer with a specification that says: LO drive level (50 ohm) =3D -16 dBm And I have a LO source that give me an output of 2.5Vpp to a capacitive load of 5pF at 40MHz. How can I relate both items and design a circuit to connect LO source to RF mixer ? If you have some web source to study about this items, I'll be glad to hear about it. Thanks Hern=E1n S=E1nchez |
nanchez wrote:
Hi. I'm doing some RF experimentation and I need to know the "relation" between dBm specificatons and voltage level for a signal. I have a RF mixer with a specification that says: LO drive level (50 ohm) = -16 dBm And I have a LO source that give me an output of 2.5Vpp to a capacitive load of 5pF at 40MHz. How can I relate both items and design a circuit to connect LO source to RF mixer ? If you have some web source to study about this items, I'll be glad to hear about it. Thanks Hernán Sánchez Maybe this will help explain... http://zone.ni.com/devzone/conceptd....256811004DD454 |
On Fri, 03 Jun 2005 18:07:51 -0400, -ex- wrote:
nanchez wrote: Hi. I'm doing some RF experimentation and I need to know the "relation" between dBm specificatons and voltage level for a signal. I have a RF mixer with a specification that says: LO drive level (50 ohm) = -16 dBm And I have a LO source that give me an output of 2.5Vpp to a capacitive load of 5pF at 40MHz. How can I relate both items and design a circuit to connect LO source to RF mixer ? If you have some web source to study about this items, I'll be glad to hear about it. Thanks Hernán Sánchez Maybe this will help explain... http://zone.ni.com/devzone/conceptd....256811004DD454 This is pretty good, but errs when it states that 0 dBm is 1 mW in a 50 ohm system. This is the usual case, but it could just as well be 70 ohm, 600 ohm or 6 3/8 ohm. |
"nanchez" wrote in message ups.com... Hi. I'm doing some RF experimentation and I need to know the "relation" between dBm specificatons and voltage level for a signal. I have a RF mixer with a specification that says: LO drive level (50 ohm) = -16 dBm And I have a LO source that give me an output of 2.5Vpp to a capacitive load of 5pF at 40MHz. How can I relate both items and design a circuit to connect LO source to RF mixer ? If you have some web source to study about this items, I'll be glad to hear about it. Thanks Hernán Sánchez To rejoin the real world, take the "16" figure and divide it by 20. get "0.8" Then find the antilog of that 0.8 [use normal 'base10' logs] get "6.31" This number is a multiply or divide factor that is applied to a 50 ohm 0dBm reference voltage. So what is this god like reference voltage?. The 50ohm 0dBm reference voltage is in actual fact 0.223Vac. The original number was "-"16 dBm. Just read the minus sign as meaning a voltage less than the 0dBm reference voltage. So that 0.223Vac reference value is divided by your 6.31 factor. get "0.035" Vac. So "-16dBm" is really 35mVac. This means you have more than enough drive voltage available from your 2.5Vpp (900mVac) local oscillator signal. Be very wary whenever you come across dBm figures. There is a minefield of disinformation out there. In many cases they are intended purely to obfscutate the reader and prevent them clearly seeing that the described circuit is junk. In many other cases they are purposely used as an extra level of abstraction to sort out the 'RF men' from the 'boys'. Manufacturers still use the dB concept for historical reasons. It doesn't effect their sales as the RF people buying their kit carry in their heads instant dB-V conversion tables. Don't know about everyone else but all my scopes and signal generators and sources and dc-ac-voltmeters and DVMs and signal probes etc, are marked in Volts and Amps. So that's what I use. (Someday I'll get round to building a real world 1:2:5:10 50ohm attenuator. I certainly can't buy one :-) regards john |
On Sat, 4 Jun 2005 14:53:43 +0100, "john jardine"
