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Differential amplifiers?


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8/1/1997 9:47 AM
Erik M. Wood
Differential amplifiers?
I am curious about tube differential amplifiers. I discovered a website which describes some pretty interesting designs for tubes. Included are your basic single ended design, balanced differential, cascode and bridge circuits, all of which are touted for their relative benefits and deficiencies.  
What I do not understand, is how when you have two parallel triodes, cathodes tied together, then both of them in series with a single resistor to a negative power supply (B- equal to B+), as in a balanced differential design, how an input signal to either grid will produce (as they describe) equal but out of phase outputs at the two plates (tied to B+)? Could someone please explain to me how one of this identical pair of triodes will be in phase with it's input while the other will be out of phase?  
I am of course assuming that in a single stage triode, the plate output is 180 degrees out of phase with the grid input signal, no?  
By the way the adress to this website is:  
 
http://www.atma-sphere.com/  
-Thanks,  
-Erik.
 
8/2/1997 8:59 AM
R.G.

>> ... how when you have two parallel triodes, cathodes tied together, then both of them in series with a single resistor to a negative power supply (B- equal to B+), as in a balanced differential design, how an input signal to either grid will produce (as they describe) equal but out  
of phase outputs at the two plates (tied to B+)?<<  
 
Look at it this way - the triode that is getting the input signal produces a changing current through its anode and cathode resistors. The shared cathode resistor voltage changes, and if the second triode's grid voltage is held constant, the changing cathode voltage at the shared resistor is in effect changing the grid-cathode voltage on the second tube by moving the CATHODE around.  
 
The second tube has an input signal, a signal that comes in through the cathode, not the grid. This means that it is operating as a common-grid stage, and it too, then has a signal that appears on its anode.  
 
If the input signal to the first triode is positive going, the first triode anode goes down, and the cathode goes up by an amount about equal to the movement of signal on the grid. This change means that the second triode cathode is moving up with respect to the grid, effectively a negative going signal at the grid, so the second triode tries to turn off, and the anode of the second triode moves up. Given equal gains, the changes on the anodes are approximately equal. With a current source on the cathodes, the signals on the anodes get VERY nearly equal and opposite.
 
8/4/1997 3:39 PM
Erik M. Wood

R.G.,  
Ok, that all makes more sense, now. My next curiosity is with the common mode rejection of this circuit using balanced inputs. Assuming the cathode signals were identical but of opposite phase, they would cancel out at their mutual cathode connection to B- except for the common mode noise and RF picked up through the mic cable, say. This stays as a minor voltage fluctuation on the cathodes, in phase with the signals on the grids. Other than this the B- varies very little. Ok so far? Next, because of what you said about the rising cathode potential looking to the circuit like a sinking grid potential, and the same but opposite for a sinking cathode potential, the common mode noise on the cathode would then cancel out the common mode noise on the grids, since the common mode peaks would effect a rise in cathode potential in time with the common mode noise causing an equal rise in grid potential, effectively creating the difference between two out of phase but identical noise signals. In other words, the grids would be going positive while the cathodes would be pulling them back down an equal amount, correct? This would leave nothing but the intended music signal -equal, but still balanced (out of phase)- at the output. Is this right, or is the noise cancellation achieved in another part of the circuit (e.g. the output)? I have had it explained to me that it is not necessary to flip the two signals back into phase in order for this particular design to offer Common Mode rejection. I'm just trying to understand how this works.  
I hope this is clearer than it looks to me.  
-Erik.
 
8/4/1997 5:26 PM
R.G.
For common mode rejection: Assuming both triodes are identical, if both grids go up by the same small amount, the transconductances of the triodes lead each to turn on by an equal amount, pulling both anodes down by an equal amount. Although the anode voltages change, they both change by the same amount, and the difference between the anodes remains constant, zero in this case.  
 
If B+ changes, it affects each tube the same amount, and the difference between the two remains the same.  
 
An unbalanced signal turns on one tube and turns the other off. This leads to one anode going up and the other going down by the same amount.  
 
On the cathodes, you have it correct, the "up" going cathode and the "down" going cathode effectively cancel, and the cathode moves almost none, with the minor junk as imperfections, as you note.  
 
Anything affecting both grids (or cathodes or anodes, for that matter) equally has little effect on the DIFFERENCE between the two anodes. Obviously, both anodes can rise and fall together, so another stage following the differential amplifier may see the common mode anode motion as signal unless it, too rejects the common mode portion of the signal.  
 
That can be a problem, and in fact is the problem addressed by current source biasing of the cathodes. To the extent that the cathodes are both connected to a perfect current source, the total current through both tubes is held constant. The current source fixes the cathode currents, but allows the actual cathode voltages to float to wherever they need to float to make the total current come out right. That means that In the case of the rising B+, the cathodes both float down a little to keep the Vgk just right to keep exactly the same current through the anodes, and the voltages on the anodes not only track one another, they stay at the same voltage with respect to B+.  
 
In the case of a common mode voltage impressed on both grids, the anodes not only stay the same relatively, they don't move compared to ground, as both cathodes float to a voltage that keeps the anodes at a constant current, and hence a constant voltage with respect to B+ (which we're assuming doesn't move for this one example) and ground.  
 
