An obscure question about triodes, impedance, clipping, and gain - diyAudio
 An obscure question about triodes, impedance, clipping, and gain
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 2nd September 2015, 05:24 AM #1 diyAudio Member   Join Date: Sep 2015 An obscure question about triodes, impedance, clipping, and gain Hello, I am rather new to design, but I am working on a design for a guitar amp with several unusual features. In my studies, I recently learned that the output impedance of the stage versus the input impedance of the next stage will affect gain. My design is rather unusual so I am following each triode with an audio transformer. I read that if the output impedance of the tube stage matches the impedance of the transformer primary, maximum power is transferred, but voltage gain will be half of what the unloaded voltage gain calculations indicate. Voltage gain increases if there is impedance bridging, and voltage gain reduces further from half if the tube stage output impedance exceeds the xfmr primary impedance. Current is reciprocal to voltage in all this. Ok, got it. Here is the question. Lets say that B+ is 100 volts and the input voltage on the grid is 10 volts peak to peak max. Ok that means that at a voltage gain of 10, the tube will just begin to clip at max input level, ignoring output impedance issues. Lets say that the gain is set to 15. OK now match the triode gain stage with the transformer, and make the impedance ratio 1:1. That cuts the voltage gain of the stage down to 7.5, not 15. Does that kill the clipping? Take that 10v p-p input signal. At 15 gain it clips like hell. At 7.5, no clipping. So, does the tube actually operate at 15 gain and then have that gainy signal attenuated by the xfmr? OR, does the tube actually operate differently because it is influenced by the xfmr, such that it does not generate an internal gain of 15, and so no clipping? Is this GAIN REDUCTION or GAIN FOLLOWED BY ATTENUATION? Thanks for your help
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Location: Lancashire
Quote:
 Originally Posted by Fractalist At 15 gain it clips like hell. At 7.5, no clipping. So, does the tube actually operate at 15 gain and then have that gainy signal attenuated by the xfmr? OR, does the tube actually operate differently because it is influenced by the xfmr, such that it does not generate an internal gain of 15, and so no clipping?
Your question is somewhat loaded by other considerations, but in simple terms, the tube operates differently because you attached the transformer. Another way of looking at it is to say that it does provides a gain of 15 (or whatever) but only 'internally'. The large voltage swings that would normally cause clipping never actually reach the outside world, because they are attenuated by the internal resistance of the tube in combination with the transformer load. Equivalent circuits, and all that.

Last edited by Merlinb; 2nd September 2015 at 07:10 AM.

 2nd September 2015, 07:53 AM #3 diyAudio Member   Join Date: May 2007 You won't get an output swing of 100V peak-peak AC from a 100V DC supply rail. Valves need a bit of voltage across them, unlike BJTs. Generally, in audio design, the last thing you want is impedance matching. A severe mismatch will usually give the best results. Transformers were used in the early days of valves for inter-stage coupling because valves then had such low voltage gains - not needed today, unless you want to restrict the bandwidth and add some distortion (which may be the aim in a guitar amp, of course).
diyAudio Member

Join Date: Sep 2015
Quote:
 Originally Posted by Merlinb Your question is somewhat loaded by other considerations, but in simple terms, the tube operates differently because you attached the transformer. Another way of looking at it is to say that it does provides a gain of 15 (or whatever) but only 'internally'. The large voltage swings that would normally cause clipping never actually reach the outside world, because they are attenuated by the internal resistance of the tube in combination with the transformer load. Equivalent circuits, and all that.
Ok, if a gain of 15 is being applied 'internally', that would cause clipping. The gain exceeds the supply rail.

So you are saying that, for clipping purposes, it operates as if the gain were 15, and then this signal is subsequently attenuated....
Right?

diyAudio Member

Join Date: Sep 2015
Quote:
 Originally Posted by DF96 You won't get an output swing of 100V peak-peak AC from a 100V DC supply rail. Valves need a bit of voltage across them, unlike BJTs. Generally, in audio design, the last thing you want is impedance matching. A severe mismatch will usually give the best results. Transformers were used in the early days of valves for inter-stage coupling because valves then had such low voltage gains - not needed today, unless you want to restrict the bandwidth and add some distortion (which may be the aim in a guitar amp, of course).
Transformers are tonal circuit elements, not any kind of utilitarian link. That's what opamps are for. I am trying to add distortion, and also using the xfmr to perform some voltage reduction and current gain needed for subsequent circuitry.

Any opinion about the question I asked would be appreciated. This is a theoretical question, not a question about standard practices.

