Deriving the transfer function of a composite amplifier on standard form - Page 1
Forum section: electronics / beginners

author: vcvs
author_rank: Contributor
timestamp: May 12, 2025, 06:29:41 pm
content: I am trying to derive the transfer function of this circuit from .Assuming simple models for the amplifier and the buffer I was able to get a very complicated and messy expression with a bunch of KVL/KCL (not really worth including, 8 pages of probably wrong equations leading to an expression that not even gives a plot similar to what is expected). But I get stuck on making that into standard form for making a bode plot or getting intuition about the poles and zeros.Any suggestions for techniques to simplify the problem is appreciated. I was thinking about whether it would be possible to approximate poles/zeros by inspection, but not sure how. Or with a symbolic algebra program such as SymPy.Thank you!:https://www.ti.com/lit/ta/sszt102/sszt102.pdf

author: jwet
author_rank: Frequent Contributor
timestamp: May 13, 2025, 06:06:12 am
content: The standard way is to note that there are really two circuits here that have little to do with each other.The top path that goes though the HF cap, into the JFET and the follower is just a high impedance buffer. Assuming its biased correctly, just do AC analysis. It looks like it has a gain of 1 first order.The bottom path with the Op-Amp is a DC servo for biasing- all very low frequency. Its has no effect on the signal path, its just picks off DC and keeps the buffer bias happy.

author: jwet
author_rank: Frequent Contributor
timestamp: May 15, 2025, 02:48:53 pm
content: Did this help at all?You can calculate the poles and zeroes by inspection I think. I didn't go through the numbers but there are two crossovers, one at the RC(HF) and the other is DC to the integrator pole which is 1/RC of the integrator. The R is the equivalent series R at the input, the C is the feedback C.

author: Terry Bites
author_rank: Super Contributor
timestamp: May 15, 2025, 03:06:31 pm
content: calculate the currents in the indvidiual nets one at a time and sum them up for each node.

author: vcvs
author_rank: Contributor
timestamp: May 15, 2025, 03:56:02 pm
content: Quote from: jwet on May 15, 2025, 02:48:53 pmI think I understand the circuit intuitively – one part is used for low frequencies (and uses the DC precision amplifier in the lower stage to achieve DC precision) and the other for high frequencies, so approximately LP and HP filters. To achieve a flat response we need to place the HP dominant pole before the LP dominant pole, and make sure to use the resistor network to match the DC gain with the gain of the buffer.But I can’t really figure out how to determine the poles. For example, I could guess that the upper capacitor affects the pole placement of both signal paths (the interdependency they talk about in the datasheet) but do not really have the theory/metholodgy to determine which components are relevant for the poles, how many there are and how to divide the circuit to find them..

author: vcvs
author_rank: Contributor
timestamp: May 15, 2025, 04:00:43 pm
content: Quote from: Terry Bites on May 15, 2025, 03:06:31 pmHow is that different from a bunch of KVL/KCL? I can setup a bunch of equations for all nets, but getting that to a readable expression where the poles/zeros are clear (i.e. bode’s normal form) is another question

author: jwet
author_rank: Frequent Contributor
timestamp: May 15, 2025, 04:13:53 pm
content: There are only two breakpoint.The integrator breaks at 1/2piRC- low pass, the HF breaks at 1/2piChf x R hf- hi passLiterally analyze them as two seperate circuits- they operate in completely different ranges.

author: vcvs
author_rank: Contributor
timestamp: May 15, 2025, 04:15:51 pm
content: Quote from: jwet on May 15, 2025, 04:13:53 pmI’ll give it a try

author: vcvs
author_rank: Contributor
timestamp: May 28, 2025, 07:55:46 am
content: I have had time to think about this again. And not surehowto decompose it into two separate circuits. How do I know which parts of the circuit is relevant for LF vs HF when the capacitors should still be included?

author: dietert1
author_rank: Super Contributor
timestamp: May 28, 2025, 02:28:23 pm
content: The next level of analysis would probably consider three frequency regions.1) A high frequency region where the opamp does nothing, maybe above 10 KHz. Buffer gain G will be a little less than 1 (maybe 0.97 or so). Input impedance is Ra1+ Ra2parallel to RHF. Considering the opamp input capacity Ra2should be much smaller than Ra1.2) The other easy to understand frequency range is below the CHF* RHFlimit, let's say 100 Hz. Here the buffer amplifier loses direct drive from the input and instead becomes part of the opamp. For the output to track the input with the same gain G as the RF buffer, one needs roughly G * Ra1/ Ra2= Rb1/ Rb2. The capacitor CFcan't be large. It isn't meant as integrator but to keep a unity gain stable opamp stable when adding the output buffer into the loop.3) In the overlap frequency range, let's say 100 Hz to 10 KHz, both circuits operate together. They can still provide a frequency independent output amplitude and phase due to the built-in opamp control loop that applies smaller and smaller corrections via RHF.The frequency and phase behavior of a scope frontend is simple. It cannot have any interesting features, as you don't want to see ringing. The circuit behaves as a 1st or 2nd order low pass, with the upper bandwidth limit defined inside the buffer amplifier transistors.I think the aim of this circuit is noise reduction. High frequency transistors tend to be small and noisy at low frequencies, while a low frequency amplifier can be made with larger transistors that behave better at low frequencies.Regards, Dieter