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I want to find the transfer function of a differential BJT pair. Like internally compensated op-amp can be assumed as a single pole system, is there any method to write the transfer function? I plotted bode plot in LTSPICE and I dont think it is a single pole(differential BJT pair).

enter image description here enter image description here enter image description here

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  • \$\begingroup\$ Show the bode plot please (amplitude and phase). \$\endgroup\$ – Andy aka Feb 23 at 10:47
  • \$\begingroup\$ Can you please check the question now? is the plot visible ? \$\endgroup\$ – Hari Krishna Feb 23 at 10:55
  • \$\begingroup\$ Start the plot at 1 kHz (not a pointless low frequency) and indicate with different colours what the two lines are and, be unambiguous about those lines i.e. don't say something meaningless like V(vo+)-V(vo-) because that just wastes people's time trying to figure out what it is meant to mean. From what I can tell you haven't made the compensation correctly i.e. there does appear to be a dominant 2nd order effect. Also, use a white background. \$\endgroup\$ – Andy aka Feb 23 at 10:59
  • \$\begingroup\$ Oh sorry for wasting your time, i will make the changes and update it here \$\endgroup\$ – Hari Krishna Feb 23 at 11:00
  • \$\begingroup\$ @Andyaka I was unable to change the expression , so I have added two different images. Is this okay? \$\endgroup\$ – Hari Krishna Feb 23 at 11:20
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Other than what @Andy Aka has written, I want to write more about the mathematical steps. Like, how to get an approximate TF during design phase of simple amplifier circuits.

You want to calculate the TF of BJT Diff pair. The only way to do that is to take you actual circuit, replace every thing with s-domain high frequency model of them and then using network theorems and other mathematical concept find Vout wrt. Vin and that would be the TF (I wish you good luck for that ;)). Any thing other than that is approximation.

Now, what do you use a TF for? there are a number of reason and one of them is to determine the bandwidth of the amplifier and variation of gain with frequency. But do we really need a TF for that? Strictly speaking, yes! But during the design phase, we can use approximations to approximate and bottom down to some design parameters and then actually testing the amplifier circuits using simulations or making the practical circuits. If actual measurements are way off than approximation then people employ more sophisticated techniques to get better approximations. Right now, I might be sounding like as if we only need approximations, but this is not true. In simple amplifier circuits, it is easy to determine the whole TF, for example a source/emitter follower but in complicated designs it is good to first stick with approximations.

So if those approximations are good enough, then using those we can write approximate TF. for start and usually for relatively simple circuit, we assume that the bode plote of the TF looks like the following:

enter image description here

There is a lower cutoff frequency, higher cutoff frequency and a mid-band gain and usually we obtain a TF to get these parameters(However, if someone is working in a feedback systems, parameters such as gain and phase margins and variation of phase with frequency becomes important, but if someone working at such high level would be educated enough to know what assumptions to take in the beginning). But what if we have methods to determine those parameters before hand i.e. without finding out the TF and then use those values to approximately calculate the TF. And this is what I am going to write about.


In a differential amplifier, there are two curves(not the only two) (a) variation of common-mode gain (b) variation of differential gain and subtracting (a) from (b) gives you variation of Common mode rejection ratio(CMRR) with frequency. And for now, I am assuming that you are talking about variation of differential gain.

Now, in order to find the differential gain, Assume that you have the following circuit: enter image description here

Note: I have not shown the signal source resistance in here, but it will be always present

C1, C2, C3, C4 are used to not disturbed the DC levels of this circuit, however, it is not necessary to have any of these. Infact in IC, you won't even find the resistor and the above mentioned caps. However, in discrete implementation they are necessary. So the problem that above mentioned capacitor poses is that they increases the lower value of frequency of the amplifier bandwidth and one good thing about them is that they do not affect the upper cutoff frequency of bandwidth. The upper cutoff frequency is determined by the parasitic capacitances of the bjt that is being used.

Now, in order to calculate the frequency response of this bjt diff pair, makes it's half circuit. That would look like as following: enter image description here

Remember that,

The above differential half circuit (common emitter amplifier) can be used to determine the frequency dependence of differential gain. The gain function \$A_d(s)\$ of the differential amplifier will be identical to the transfer function of this common emitter amplifier [1].

For this half circuit calculate the mid-band gain which is normally the gain we calculate using low frequency hybrid-pi model and replacing all power supply with ground and replacing caps with short circuits and for a common emitter it is given by \$A_{mg}=g_mR_d\$.

How Determine the Lower Cutoff Frequency?