 2nd September 2015, 09:32 AM #6 diyAudio Member   Join Date: Feb 2010 Location: Helsinki Why not use the transformers as plate loads? Then you could simply draw a load line for the tube and know exactly what the gain is.
diyAudio Member

Join Date: May 2007
Quote:
 Originally Posted by Fractalist Ok, if a gain of 15 is being applied 'internally', that would cause clipping. The gain exceeds the supply rail. So you are saying that, for clipping purposes, it operates as if the gain were 15, and then this signal is subsequently attenuated.... Right?
Wrong. The point where clipping occurs is at the anode, not the internal fictitious voltage source. Hence reducing gain by adding anode loading will change the input point at which output clipping occurs.

Quote:
 Any opinion about the question I asked would be appreciated. This is a theoretical question, not a question about standard practices.
You will find that on this forum people do not restrict themselves to simply providing answers to questions. They also comment on the question, and discuss issues surrounding it. Some newcomers may find this disconcerting at first.

diyAudio Member

Join Date: Sep 2015
Quote:
 Originally Posted by MrCurwen Why not use the transformers as plate loads? Then you could simply draw a load line for the tube and know exactly what the gain is.
Because I am trying to take advantage of the tonal differences caused by impedance mismatching and because I am using a pot for the load resistor so that both gain and output impedance can vary.

My understanding is that the character of the distortion differs with different gain amounts. Even if the amount of clipping and THD/IMD are exactly equal among 2 triode stages (with different input levels), if these are set to different gains they will exhibit a different harmonic structure and their dynamic timbral characteristics differ as well. So this is the basis for using variable gain.

Also, there are tonal differences caused by impedance matching/mismatching. Not just frequency response. TubeZ<XfmrZ causes 'tightness' and TubeZ>XfmrZ causes 'looseness'.

Also, a subsequent optical compressor circuit is sensitive to current levels. At low gain with TubeZ<XfmrZ, the current post-Xfmr is less. At high gain with TubeZ>XfmrZ, the current post-Xfmr is more. More current causes the optical vactrols to have faster attack and release. So the gain setting also controls attack/release of the subsequent compressor.

diyAudio Member

Join Date: Sep 2015
Quote:
 Originally Posted by DF96 Wrong. The point where clipping occurs is at the anode, not the internal fictitious voltage source. Hence reducing gain by adding anode loading will change the input point at which output clipping occurs. You will find that on this forum people do not restrict themselves to simply providing answers to questions. They also comment on the question, and discuss issues surrounding it. Some newcomers may find this disconcerting at first.
Thank you for your reply. I believe what you are saying is that there is no gain of 15, internal or external. The transformer modulates the gain.

Regarding off-topic replies, I believe that when a person asks for something, that is what they seek. When I put coins in a vending machine and select cola, I rarely wish to be given tomato juice instead. If I am asking for people to recommend a beverage, then all answers are welcome. If I am asking for the calorie content of a can of Coke, I don't want people asking me why I want that or telling me that Pepsi would be better for me. I hope this response is not too disconcerting for you.

 4th September 2015, 10:55 PM #10 diyAudio Member     Join Date: Jan 2003 Location: Milwaukee, WI Ah, but what could the input impedance be? At audio frequencies, this is essentially a useless question: the actual input impedance might be several megohms, so that if you could get a transformer of such high impedance, yes, your power gain could be considerable! But if such a transformer is even possible, it likely doesn't have much bandwidth, so that your power match will only be true at a narrow range of frequencies, perhaps 1-2kHz. This is a very valid, and important, question at RF, where the equivalent input resistance might be on the order of only 1kohm! (The capacitive reactance might be 100-200 ohms, so that the grid appears to the circuit as a lossy capacitor.) This is typical of e.g. 6AU6 at 200MHz or so, and limits the SNR and gain of an RF/IF amplifier. For wideband applications, the loss component is essentially invisible, because again, capacitance is already dominant by a sizable factor; and at the frequencies available, even more so (the loss resistance goes as 1/F^2, so that an RF amplifier with 10MHz bandwidth at a 200MHz center frequency has a much harder time than a DC coupled oscilloscope amplifier at 0-10MHz). The reason an audio frequency transformer is impractical is, at low frequencies, the inductance of the winding presents a low reactance, shunting the signal; this can be solved by using more turns, of finer wire, but more wire means more capacitance, shunting the signal at high frequencies. The core material also contributes loss, so that at modest frequencies (100Hz-10kHz?), the impedance continues to rise, but not as fast as a true inductance should (the core presents quite a lot of resistance and not much inductance in this range). For a given circuit impedance, you only get as much bandwidth as the range where transformer impedance remains higher than circuit impedance. So for a number of reasons: impracticality, cost, bandwidth and so on, transformer coupling is rarely used. On the other hand, if you have an application that needs grid current, the impedance is suddenly considerably lower, and a transformer drive may be quite desirable (though cathode followers, or other assistance, may still be better). Tim __________________ Seven Transistor Labs, LLC Projects and Resources / Electronic Design and Consultation

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