wrote: "nanchez" wrote in message oups.com... Hi. I'm doing some RF experimentation and I need to know the "relation" between dBm specificatons and voltage level for a signal. I have a RF mixer with a specification that says: LO drive level (50 ohm) = -16 dBm And I have a LO source that give me an output of 2.5Vpp to a capacitive load of 5pF at 40MHz. How can I relate both items and design a circuit to connect LO source to RF mixer ? If you have some web source to study about this items, I'll be glad to hear about it. Thanks Hernán Sánchez To rejoin the real world, take the "16" figure and divide it by 20. get "0.8" Then find the antilog of that 0.8 [use normal 'base10' logs] get "6.31" This number is a multiply or divide factor that is applied to a 50 ohm 0dBm reference voltage. So what is this god like reference voltage?. The 50ohm 0dBm reference voltage is in actual fact 0.223Vac. The original number was "-"16 dBm. Just read the minus sign as meaning a voltage less than the 0dBm reference voltage. So that 0.223Vac reference value is divided by your 6.31 factor. get "0.035" Vac. So "-16dBm" is really 35mVac. This means you have more than enough drive voltage available from your 2.5Vpp (900mVac) local oscillator signal. Except neither you or Hernan can be sure of this. His source is not specified to work into a 50 ohm load or present a 50 ohm source impedance to the mixer (what is really needed). Who knows what the delivered voltage will be when driving the mixer port? A measurement is in order. Terminate the source in 50 ohm and measure the power and/or voltage. If it exceeds -16 dBm, attenuate accordingly. Be very wary whenever you come across dBm figures. There is a minefield of disinformation out there. In many cases they are intended purely to obfscutate the reader and prevent them clearly seeing that the described circuit is junk. Spoken like a real expert on bafflegab. In many other cases they are purposely used as an extra level of abstraction to sort out the 'RF men' from the 'boys'. Manufacturers still use the dB concept for historical reasons. It doesn't effect their sales as the RF people buying their kit carry in their heads instant dB-V conversion tables. Don't know about everyone else but all my scopes and signal generators and sources and dc-ac-voltmeters and DVMs and signal probes etc, are marked in Volts and Amps. So that's what I use. (Someday I'll get round to building a real world 1:2:5:10 50ohm attenuator. I certainly can't buy one :-) regards john |
It is worse than that, Wes. I fail to understand how a device can be
specified as X dB into a capacitive load. Last time I looked, a capacitive load couldn't dissipate ANY power. Jim This is pretty good, but errs when it states that 0 dBm is 1 mW in a 50 ohm system. This is the usual case, but it could just as well be 70 ohm, 600 ohm or 6 3/8 ohm. |
This may help - url:
http://www.hardware-guru.com/LabStuff%5CdBm.htm -- CL -- I doubt, therefore I might be ! roups.com... Hi. I'm doing some RF experimentation and I need to know the "relation" between dBm specificatons and voltage level for a signal. I have a RF mixer with a specification that says: LO drive level (50 ohm) = -16 dBm And I have a LO source that give me an output of 2.5Vpp to a capacitive load of 5pF at 40MHz. How can I relate both items and design a circuit to connect LO source to RF mixer ? If you have some web source to study about this items, I'll be glad to hear about it. Thanks Hernán Sánchez |
On Sat, 4 Jun 2005 08:56:56 -0700, "RST Engineering"
wrote: It is worse than that, Wes. I fail to understand how a device can be specified as X dB into a capacitive load. Last time I looked, a capacitive load couldn't dissipate ANY power. Jim, the OP said: "And I have a LO source that give me an output of 2.5Vpp to a capacitive load of 5pF at 40MHz." No dB or dBm mentioned. This sounds like a CMOS device with limited drive capability. That's why I suggest terminating it in 50 ohm, assuming this doesn't destroy it, and see what kind of power it can deliver. Most likely, a buffer will be needed, although -16 dBm isn't much and at that level it suggests that this is an active, not passive, mixer and that it might be driven by a higher source Z okay. Wes Jim This is pretty good, but errs when it states that 0 dBm is 1 mW in a 50 ohm system. This is the usual case, but it could just as well be 70 ohm, 600 ohm or 6 3/8 ohm. |
Actually, if it is a CMOS/HCMOS output, it might be better to terminate
