Please check my derivation of the gain of single ended diff amp

[yungman]

I am having a very difficult time verifying the equations for full differential amplifier. I literally read all the articles I can find from Google. Here is one of the question I think the paper by Jim Karki is wrong. The equation is equation (9). I verified the rest of the equation already, only the (9) is wrong. But I want to verify this here first. This is the paper I am referring to:

http://www.ti.com/lit/an/slyt310/slyt310.pdf
http://www.ti.com/lit/an/slyt310/slyt310.pdf
This is my work to proof the article is wrong.

I go step by step using Thevenin to get the gain of V_OD/V_SIG is what the paper claim is V_OD/V_IN. Please help me verifying.

Thanks

 

[The Electrician]

I go step by step using Thevenin to get the gain of V_OD/V_SIG is what the paper claim is V_OD/V_IN. Please help me verifying.

Thanks

I agree with your result for V_OD/V_SIG. Also note that the paper's Eq (9) has an extra factor of 2.

Did you look at the referenced paper by TI: http://www.ti.com/lit/an/sloa054e/sloa054e.pdf which analyzes the full differential amplifier?

 

[yungman]

I agree with your result for V_OD/V_SIG. Also note that the paper's Eq (9) has an extra factor of 2.

Did you look at the referenced paper by TI: http://www.ti.com/lit/an/sloa054e/sloa054e.pdf which analyzes the full differential amplifier?

Thanks for your response.

Yes, I read about everything I can find on Google search. Calculating the gain and input impedance is a lot more difficult and un-obvious than I expected. It is actually quite difficult and people still resort iteration method to find the value. I gone through in very detail deriving the article you suggested already. This is my notes that started with the SLOA054E and worked into the second article where I find the problem.

There are other articles that use the closed loop gain K to calculate. But I did simulation with LTSpice, the gain is not even close to as simple as Rf/R1 type. I can't even trust those articles. Most of the articles relate to Fully differential amplifiers are written by Ti, I think I read them all and tried to derived them with no luck. This is the closest I get in calculating the input impedance and gain. You can see 4 pages of notes just to get the input impedance. I still am working on the gain.

Things just got very hard when you go into single ended input driving and differential output. I spent like 3 hours a day for the last 3 weeks on this already. I just don't trust any articles until I can verify every single step of it. And here I am!

Thanks

 

[The Electrician]

You might find this useful: https://forum.allaboutcircuits.com/threads/fully-differential-op-amp-expression.26029/

Also this thread: https://forum.allaboutcircuits.com/threads/output-impedance-for-differential-amp-error-in-book.44062/

Your problem could use Shekel's method discussed here: https://forum.allaboutcircuits.com/threads/two-stage-bjt-amplifier-with-feedback.26710/

 

[yungman]

Thanks

I scan through the links quickly, they are really not related to my problem. One talked about differential input which is a lot more obvious and easier. If you look at page 3 of my pdf file, you can see I am working on a single end to differential output circuit driven by a triode. The closest circuit of this kind is the high speed differential ADC driver circuit with matching terminating impedance from the source. If you look at some of the datasheet, the gain is not obvious at all.

If you look at the values on page 23, Table 3 of the datasheet of THS4500 http://www.ti.com/lit/ds/symlink/ths4500.pdf
You will find the values of the resistors to set the gain is not obvious at all. Nothing like the balance differential input that the gain is simple Rf/Rg.

 

[The Electrician]

Also have a look here for info on using admittance matrices with opamps: https://forum.allaboutcircuits.com/threads/admittance-matrix-for-fliege-notch-filter.112366/

 

[The Electrician]

Here's the solution to your Figure 2 in the pages of your work attached to post #3. I did this rather quickly so I could have made a mistake, but I think it's correct. It shows the method of Shekel and its power for solving problems of this type. The formulation of the problem actually uses a finite gain for the amplifier, but I allowed it to become infinite to simplify the final expressions for gain and input impedance.

 

[The Electrician]

I compared my result to the result you get in your Figure 2 work and it's exactly the same expression for Vod/Vin.

 

[yungman]

Thanks Electrician

I have to look at your matrix tomorrow and likely have more question on how you do it.

But at the mean time, I figured out my question already from you pointing out there is a "2" in the equation. The equation (9) is correct. It assume impedance matching between the source Rs resistance is matching to the termination resistance R_T//R_IN. So V_IN is exactly 1/2 of V_SIG. So the gain of V_OD/V_IN = 2 X V_OD/V_SIG.

But this does not help me because I am not doing matching impedance in transmission line. This is a tube circuit. the R_S is really the internal plate resistance r_p of the tube. R_T is really the external plate resistance of the tube. For tube circuit, you cannot measure V_SIG, it's not accessible. You can only measure the plate voltage which is V_IN in my drawing.

