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Was going through this webpage and noticed that a trans-impedance amplifier using single supply. However, I am unable to understand how it would work?

How would the opamp see inverting input(photodiode - on non inverting) current and reference it(non inverting in ground)? Is there any advantage of using single supply in this case?

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Ahh.. found a few posts here which explains this.

No, it's not using a single supply as per the question linked in your comment, the Maxim circuit is generating a mid-rail - there is an important distinction: -

enter image description here

R1 and R2 bias the non-inverting input at Vcc/2. This sets the inverting input at Vcc/2 and the output of the amplifier also at Vcc/2 when zero photodiode current is flowing.

Basically you can make a single supply TIA with inputs referenced to the most negative rail providing the input current to the TIA is negative i.e. as from a photodiode with the cathode at the input. The photodiode current flows into the cathode and out of the anode and this means the op-amp output will rise above 0V (most negative rail) to re-balance the situation. The re-balance is the output.

Clearly, rail-to-rail op-amps are required for this to work effectively. Take a look at this picture below. It doesn't show power rails but think about the signals. Light generates a photodiode current that flows away from the inverting input. This has to make the output rise higher than the non-inverting input and here lies the basis for a true single-supply, ground referenced TIA. The only caveat is that there will be a small (a few mV) dead-band until sufficient current flows in the photodiode.

enter image description here

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  • \$\begingroup\$ Thanks for the explanation. What would happen if the non inverting input of the opamp is tied to ground (like a classic circuit)? \$\endgroup\$ – Anuj Purohit Apr 1 '15 at 9:01
  • \$\begingroup\$ See my additions \$\endgroup\$ – Andy aka Apr 1 '15 at 9:05
  • \$\begingroup\$ Thanks. I am thinking to use MCP TC913A (Chopper auto-zero) opamp as TIA and realized later that it does not accept +12/-12v power supply therefore cannot directly replace my AD712. I am thinking to tie -12 to ground to test my circuit but I am not sure if it will give me some advantage. (I am not keen on using 7806,7906 to make my +/-12 v for +/-6V) \$\endgroup\$ – Anuj Purohit Apr 1 '15 at 9:17
  • \$\begingroup\$ Nice (+1), The mid rail bias (R1/R2) also reverse biases the photodiode which will reduce it's capacitance and speed up the circuit somewhat. \$\endgroup\$ – George Herold Apr 1 '15 at 12:35
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    \$\begingroup\$ I believe that to be the case! Advantages are not profound - you could feed Vcc/2 and the output to a differential ADC and not bother with a buffer. \$\endgroup\$ – Andy aka Apr 1 '15 at 17:04
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The resistor divider on the non inverting input creates a margin between the signal and ground. Note the Vcc connection; the output of the photodiode is on the inverting input in that figure. In the third figure, a first stage opamp is merely creating a reference other than ground for the circuit.

The only real advantage is that this method can allow you to work with a single supply. Some opamps - such as the one in that figure - are better suited for this than others. When selecting an opamp you have to make sure your choice of voltage supplies won't cause the output to clip - unless that's the point of the circuit of course.

Always bear in mind that when designing a sensor circuit, such as this would be used for, you are ultimately in control of how the output behavior is manipulated. The transfer equations in those three figures would be different, but it is of little consequence to the designer, as long as the opamp selection supports the voltages being used.

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  • \$\begingroup\$ Thanks for the explanation. What would happen if the 1st opamp is removed and the non inverting input of the second opamp is tied to ground (like a classic circuit)? \$\endgroup\$ – Anuj Purohit Apr 1 '15 at 8:57

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