it into a voltage divider of, say, 270 ohms from the osc output to 56 or 62 or 68 ohms to ground. Many of the common clock oscillators are not intended to directly drive a 50 ohm load. Then an output taken across the resistor to ground will look like it's from a nominally 50 ohm source. You could use a larger voltage divider ratio to get the output down further if desired. If the osc has square wave output, that's likely OK for a mixer input, but if it's not 50% duty cycle, you might benefit from cleaning it up a bit with a tuned circuit, for example. And if the oscillator has a TTL output (rather than HCMOS), you might benefit from returning the voltage divider to the osc power supply, presumably 5V. Beware when calculating the power delivered from the voltage divider to a (say) 50 ohm load; put that load in parallel with the output resistor of the divider to calculate the net division ratio. So the suggested resistors, e.g. with 68 ohm output R, might give you about -8dBm, if the osc delivers 2.5V P-P into the divider, yielding roughly .24V P-P output into a presumed 50 ohm load. -- Also, the oscillator probably has DC on its output, and you might benefit from a blocking capacitor. 1000pF would be adequate. And beware that the osc might deliver noticably higher voltage into a resistive load. Perhaps the OP could provide a bit more detail... Also, I'd ask if that -16dBm is accurate...that's pretty low even for active mixers. Finally, RFSim99 is a nice little free program for playing with linear RF circuits, and includes an "RF calculator" which has a tab for signal levels, converting among dBm, watts, volts-RMS, and volts-P-P for a user-specified impedance level. (I wish that tab had "locks" on the values like the resonance one does, so you could see what power level you get when you load a fixed voltage with various resistances, but you can always just copy-and-paste the voltage to "remember" it over a resistance change.) I always appreciate that RFSim99 has a lot of tools all in one place, and I don't have to remember a whole bunch of different programs, each of some very limited scope. Cheers, Tom |
From: "nanchez" on Fri 3 Jun 2005 14:59
I'm doing some RF experimentation and I need to know the "relation" between dBm specificatons and voltage level for a signal. I have a RF mixer with a specification that says: LO drive level (50 ohm) = -16 dBm And I have a LO source that give me an output of 2.5Vpp to a capacitive load of 5pF at 40MHz. How can I relate both items and design a circuit to connect LO source to RF mixer ? A basic definition that is industry-wide, government-wide, has "dbm" as decibels of "0 dbm" related to a power level of 1.0 milliWatts in a "50 Ohm system." That has become so widespread that specification writers don't always include those words. It is implicit when referring to RF components. The RMS voltage can be quickly calculated from some identities on the basic formula for Watts: P = E x I. Knowing R (50 Ohms) one can substitute Ohm's Law of Resistance of I = E / R into that to get E = SquareRoot (P x R). For 1.0 mW in a 50 Ohm system, P x R = 0.050 and the square root of that is 0.2236 so 0 dbm has an RMS voltage of 223.6 milliVolts. In your mixer specification, -16 dbm is equal to 35.44 mV RMS across a resistance of 50 Ohms. You can't DIRECTLY use your 40 MHz source value of 2.5 V peak-peak across a 5.0 pFd capacitance because it does not include the characteristic RESISTIVE impedance of 50 Ohms. Power in Watts must be related to the impedance of a load in order to perform "work." [a basic definition of power in Watts is "a unit of work"] Capacitance across a load will vary its impedance depending on the frequency. For that reason the electronics industry has long relied on a basic resistive impedance to measure and characterize RF components. The result is the very common "dbm" referred to 1.0 mW across a resistive 50 Ohm load...or the characteristic impedance of the measurement system, both source and load impedance. To relate your mixer specification to your RF source, you will have to put a 50 Ohm load across the source and measure that. If you have some stray capacitance across that load (inevitable) and know approximately what that is, you can calculate its effect across a resistance. At 40 MHz a 5.0 pFd capacitance has a reactance of 796 Ohms. That is not much but it changes the magnitude of the parallel R-C from 50.0 Ohms resistive to 47.0 Ohms slightly capacitive. That's a small change and can generally be neglected for experimental bench work. If you have some web source to study about this items, I'll be glad to hear about it. It's in practically every textbook on the subject of RF electronics. What can confuse newcomers to RF is the implicit "standard" which is not always included in specification sheets. The definition of "dbm" is arbitrary and probably picked (way back in time) for sake of convenience in measurement by all concerned. The reason for picking "50 Ohms" in a "system" is more obscure and ties into the physics of power transfer in coaxial cables. That's a curiosity that some can look up if they are interested but does not apply to how to USE the "dbm" specifications. To use "dbm" one only needs to remember the definition and apply simple forumulas for Power and Ohm's Law of Resistance. I hope that was of some help to you. |
