How would I go about finding the gain of this circuit? Here's a link to the datasheet.

Using our test apparatus to create a PWM on the diode, I get 1.16V out of pin 1. I can provide more information if needed.

Thanks, Luke

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  • \$\begingroup\$ This type of amplifier is commonly called "transimpedance". So, a quick search for "transimpedance amplifier gain" should get you where you need. \$\endgroup\$ – scld May 15 '15 at 14:36

The BP104 has a light sensitive area of 7.5 sq mm. If you look in the data sheet, an irradiance of 1mW per sq cm produces a current of about 33uA.

The light power hitting the 7.5 sq mm can be calculated as simply: -

Irradiance = 1mW per 100 sq mm therefore power hitting sensitive area (7.5 sq mm) is 75uW.

So now we have 75uW producing 33uA into the TIA (trans impedance amplifier). The gain of the TIA at low to medium frequencies (ignoring the effect of C4) is 5.1Mohms.

Therefore the gain of the circuit (volts out to light power in)

= 5.1M * 33uA / 75uW = 2.244 volts per micro watt and this assumes the incident light is at 950nm or thereabouts.

I'll also add that because the op-amp is a FET input type, the resistor R13 is not needed. Why is it there because it cannot possibly hope to counteract any leakage bias currents in the device.

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  • \$\begingroup\$ To quickly answer your last question - I'm picking up the project someone else started. The reason for it is unknown to me. I thought it was to prevent a floating node? \$\endgroup\$ – Luke May 15 '15 at 15:24
  • \$\begingroup\$ Excellent answer by the way. This is essentially what I was looking for. I appreciate you taking the time to write out the process. I didn't consider to go through this process to find the output voltage of the photodiode and use that as the input voltage to find the Vo/Vi gain. \$\endgroup\$ – Luke May 15 '15 at 15:33
  • \$\begingroup\$ @Luke - I didn't find the output voltage of the photodiode, I found the output current for a given light input. Neither is the gain Vo/Vin - it is Vout / watts in. A TIA holds (forces) the inverting input (via feedback) to be at the same potential as the non-inverting input. This means (theoretically) Vin = zero aka the input dynamic input impedance is zero. \$\endgroup\$ – Andy aka May 15 '15 at 15:50

The opamp is in current to voltage configuration. All the current provided at the negative input node flows through R15 and becomes a voltage, the output voltage is then: $$ V_{out} = R_{15}I^- $$ Where \$I^-\$ is the current flowing into the negative input node. Please note that no current flows into the amplifier negative input, but some current flows into the associated node (and through R15 then).

Your gain is then:

$$ A = 5.1M\Omega $$

I am assuming that your signal is at frequencies low enough to neglect C4, that guy is there to improve stability.

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  • \$\begingroup\$ So the gain is 5.1M? The ohm sign is confusing me a bit. \$\endgroup\$ – Luke May 15 '15 at 14:55
  • \$\begingroup\$ No, the gain is 5.1M\$\Omega\$. Your photodiode will have some current/light_intensity gain, if you multiply the two gains you get a volt/light_intensity gain. Look at fig.4 of the datasheet. \$\endgroup\$ – Vladimir Cravero May 15 '15 at 15:02
  • \$\begingroup\$ Thanks Vladimir. I appreciate your post and answering my second question. Andy expanded on this comment in his post. \$\endgroup\$ – Luke May 15 '15 at 15:26

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