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I am trying to simulate the circuit below (from this paper) at the transistor level using a 0.18um, 1.8V process. The circuit serves as an amplifier for recording neural activities, and uses 2 capacitors (C1) to block DC offset and set the gain.

enter image description here

The OTA is implemented as below:

enter image description here

However, C1 also blocks the DC bias voltage that I set in the signal source, and keeps the input common-mode voltage close to zero. This makes the two input PMOS (M1, M2) work in triode region in my case (Vs is about 750mV), and significantly reduces the Gm of the OTA. The voltage gain is even less than 1 in my simulation.

I'm wondering if this is a problem of the OTA. Should the OTA be designed to operate at fairly arbitrary input common-mode voltages?

Also, is there a way to define the input common-mode voltage as I want (around 350mV) if I keep this configuration?

By the way, I just connected VSS to ground and VDD to 1.8V in my simulation, while in that paper they used \$ \pm 2.5V \$ supply. Don't know if this causes the problem.

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1 Answer 1

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The opamp structure works in the paper because they use +/-2.5V supply for the opamp. If VSS becomes -2.5V for the OTA, the drain of M1 and M2 will move lower compared to the case where VSS = 0V. This will bring M1 and M2 out of triode and opamp will start working.

Edit: If you want to use 1.8V/0V for the opamp, you can try the following

enter image description here

Note that I have added 700mV reference instead of the 350mV that you mentioned. Reason is that if Vout = 350mV, you may have trouble biasing the output cascode and M6 transistors in your opamp.

In the real world, the 700mV will be from a bandgap reference which will have noise and one needs to simulate and check how much noise it contributes at the output i.e., Vout.

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  • \$\begingroup\$ Thanks for your answer. Actually I also have some trouble understanding the top-level circuit (eg. why Vout will be 700mV if we set Vin, cm = 700mV). Does the golden rule of opamps still apply to OTAs? \$\endgroup\$
    – Jack Black
    Commented Apr 28, 2023 at 1:07
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    \$\begingroup\$ @GeorgeGuo, any high gain opamp when in -ve feedback has vp=vn. This rule is applicable here as well. Ma, Mb are to be treated as very high resistance because diode resistance is very high when there is zero current in it. At DC, when you remove the caps, the config is a unity gain buffer hence Vout = 700mV. As a word of caution, since we are dealing with very high resistances, small leakage currents can change the Vin, cm and Vout values and you need to have simulator settings like gmin to very low value such that it does not impact simulation outcome. \$\endgroup\$
    – sai
    Commented Apr 28, 2023 at 1:21

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