From: "nanchez" on Fri 3 Jun 2005 14:59
I'm doing some RF experimentation and I need to know the "relation" between dBm specificatons and voltage level for a signal. I have a RF mixer with a specification that says: LO drive level (50 ohm) = -16 dBm And I have a LO source that give me an output of 2.5Vpp to a capacitive load of 5pF at 40MHz. How can I relate both items and design a circuit to connect LO source to RF mixer ? A basic definition that is industry-wide, government-wide, has "dbm" as decibels of "0 dbm" related to a power level of 1.0 milliWatts in a "50 Ohm system." That has become so widespread that specification writers don't always include those words. It is implicit when referring to RF components. The RMS voltage can be quickly calculated from some identities on the basic formula for Watts: P = E x I. Knowing R (50 Ohms) one can substitute Ohm's Law of Resistance of I = E / R into that to get E = SquareRoot (P x R). For 1.0 mW in a 50 Ohm system, P x R = 0.050 and the square root of that is 0.2236 so 0 dbm has an RMS voltage of 223.6 milliVolts. In your mixer specification, -16 dbm is equal to 35.44 mV RMS across a resistance of 50 Ohms. You can't DIRECTLY use your 40 MHz source value of 2.5 V peak-peak across a 5.0 pFd capacitance because it does not include the characteristic RESISTIVE impedance of 50 Ohms. Power in Watts must be related to the impedance of a load in order to perform "work." [a basic definition of power in Watts is "a unit of work"] Capacitance across a load will vary its impedance depending on the frequency. For that reason the electronics industry has long relied on a basic resistive impedance to measure and characterize RF components. The result is the very common "dbm" referred to 1.0 mW across a resistive 50 Ohm load...or the characteristic impedance of the measurement system, both source and load impedance. To relate your mixer specification to your RF source, you will have to put a 50 Ohm load across the source and measure that. If you have some stray capacitance across that load (inevitable) and know approximately what that is, you can calculate its effect across a resistance. At 40 MHz a 5.0 pFd capacitance has a reactance of 796 Ohms. That is not much but it changes the magnitude of the parallel R-C from 50.0 Ohms resistive to 47.0 Ohms slightly capacitive. That's a small change and can generally be neglected for experimental bench work. If you have some web source to study about this items, I'll be glad to hear about it. It's in practically every textbook on the subject of RF electronics. What can confuse newcomers to RF is the implicit "standard" which is not always included in specification sheets. The definition of "dbm" is arbitrary and probably picked (way back in time) for sake of convenience in measurement by all concerned. The reason for picking "50 Ohms" in a "system" is more obscure and ties into the physics of power transfer in coaxial cables. That's a curiosity that some can look up if they are interested but does not apply to how to USE the "dbm" specifications. To use "dbm" one only needs to remember the definition and apply simple forumulas for Power and Ohm's Law of Resistance. I hope that was of some help to you. |
From: "nanchez" on Fri 3 Jun 2005 14:59
I'm doing some RF experimentation and I need to know the "relation" between dBm specificatons and voltage level for a signal. I have a RF mixer with a specification that says: LO drive level (50 ohm) = -16 dBm And I have a LO source that give me an output of 2.5Vpp to a capacitive load of 5pF at 40MHz. How can I relate both items and design a circuit to connect LO source to RF mixer ? A basic definition that is industry-wide, government-wide, has "dbm" as decibels at "0 dbm" related to a power level of 1.0 milliWatts in a "50 Ohm system." That has become so widespread that specification writers don't always include those words. It is implicit when referring to RF components. The RMS voltage can be quickly calculated from some identities on the basic formula for Watts: P = E x I. Knowing R (50 Ohms) one can substitute Ohm's Law of Resistance of I = E / R into that to get E = SquareRoot (P x R). For 1.0 mW in a 50 