The key here is "current source" - a constant current that does not change. For tube circuits, this is approximated by a large resistance, sometimes tied to a large negative voltage. The larger this resistance is compared to the changes the circuit sees, the better it approximates a current source, and the better the CMRR is for the amp.  
 
Note that a diff-amp has multiple, sometimes different rejection ratios. The CMRR to B+ is one - the anodes go up and down 1:1 with B+, but do not amplify the common signal. The CMRR with respect to the grids is very high with a current source in the cathodes and decreases as the "current source" is not as perfect.  
 
When you're trying to understand diff-amps, the best approach is to look at them as current steering amplifiers- the grid signals steer current to the side of the diffamp with the highest grid, and away from the side with the low grid. No steering - the currents stay (approximately) the same.
 
8/5/1997 9:56 AM
Erik M. Wood

RG,  
Thanks for the reply. I am beginning to grasp the concept of differential amplifiers. In essence, they look to be the same as the push pull amplifier stages we see so often in power amplifiers. The only difference is the biasing, it seems. The principles seem to be very similar. Now, I was planning to dabble with differential stages for a mic preamp I am making out of an old Bogen PA amplifier, but I find that I must now reconsider the idea of fully balancing it end to end. If the differential amplifier cancels out asymmetrical distortions (e.g. even harmonics), it will significantly cut down on what makes single ended triodes sound so desirable. Overload of a single ended stage contributes to that tube sound we know and love. I don't know if you've read it, but there is an article which was presented to the AES back in 1971 by recording engineer Russel O. Hamm. In it he examines the electrical causes of and physiological responses to tube distortion and transistor/op-amp distortion. In it, he finds that the overload characteristics of a tube in a mic preamp (guitar amps, which are designed to be overdriven are another story) are usually found where there is a mechanical-electrical interface, such as with a transducer (microphone, pickup, speaker, etc...) and a tube stage. This makes sense as real world dynamic ranges of sound as they are produced live, are much wider than the capabilities of the amplifier to accurately reproduce them. Thus, we get squashing of an overloaded input signal, say from a mic. This squashing of the signal accompanied by the asymmetrical clipping and the shifted duty cycle characteristic of a single ended tube stage would not occur with a differential input stage. I'm sure you already know all this, I'm just typing out loud. So, what I think I'm going to do is balance the input with a transformer into a single ended amplifier stage, then run the single ended output into a differential stage since it would not experience the same dynamics as the previous one. The signal would already have the desired squashed, asymmetrical, and shifted duty cycle characteristics, so the benefits of the differential amp would be harmless to the good stuff while still cancelling out further Common Mode signal.  
In case you were wondering, the address to that article is:  
http://www.access.digex.net/~mmilbert/tstxt.htm  
Thanks again for your reply, RG.  
-Erik.
 
8/6/1997 3:17 AM
Dave Harris

"Squashed, asymmetrical and shifted duty cycle". I donít fancy singing through that ! I thought the tube mic pre amps sounded better because in normal use they are likely to produce LESS distortion than an IC pre amp because of their greater dynamic range. The ICs run off 15V and the tubes off 300 so the tubes are less likely to be overdriven. Also the tubes donít have F/B so any (unintentional) distortion produced will have less high order odd harmonics and therefore sound better.  
 
I can only think that you would want the signal "Squashed, asymmetrical and shifted duty cycle" if you were using going to mic up a guitar amp through a PA with no tweeters. "Shields to maximum Mr Scott".
 
8/6/1997 9:45 AM
Erik M. Wood

Dave, I know it's strange to think that the "squashed asymmetrical duty cycle" is counterproductive to a good sounding signal, but it really has to do with the king of distortion produced by the relative technologies. If you were to measure the THD of a tube preamp and a transistor preamp, very often you will find that the measurements are not that far off from one another. The real difference is in what kind of distortion is produced by these two technologies, and that is simply inherent in their nature. Single ended tube triode stages overload so as to generate a predominance of even harmonics. Each harmonic relative to the fundamental frequency is specifically responsible for a particular characteristic of the timbre of the sound you hear . Distortion of the even harmonic variety is much more consonant than odd harmonic distortion -that which is predominantly generated by transistors. THis is because the lower order even harmonic series is more closely musically related to the fundamental than the odd harmonic series. It's easy to prove this by simply working out the harmonic series above any arbitrary frequency. All you have to do is work out the ratios of the diatonic notes of a scale to your fundamental note and compare the accepted consonances and dissonances of the notes represented by each harmonic. As a result, to our ears, a tube amp in the early stages of overload actually sounds cleaner and more pleasing than a transistor amp distorted the same amount (THD measurement). The shifted duty cycle and assymetrical clipping are what produce even harmonics. In a transistor, these two actions do not occur. Instead, the duty cycle is nearly perfect, and the clipping occurs the same amount, top and bottom. This is responsible for the predominance of odd harmonics. Now this is not to say that tubes only produce even harmonics while transistors only produce odd harmonics. It is simply the fact that each technology in it's most basically designed circuit will produce a greater amplitude of one kind over another. To go into the details of it all is beyond the scope of what I have time to type. And I'm sure I've not been exceedingly clear with what I mean, but look up the website I mentioned and it will explaing it better than I can.  
-Erik.  
 
The address to the website article is:  
http://www.access.digex.net/~mmilbert/tstxt.htm
 

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