Adjusting gain in a transimpedance amplifier circuit

[pacman22]

I reverse engineered a VOLTCON board that converts current from photo diode to voltage and drew the following schematic:

As per the data sheet (https://www.farnell.com/datasheets/3797110.pdf), POT1 varies the gain of the opamp. How does this change the gain when it is placed in parallel with the output of the opamp?

 

[Easy peasy]

because the output is a current - not a voltage

 

[pacman22]

because the output is a current - not a voltage

Isn't the output suppose to be voltage? Isn't that what a transimpedance amplifier does?

 

[KlausST]

Hi,

I don´t think the schematic of post#1 matches the VOLTCON circuit.

Klaus

 

[pacman22]

Hi,

I don´t think the schematic of post#1 matches the VOLTCON circuit.

Klaus

Do you mind telling me where I am wrong? I have checked again, it is similar to the VOLTCON module

 

[danadakk]

The output of the schematic / OpAmp is a V. I concur with Klaus, schematic
not right. This basically a single ended output solution, and that pot in
rendered schematic does nothing to G due to schematic error.....

 

[KlausST]

Hi,

Your GAIN setting pot just loads the output. In worst case it causes a short circuit. It does nothing in the meaning of gain setting.

According datasheet there is Rf and Cf for gain setting.
Where do you think they are in your circuit? Name them accordingly in your circuit, so everyone knows.

The datasheet shows two SO-8 ICs ... I can only find only one in your circuit.

Where is the input diode connected in your circuit?

There are at least three capacitors in the datasheet, your circuit shows only 2.

There are 6 resistors in your circuit .. I can see only 2 in the datasheet.

There is a small signal diode in the datasheet, ... missing on your schematic

What about the black part (diode? capacitor?) of the datasheet near the pot?

maybe more...

***
It seems you either posted the wrong schematic or the wrong datasheet link....


Klaus

 

[D.A.(Tony)Stewart]

An "Integrated OTA" has current sources on the output, although Wiki calls it the Operational Transconductance Amplifier (OTA).
An OTA equivalent circuit may use a standard Op Amp as a voltage source output to match the input current by nulling the input error voltage. This is true whenever the output is in the linear range between the supply rails and performs the transconductance of the input current.

Vout = feedback impedance * (- feedback current).

Using the image from Wikipedia, I inverted the current source so that a single +ve supply CMOS Op Amp may be used.
This would be the cathode to -Vin and anode grounded in photoconductance mode.
This also works with LEDs as PDs.

 

[KlausST]

An "Integrated OTA"

Why do you talk about an OTA?

OTA inside an OPAMP or not: A TIA usually is explained as a usual OPAMP circuit.
And due to the feedback the output appears like a voltage output.

Klaus

 

[danadakk]

The second SMD part is a 78L05 V regulator on the board.

 

[KlausST]

The second SMD part is a V regulator on the board.

How do you know?

My guess is that it is a two stage circuit:
Current input --> TIA (range setup) --> Variable gain (pot 0...100%) --> output

Klaus

 

[danadakk]

Loaded board into image viewer and read the part number.

Guessing, and I have been guilty of it, is not good engineering practice.

Knight

 

[D.A.(Tony)Stewart]

Why do you talk about an OTA?

OTA inside an OPAMP or not: A TIA usually is explained as a usual OPAMP circuit.
And due to the feedback the output appears like a voltage output.

Klaus

TY for the error correction.

Transconductance is defined as a transform from V to I or VCCS performed by OTAs.
Transimpedance is defined as a transform from I to V or CCVS performed by -Vin/Zfb of Op Amps.

Yes there is a 78L05A on board.

 

[KlausST]

Guessing, and I have been guilty of it, is not good engineering practice.

I totally agree. I don't like guessing either, so I avoid it wherever possible.

That's why I didn't write down my guess before your post.
I still can't read a part number.

@all:
I clearly marked it as “guess” so everyone knows that this information is not proven.
Some posters miss to do so.

Klaus

 

[D.A.(Tony)Stewart]

How do you know?

My guess is that it is a two stage circuit:
Current input --> TIA (range setup) --> Variable gain (pot 0...100%) --> output

Klaus

Which info are you doubting?

 

[KlausST]

Which info are you doubting?

I don´t doubt it now,
but before I did not have a high resolution photo of the PCB.

Klaus

 

[FvM]

Remaining open point is connection of gain potentiometer. Post #1 schematic is either incorrect or makes the trimpot useless.

 

[Bob19]

If POT1 is parallel with the opamp output, it changes the output impedance. By decreasing POT1, you lower the impedance, which increases the current through the feedback resistor, thus increasing the gain. It's crucial to balance this, as too low impedance can worsen the circuit's stability.

Remember to check the values on the datasheet and choose the components accordingly. You can experiment at this stage, but do so carefully! Good luck!

Bob

 

[KlausST]

If POT1 is parallel with the opamp output, it changes the output impedance. By decreasing POT1, you lower the impedance, which increases the current through the feedback resistor, thus increasing the gain. It's crucial to balance this, as too low impedance can worsen the circuit's stability.

Remember to check the values on the datasheet and choose the components accordingly. You can experiment at this stage, but do so carefully! Good luck!

Bob

While not impossible ..
it makes as much sense as: fully pushing the gas pedal on your car ...and control the speed with the brake pedal.

It´s expected that something will get damaged soon.

Klaus