Ohm system, P x R = 0.050 and the square root of that is 0.2236 so 0 dbm has an RMS voltage of 223.6 milliVolts. In your mixer specification, -16 dbm is equal to 35.44 mV RMS across a resistance of 50 Ohms. You can't DIRECTLY use your 40 MHz source value of 2.5 V peak-peak across a 5.0 pFd capacitance because it does not include the characteristic RESISTIVE impedance of 50 Ohms. Power in Watts must be related to the impedance of a load in order to perform "work." [a basic definition of power in Watts is "a unit of work"] Capacitance across a load will vary its impedance depending on the frequency. For that reason the electronics industry has long relied on a basic resistive impedance to measure and characterize RF components. The result is the very common "dbm" referred to 1.0 mW across a resistive 50 Ohm load...or the characteristic impedance of the measurement system, both source and load impedance. To relate your mixer specification to your RF source, you will have to put a 50 Ohm load across the source and measure that. If you have some stray capacitance across that load (inevitable) and know approximately what that is, you can calculate its effect across a resistance. At 40 MHz a 5.0 pFd capacitance has a reactance of 796 Ohms. That is not much but it changes the magnitude of the parallel R-C from 50.0 Ohms resistive to 47.0 Ohms slightly capacitive. That's a small change and can generally be neglected for experimental bench work. If you have some web source to study about this items, I'll be glad to hear about it. It's in practically every textbook on the subject of RF electronics. What can confuse newcomers to RF is the implicit "standard" which is not always included in specification sheets. The definition of "dbm" is arbitrary and probably picked (way back in time) for sake of convenience in measurement by all concerned. The reason for picking "50 Ohms" in a "system" is more obscure and ties into the physics of power transfer in coaxial cables. That's a curiosity that some can look up if they are interested but does not apply to how to USE the "dbm" specifications. To use "dbm" one only needs to remember the definition and apply simple forumulas for Power and Ohm's Law of Resistance. I hope that was of some help to you. |
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"Wes Stewart" wrote in message ... On Sat, 4 Jun 2005 14:53:43 +0100, "john jardine" wrote: "nanchez" wrote in message oups.com... Hi. I'm doing some RF experimentation and I need to know the "relation" between dBm specificatons and voltage level for a signal. I have a RF mixer with a specification that says: LO drive level (50 ohm) = -16 dBm And I have a LO source that give me an output of 2.5Vpp to a capacitive load of 5pF at 40MHz. How can I relate both items and design a circuit to connect LO source to RF mixer ? If you have some web source to study about this items, I'll be glad to hear about it. Thanks Hernán Sánchez To rejoin the real world, take the "16" figure and divide it by 20. get "0.8" Then find the antilog of that 0.8 [use normal 'base10' logs] get "6.31" This number is a multiply or divide factor that is applied to a 50 ohm 0dBm reference voltage. So what is this god like reference voltage?. The 50ohm 0dBm reference voltage is in actual fact 0.223Vac. The original number was "-"16 dBm. Just read the minus sign as meaning a voltage less than the 0dBm reference voltage. So that 0.223Vac reference value is divided by your 6.31 factor. get "0.035" Vac. So "-16dBm" is really 35mVac. This means you have more than enough drive voltage available from your 2.5Vpp (900mVac) local oscillator signal. Except neither you or Hernan can be sure of this. His source is not specified to work into a 50 ohm load or present a 50 ohm source impedance to the mixer (what is really needed). Who knows what the delivered voltage will be when driving the mixer port? A measurement is in order. Terminate the source in 50 ohm and measure the power and/or voltage. If it exceeds -16 dBm, attenuate accordingly. I imagine we would both have read Hernans post the same way. I.e that he has some logic running at 40MHz and is seeing a rough 2.5Vpp sinewave on his 50-100MHz oscilloscope via a 10:1 probe. From this I'd also assume we both knew that the logic drive impedance would be a couple hundred ohms at the most and that serious mis-matching was not going to be a problem. I'd though, suggest he just connects the parts together and see what happens. He's