But I have very good idea how to get the answer. Thanks for your help. I got my answer. I'll post the equation after I derive it out.

Thanks

 

[Baluncore]

But I did simulation with LTSpice, the gain is not even close to as simple as Rf/R1 type.

Can you post the LTspice.asc.txt file, the extra .txt extension allows it to be attached to a PF post.
What model FDA did you use? Please include device.asy.txt and device.sub.txt files.

The factor of half, due to the double-ended line matching is going to bite you if you ignore it.
Only when the line is correctly matched does the situation reduce to a simple ratio.
Any (Vsig source resistance) + Rs needs to be equal to Zo.
Termination matching requires that; Rg // Rt = Zo.
Line input series termination of Vin by Rs gives; Vin = Vsig / 2.
The output voltage will then be; –Vin * Rf / Rg,
Which becomes; Vout / Vsig = – Rf / ( 2 * Rg ) that is eqn; (2).
The open loop gain of the FDA and any mismatching at the ends of the transmission line will make it very complex.

 

[The Electrician]

Can you post the LTspice.asc.txt file, the extra .txt extension allows it to be attached to a PF post.
What model FDA did you use? Please include device.asy.txt and device.sub.txt files.

The factor of half, due to the double-ended line matching is going to bite you if you ignore it.

As he said in post #9: "I am not doing matching impedance in transmission line."

The transmission line shown in the TI document is not relevant to the OP's situation; he won't have any such line.

 

[Baluncore]

The transmission line shown in the TI document is not relevant to the OP's situation; he won't have any such line.

That is OK, I just wanted the same FDA model he used so that we were not running different models. Adding a transmission line and matching is trivial.

 

[The Electrician]

I've verified your derivations associated with Figure (5). I used b1 and b2 to represent your feedback factors β+ and β-. It would be trivial to get the expressions for the case where A is finite, including where A has a rolloff with frequency, such as a single pole rolloff.

 

[yungman]

Good morning guys.

I attached the LTSpice simulation. I just use a differential amp from LT so I don't have to include the model file. Just any diff amp will do. I put in a few examples of the values presented in Table 3 in page 23 of the THS4500 datasheet in post #5. I change the values of Rs and RT etc and simulate the result. You can see some comments I have whether those change the gain or Rin.

I am a retired EE that still have a lot of passion in electronic design. I spent a year designed and built two tube guitar amps, then I moved to design audiophile amplifiers. I designed and built 2 SS power amps, now I am designing an audiophile tube power amp. This is part of the design.

This is really a part of my tube amp design nothing to do with differential amps at all. I don't match impedance at all, what's important for me is getting as much gain as possible from the driving tube as shown in Fig. 4 BUT keep the Rf resistance low enough so the roll off frequency is as high as possible. Originally, I was thinking to go with the simple design shown in Fig.2, but that entails using a low impedance driver like a cathode follower ( like emitter follower). That will waste half of the tube. Unlike transistor, tube is a whole lot more expensive...and more importantly, take up valuable space on the chassis. I have only room for one twin triode, I cannot afford to give up one to lower the output impedance to drive the diff amp in Fig.2. That's what complicates the derivations.

My constrain I put is R₁ = R₂ + r_p1//R_p1 to make the circuit balance to lower the even harmonics and eliminate DC offset. that's the reason it is so much more complicated than in Fig.2 and even the case of matching source and termination impedance shown in the datasheet.

I am working on the derivation of the gain for my design, I'll post it after I double checked the formulas. It's a whole lot more complicated than equation (9) for match source and terminating impedance.

I'll be back.

 

[yungman]

I've verified your derivations associated with Figure (5). I used b1 and b2 to represent your feedback factors β+ and β-. It would be trivial to get the expressions for the case where A is finite, including where A has a rolloff with frequency, such as a single pole rolloff.

View attachment 217480

Hi Electrician

Thank you for verifying my other equations on Rin and Rinp. Now I feel a lot better. I never studied linear algebra, so it'll take me a little time to read your matrix.

Alan

 

[The Electrician]

Hi Electrician

Thank you for verifying my other equations on Rin and Rinp. Now I feel a lot better. I never studied linear algebra, so it'll take me a little time to read your matrix.

Alan

Here's a good exposition on how to set up an admittance matrix by inspection: http://web.ecs.baylor.edu/faculty/grady/EE411_Fall2011_Week_02.pdf

Here's the step-by-step setup using opamps: https://forum.allaboutcircuits.com/threads/admittance-matrix-for-fliege-notch-filter.112366/#post-869823

 

[The Electrician]

Shekel's method using an admittance matrix as I've shown in this thread doesn't require deriving expressions for impedances at various internal nodes. The transfer function just falls out naturally.

