Building a voltage amplifier for the implementation of a transimpedance amplifier

[VisheshG24]

I am trying to build a transimpedance amplifier with the following specifications:

1. The input current range is 100nA-10uA.
2. The bandwidth requirement is 100Hz.
3. The output voltage should be from 0V-4V.
4. I have the vdd=+5V and vss=-5V.

The following figure implements the transimpedance amplifier using an opamp:

According to opamp theory, the transimpedance gain of the above circuit will be -R.

I want to replace the opamp in the above circuit with a transistor voltage amplifier implementation as to take the whole circuit to an ASIC design.

I tried making the opamp using mos differential pair with single output as follows:

The problem that I am facing with this voltage amplifier implementation is that the M1 transistor is going in the linear region when Iin goes from 5uA to 10uA and the M4 transistor is always in linear region, which makes it a bad current source. Which ultimately gives a very variable transimpedance gain.

If someone can help with either improving my voltage amplifier implementation or can suggest other simple voltage amplifier circuits that people use, I will really be grateful. Or even suggest any other method to implement the TIA that can match my specs, I will really appreciate this.

Thanks

 

[arthorios]

Several things to consider here.
First of all, what is the voltage at the OTA positive input? This will determine the voltage at the output when I_n=0. Ideally, it should be (vdd-vss)/2 to allow maximum swing. This is 0V for dual supply.

Next, what is the current through M4, and what is the value of R? You need to consider that R is effectively loading your OTA output, and so the OTA (which is a class-A stage) needs to be able to provide enough current to R for the maximum Iin. By the same token, loading your relatively high output impedance OTA with a low R will also degrade your DC loop gain. That's usually solved by buffering the OTA from R using a source-follower (for example).

 

[VisheshG24]

Arthorios, thanks a lot for the reply. The positive input has a DC reference voltage. It will be the reference voltage, which can be assumed ground for now.
M4 is in a linear regime, so its current depends on the Vds that exist across it (I am sorry if I didn't understand what you want). The R is 500K ohms.

Isn't the Rin and Rout being reduced by the loop gain because of the feedback, and thus will have less loading impact? Or do these feedback concepts only apply to small signals, and this is a large signal?

I would be really grateful if you could share any details explaining the implementation of the voltage amplifier. I am new in analog designing and wanted to really learn.

 

[KlausST]

Hi,

Isn't the Rin and Rout being reduced by the loop gain because of the feedback, and thus will have less loading impact? Or do these feedback concepts only apply to small signals, and this is a large signal?

On a transimpedance amplifier both input and output are considered very low ohmic.

The TIA forms current (without changing the input voltage) into a voltage (independent of load current)
The benefit is:
* the input source (to the TIA) is a current source, it is considered high impedance.
* due to the source´s high impedance it is prone to cause errors on capacitive currents
* the TIA keeps input voltage constant, thus there is no capacitive current. Low error
* the output of the TIA is independent of current, thus it can drive rather low ohmic loads .. while the current causes no error

Klaus

 

[BradtheRad]

I want to replace the opamp in the above circuit with a transistor voltage amplifier
Or even suggest any other method

This isn't necessarily a complete circuit although it shows what is possible. NPN transistor operated via common base mode. All adjustments are delicate. Similar to old-fashioned VOM's it should be zeroed before taking readings.
Arbitrarily the sense resistor generates 200mV as 10uA passes through it.

 

[arthorios]

Since you are new to analog design, I suggest you check books and do simulations to understand your target specifications better.

Arthorios, thanks a lot for the reply. The positive input has a DC reference voltage. It will be the reference voltage, which can be assumed ground for now.
M4 is in a linear regime, so its current depends on the Vds that exist across it (I am sorry if I didn't understand what you want). The R is 500K ohms.

To simplify your sims, first check what are the DC operating points when your OTA is in unity gain. Is M4 still in triode? If so, the sizing of your other transistors (or even M4) might be the problem.

Note that you have R=500kOhm. With a 10uA input current, the TIA output voltage is already at -5V. That's the reason why you see M1 going to triode. You should select R based on the maximum swing (positive and negative) your OTA can handle.

As I said before, your OTA is DRIVING the feedback resistor (R). And so it must be able to provide the current that runs through it.

Isn't the Rin and Rout being reduced by the loop gain because of the feedback, and thus will have less loading impact? Or do these feedback concepts only apply to small signals, and this is a large signal?

Rin is defined by R and the loop gain, while the OTA rout and the loop gain define Rout. Keep that in mind.