experimenting. Monitoring the results (good or bad) is just part of the due process. Doesn't look like he's got a spectrum analyser, so how can he validate a 50ohm test measurement as exceeding -16dBm?. His sticking point was specifically about the link between Dbs and Volts. A common, basic electronics question yet surprisingly badly answered by the original suggested link, which started off with the concept of a "dimensionless gain" and went downhill from there using 3 pages of sums. How many radio hams talk to each other about their TX powers in terms of (ISO standard) units of dBs wrt one watt?. Why do the filter tables tell me a Cheb' filter ripple is in dBs when all I want is percent values. Why do my function generators handbooks tell me the sine THD is 0.2% upto 200kHz but beyond that the distortion suddenly becomes an obscure "-30db wrt the fundamental". A recent EW magazine article for a AC mV meter said the response was flat within 0.0024dB. How flat is that?. Be very wary whenever you come across dBm figures. There is a minefield of disinformation out there. In many cases they are intended purely to obfscutate the reader and prevent them clearly seeing that the described circuit is junk. Spoken like a real expert on bafflegab. Indeed, the word was intended that way (UK English). regards john |
From: Roy Lewallen on Sat, 04 Jun 2005 14:06:41 -0700
wrote: A basic definition that is industry-wide, government-wide, has "dbm" as decibels of "0 dbm" related to a power level of 1.0 milliWatts in a "50 Ohm system." That has become so widespread that specification writers don't always include those words. It is implicit when referring to RF components. . . . That's common in the RF industry, but many others also use dBm -- for example, it's often used in video systems where the standard impedance is 75 ohms, and others where the standard is 600 ohms. In all those applications, dBm is universally defined and understood to mean dB relative to 1 mW, regardless of the impedance. Quite true, Roy. :-) I restricted myself to "50 Ohms" for a couple of reasons: As far as "pure" RF components go, 50 Ohms is the Z characteristic; I didn't want to complicate my explanation. The TV Cable industry is HUGE and they use 75 Ohms. However, so many radio amateurs run around snarking on "TV" that it could have raised a lot of unneccessary babbling in here. :-) TV cable is digital in some locations (ours is in the SF Valley of L.A.) and has more TV channels crammed into the same VHF-UHF space than old analog TV. Don't know the modulation of my digital TV cable signals, whether it is wider or narrower than analog channels...but my TV service crams over 400 channels into the same bandwidth. TV sure isn't the narrow-band stuff that many hams are used to. "dbv" (sometimes "dbu" but rarely) refers "0" as 1.0 microVolt, almost any characteristic. Not seen much in specifications, though. The "VU" (for Volume Unit) is an old, old one in the audio and telephone industry with "0 VU" being 1.0 mW into 600 Ohms impedance. Not only that, the "VU" industry standard used to call out the indicating meter's ballistic (needle or meter motor) characteristics! :-) "dbc" is an often-used term on component specifications but is still a relative term of db in regards to the Carrier of a signal where the noise is called out. ["C" for Carrier] "dba" is a legal term in the USA standing for "Doing Business As" in local governments that require business licenses. :-) My apologies if some ISPs show two more postings of my message. When I posted via Google on Saturday early afternoon, Google was interrupting itself with lots of users (?) or something that kept prompting "server error." I've since removed redundant posts on Google, but some other ISPs may be storing the multiples. Stuff happens. |
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I've never seen "dBv" refer to 1uv either. Always 1V.
Joe W3JDR "Roy Lewallen" wrote in message ... wrote: . . . "dbv" (sometimes "dbu" but rarely) refers "0" as 1.0 microVolt, almost any characteristic. Not seen much in specifications, though. . . . Interesting. I've many times seen dB relative to one volt as dBV, and dB relative to one microvolt as dBuV, but never dBv meaning dB relative to one microvolt, or dBu at all. Do you have any references that show these unusual usages? Roy Lewallen, W7EL |
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This page: http://decibel.biography.ms/ mentions both dBv and dBu.