Method of Short Circuit Time Constants

  1. Set the input signal source (V1) to 0
  2. Take one capacitor at a time. Say you took \$C_i\$, so short circuit rest of the capacitors in whole circuit.
  3. Find the Thevenin's equivalent resistance across the capacitor under consideration. This is basically the total resistance seen by that capacitor. Let it be \$R_i\$.
  4. Calculate \$C_i\times R_i \$ for all \$ i\$

The lower cutoff frequency (\$f_L\$) can be estimated $$f_L\approx\frac{1}{2\pi}\sum_{i=0}^{n}\frac{1}{C_i\times R_i}$$

How to determine the upper cutoff frequency?

Although one can easily use miller approximation to spilt any feedback capacitor in circuit and then can determine the upper cutoff frequency. But the following method gives better approximate answers and is more general in nature.

Method of Open Circuit Time Constants

First draw the high frequency model small signal model of the amplifier. Short circuit all the capacitors which were used in determining the lower cutoff frequency because we assume here that at higher cutoff frequency they will be acting as short circuit(this assumption allows us to make small signal model of the Amplifier …!). Now do the following for all capacitor in the small signal high frequency model of amplifier:

  1. Reduce the input signal source to 0.
  2. Select a capacitor \$C_i\$, replace all other capacitors with open circuit and find the Thevenin's equivalent resistance around \$C_i\$ and let it be \$R_i\$.
  3. Calculate \$C_i\times R_i\$

Hence high cutoff frequency is approximated by the following: $$f_H\approx \frac{1}{2\pi}\times \frac{1}{\sum_i C_i \times R_i}$$

Final words

now that you have 3 parameters, you can approximate the TF but that would be useless because it would be good only to get those parameters and don't try to get anything more out of that.

Again I am highlighting the fact that this is just an approximation. Sometimes, you can have a higher order response but that will depends in how much detail you want to analyze your circuit.

If you want to read more then go though chapter 6, 8 & 9 of the book 1.

[1] Microelectronics circuits theory and applications; Sedra Smith; 7th edition.

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  • \$\begingroup\$ Thanks for taking time to answer. \$\endgroup\$ – Hari Krishna Feb 24 at 11:15
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I want to find the transfer function of a differential BJT pair

I've taken your graphs and merged/stretched them to align their x-axes so that I could then scrawl all over them to highlight where the poles appear to be. I have assumed that the magnitude plot is |output|÷|input| rather than some obscure and meaningless name that is used in the original plot.

Here's the result: -

enter image description here

Firstly, look at the black line I drew; it corresponds with a 1st order slope of 20 dB per decade. As you can see, between 100 MHz and 1 GHz (dark blue lines), it's starting to become more than 20 dB/decade (pale blue lines) and is strongly hinting at another pole (around where the blue circle is).

The phase response; see how it never levels out at a 90° lag (purse first order system) but keeps on dropping towards a 180° lag (a bit below 1 GHz) and then keeps on dropping even lower to hint at a 3rd order response.

Having said all that, if you navigate left to right along the horizontal green line on the magnitude plot from 0 dB to where it hits the magnitude curve, then project that point downwards to the phase plot and then across the phase axis, there is going to be a case for it having marginal stability but only just. There will likely be a lot of transient overshoot if heavy feedback is applied.

Regarding the orange lines I added; I began at a 45° lag on the phase response because this is synonymous with the lowest pole in your circuit. I then projected across to the phase curve and then upwards to the magnitude curve to check that there was indeed a 3 dB reduction in magnitude at this frequency. And, of course that looks good.

So, in summary, I believe your first pole starts around 24 MHz (orange circle) and, where the blue circle is (about 200 MHz), it turns into a 2nd order system. Somewhere in the high hundreds of MHz to low GHz, a third pole is beginning to emerge.

My recommendation is that you model it with a simulator using: -

  • a gain block of 26 dB followed by
  • an RC low pass filters at 24 MHz,
  • a unity gain buffer
  • an RC low pass filter around 200 MHz
  • a unity gain buffer
  • and finally an RC low pass filter around 2 GHz.

Then, you should be able to get a reasonable match and that modelled circuit, because it is mimicking your actual circuit is basically telling you what the TF is. Solve the TF of the above model (that mimics your circuit) and you have the result you want.

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  • \$\begingroup\$ Ohkay , thank you @Andyaka \$\endgroup\$ – Hari Krishna Feb 24 at 6:09
  • \$\begingroup\$ @HariKrishna maybe you should take the 2 minute tour to find out what is expected from you when a question has received answer(s). Also, if there's anything else related to the question that you need clarifying, please leave a comment. \$\endgroup\$ – Andy aka Feb 24 at 7:58

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