The very last part of your work shown in the attachment of post #3 adds a vacuum tube with transconductance gm and plate load RL. Adding this to the circuit of Figure (5) is easy; all that is required is the addition of one row and one column. The resistor R1 is changed from R1 = R2 + rpl || Rpl to R1 = R2 + rpl || RL to maintain balance. A grid resistor Rg is added to prevent the matrix from becoming singular, but Rg doesn't appear in the output expression if the grid is driven from a voltage source.

Here's the result for the transfer function Vod/Vg when a single triode drives the FDA:

 

[yungman]

Here is my derivation of the gain of V_OD/V_IN

R_IN is looking into R₂ from the left side with the left side of R₂ open.

I just want to verify if this is correct. If so, I am in business.

Thanks

 

[The Electrician]

I see some inconsistencies in the Vod calculation. You have Vod/Vth = Rf/R1; apparently this is with R1 not equal to R2 + rp1 || Rpl. But then you have Vod/Vth = Rf / R1 = Rf / (R2 + rp1 || Rpl) which would seem to indicate R1 has been replaced with R2 + rp1 || Rpl.

In either case I don't think Vod/Vth is correct as shown.

Also, what are you using for β+ and β-?

I would find it less confusing if when calculating Vod/Vin you would show a schematic with nothing to the left of the Vin node. In other words, don't even show Vth or the resistor labeled rpl ||Rpl. Also, show your expressions for β+ and β-.

For such a circuit, if R1 is replaced with just R2, here's what I get for Vod/Vin:

If R1 is left as R1, then I get this for Vod/Vin:

And, finally, if R1 becomes R2 + rpl || Rpl, then I get:

 

[The Electrician]

If I use the circuit where the input source is Vth driving through a resistor equal to rpl || Rpl and with R1 = R2 + rpl || Rpl, I get the following:

Notice that the result for Vod/Vin here is the same as it was in the last image in the previous post. This is as it should be because everything to the left of R2 (the Vin node) doesn't matter when we're calculating Vod/Vin.

 

[yungman]

The only constrain is R₁= R₂ + r_p1//R_p1. So the gain of V_OD/V_TH is just simply R_f/R₁.

beta+ and beta- is defined in page 3 of the write up I linked in post #3

If you use V_TH to drive as shown in post #18. the differential amp is totally balanced, the gain is just simply

V_OD/V_TH= R_f/R₁ = R_f/[R₂ + r_p1//R_P1]

 

[The Electrician]

The only constrain is R₁= R₂ + R_S//R_T. So the gain of V_OD/V_TH is just simply R_f/R₁.

beta+ and beta- is defined in page 3 of the write up I linked in post #3

I don't see RS and RT in the schematic of post #18. Where are they defined?

 

[yungman]

I don't see RS and RT in the schematic of post #18. Where are they defined?

I'm sorry, I just got mixed up, the article use R_S and R_T, I use r_p1 and R_p1.

I changed post 21 already.

I defined R₁= R₂ + r_p1//R_p1 and never change through out the whole derivation. This is needed to minimize distortion and eliminate DC offset.

 

[The Electrician]

OK. To make sure we're on the same page, here is a schematic of the circuit I will be analyzing. I'll use the definitions for β+ and β- from page 3 of your attachment to post #3, but I will use b1 for β+ and b2 for β-. I've numbered the nodes in red as used in the admittance matrix for the circuit. Node 1 is row 1 of the matrix, node 2 is row 2 of the matrix, and so on.

Is this what you are using?

 

[yungman]

Yes, that's it.

Thanks

 

[The Electrician]

OK. Once I get the topology exactly the same as yours, and get the same definitions as yours, I get the same result. Here is my work:

 

[The Electrician]

To show the power of Shekel's method, suppose you wanted Vod/Vin with a finite opamp gain A. In an instant I can get that result:

Or perhaps you would like to know VP:

Look how much more complicated the expression for VP is for finite opamp gain A:

 

[yungman]

Wow, thanks Electrician.

Thanks for all your work to verify for me. When I finish my design, I really need to stop and go over the stuffs you sent me. Took me a long time to do it my way and you can do it so fast by setting up the matrix and solve it.

Is this a simulation program?

Thanks

Alan

 

[The Electrician]

No, it's just using a so-called mathematical assistant program to do the tedious algebra involved in the matrix arithmetic. I'm using Mathematica, but there are others such as Matlab, Mathcad, Maple, Macsyma, etc.

 

[jim hardy]

Just amazing...