Says in "electrical voltage" (probably with reference to audio industry), dBu and dBv both mean dB relative to 0.775V -- generally 0.775Vrms, but dB taken as 20 log(V/0.775), without reference to a particular impedance. But the same page says dBu or dB(lower-case Greek mu: "micro") as radio power is dB relative to one microvolt per square meter. Go figure. It all points out the need to be careful to define your terms if there's any chance of ambiguity. If you're not careful, your reader may think dBu refers to Dallas Baptist University, or Deutsche Billiard Union, or Duluth Business University... though the capitalization would be wrong for those. Cheers, Tom |
Thanks, Tom. This newsgroup is truly educational. I'm slowly learning
that out there, somewhere, just about every possible convention or nomenclature is used by someone for some purpose. I'll bet I'm the only kid on my block now who knows what dBu and dBv mean. And I guess Len is the only kid on his block that knows they sometimes mean dB relative to 1 uV, as well -- ah, what's a factor of 775,000 one way or the other, anyhow. dBu = dumb ******* unit. Roy Lewallen, W7EL K7ITM wrote: This page: http://decibel.biography.ms/ mentions both dBv and dBu. Says in "electrical voltage" (probably with reference to audio industry), dBu and dBv both mean dB relative to 0.775V -- generally 0.775Vrms, but dB taken as 20 log(V/0.775), without reference to a particular impedance. But the same page says dBu or dB(lower-case Greek mu: "micro") as radio power is dB relative to one microvolt per square meter. Go figure. It all points out the need to be careful to define your terms if there's any chance of ambiguity. If you're not careful, your reader may think dBu refers to Dallas Baptist University, or Deutsche Billiard Union, or Duluth Business University... though the capitalization would be wrong for those. Cheers, Tom |
Hi
Thanks to all of you... like Roy said, this newsgroup is very educational... now I have enough information to make some experimentation with this mixer and oscillator... actually, the mixer is an Analog Devices AD607 and the oscillator is a Linear LTC6903... I'm planning to build my own rig... Thank you so much. Hern=E1n S=E1nchez HJ4SZY |
Hi Hern=E1n,
Hope a couple comments will be helpful: I'd suggest you reconsider using the LTC6903. The data sheet says it has a lot of jitter, which will translate to huge phase noise... If you connect up just the LTC6903 and listen to its output on a good receiver, if it doesn't sound like a clean tone (SSB/CW mode receiver) or a noisless carrier, that's what it's going to add to whatever signal you want to listen to when using it as an LO for a receiver. There are probably some very decent low-power LO designs in the QRP community that you could adapt. May be difficult to get wideband digital control, which it looks like you might be trying to get with the LTC part. I suppose the AD607 is pretty tolerant of levels at the LO input, and it's a relatively high input impedance. You don't have to waste power by adding an external 50 ohm resistor. Get the LO input voltage level close to the suggested value or a bit more and you should be fine. Cheers, Tom nanchez wrote: Hi Thanks to all of you... like Roy said, this newsgroup is very educational... now I have enough information to make some experimentation with this mixer and oscillator... actually, the mixer is an Analog Devices AD607 and the oscillator is a Linear LTC6903... I'm planning to build my own rig... =20 Thank you so much. =20 Hern=E1n S=E1nchez HJ4SZY |
On Mon, 06 Jun 2005 11:12:03 +0000, W3JDR wrote:
I think that would be dBuv not dBv which is Ref 1V I've never seen "dBv" refer to 1uv either. Always 1V. Joe W3JDR "Roy Lewallen" wrote in message ... wrote: . . . "dbv" (sometimes "dbu" but rarely) refers "0" as 1.0 microVolt, almost any characteristic. Not seen much in specifications, though. . . . Interesting. I've many times seen dB relative to one volt as dBV, and dB relative to one microvolt as dBuV, but never dBv meaning dB relative to one microvolt, or dBu at all. Do you have any references that show these unusual usages? Roy Lewallen, W7EL -- Korbin Dallas The name was changed to protect the guilty. |
On Tue, 07 Jun 2005 01:54:14 -0700, Roy Lewallen wrote:
dBv = 1v dBu = 1 uV dBm = 1 mw dBw = 1 w dBk = 1 KW Thanks, Tom. This newsgroup is truly educational. I'm slowly learning that out there, somewhere, just about every possible convention or nomenclature is used by someone for some purpose. I'll bet I'm the only kid on my block now who knows what dBu and dBv mean. And I guess Len is the only kid on his block that knows they sometimes mean dB relative to 1 uV, as well -- ah, what's a factor of 775,000 one way or the other, anyhow. dBu = dumb ******* unit. Roy Lewallen, W7EL K7ITM wrote: This page: http://decibel.biography.ms/ mentions both dBv and dBu. Says in "electrical voltage" (probably with reference to audio industry), dBu and dBv both mean dB relative to 0.775V -- generally 0.775Vrms, but dB taken as 20 log(V/0.775), without reference to a particular impedance. But the same page says dBu or dB(lower-case Greek mu: "micro") as radio power is dB relative to one microvolt per square meter. Go figure. It all points out the need to be careful to define your terms if there's any chance of ambiguity. If you're not careful, your reader may think dBu refers to Dallas Baptist University, or Deutsche Billiard Union, or Duluth Business University... though the capitalization would be wrong for those. Cheers, Tom -- Korbin Dallas The name was changed to protect the guilty. |
Korbin Dallas wrote:
On Tue, 07 Jun 2005 01:54:14 -0700, Roy Lewallen wrote: Just for the record, no, I didn't write this: dBv = 1v dBu = 1 uV dBm = 1 mw dBw = 1 w dBk = 1 KW Roy Lewallen, W7EL |
Hi.
Thanks for your suggestions.... Yes, I'm trying to do some digital control to the frequency of LO using the LTC part. I will think in another way to do that... I'm looking for an oscillator of about 33MHz to replace the LTC part... any ideas ? About the AD607 comment, I have a voltage divider to couple the previous LO output to the level of LO input of AD607... is it what you mean ? Thanks Hern=E1n S=E1nchez |
Hi again.
One last question (for this thread) about jitter... how low it needs to be ? The LTC6903 datasheet says it's 1% (max value). Thanks Hern=E1n S=E1nchez |
nanchez wrote: Hi again. One last question (for this thread) about jitter... how low it needs to be ? The LTC6903 datasheet says it's 1% (max value). Thanks Hern=E1n S=E1nchez Because I work with high-speed ADCs, I'm most familiar with articles about jitter in sampled systems. A sampler and a mixer are pretty similar, and you should be able to learn a lot from things like Analog Devices ap notes AN-501 and AN-756, and even the data sheets for converters like the AD6644 and AD6645. For example, the SNR for an AD6644 sampling a 30MHz sinewave at 65Msamples/sec is degraded perceptably by clock jitter of 0.15psec, which is about 0.001% jitter, expressed as a percentage of the clock period. I'd post links to the pdf files for those ap notes, but the form I have them in, they are too long to reliably include in a posting. Just go to http://www.analog.com/en/index.html and enter AN-501 or AN-756 into the search box. Also, if you enter "jitter phase noise" (without the quotes) into a Google search, you'll get LOTS of references. http://www.maxim-ic.com/appnotes.cfm...te_number/3359 is an ap note on conversion between clock jitter and phase noise. Disclaimer: I have not reviewed any of these apnotes critically for accuracy. Cheers, Tom |
Yes, a voltage divider should do fine. It may be convenient to have
the voltage divider output look approximately like a 50-ohm source, but from what I can see in the AD607 data sheet, there is no requirement that the LO come from a 50 ohm source. You just want to deliver a voltage to the LO input pin which is equivalent to -16dBm across a 50 ohm resistor: in other words, about 35mV RMS or 0.1V peak-to-peak. I expect that if your level is anywhere between 0.1Vp-p and 0.2Vp-p, or even more, or perhaps even a bit less than 0.1, it should work fine. I didn't notice anything very explicit in the data sheet. about it. Two choices for digitally-controlled LO are DDS (direct digital synthesis) and PLL (phase locked loop). Each has its strong points and drawbacks. Do you want to keep the power very low? What sort of frequency resolution do you want? How simple do you need to keep the circuit? How critical is it that the LO not have any noticable spurious outputs? -- I still suspect that the QRP community has already done some nice work on LOs that might suit your need. Cheers, Tom |
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