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I was reading about the I to V converter using an op-amp.

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Doesn't the addition of the \$R_{\text{L}}\$ resistor affect the \$I_{\text{R}}\$ current? Because the same current that flows through the feedback resistor would also flow through the \$R_{\text{L}}\$ resistor, right? If that's the case, what is the usage of the \$R_{\text{L}}\$ resistor and how would it not affect the working of the I to V converter?

Null
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2 Answers2

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An ideal op amp has infinite impedance at its inputs but zero impedance at its output. That means that no current flows into its inputs, so by KCL $$I_{\text{R}} = I_{\text{in}}$$ regardless of anything else in the circuit.

However, because the op amp's output is low impedance it can either source or sink current as needed (which it will do in order to keep both its inputs at ground). Therefore, by KCL the current \$I_\text{L}\$ through \$R_\text{L}\$ is the sum of \$I_{\text{R}}\$ and the current sourced by the op amp's output -- which in general means that \$I_\text{L}\$ is not equal to \$I_{\text{R}}\$.

By Ohm's Law $$I_\text{L} = \frac{V_{\text{out}}}{R_{\text{L}}}$$

and, since the op amp's inverting input is at ground, we have

$$V_{\text{out}} = -I_{\text{in}}R$$

Therefore

$$I_\text{L} = \frac{-I_{\text{in}}R}{R_{\text{L}}} = \frac{-I_{\text{R}}R}{R_{\text{L}}}$$

The purpose of \$R_{\text{L}}\$ is to model the load resistance, which in general is not infinite. The fact that \$R_{\text{L}}\$ does not affect the operation of the transimpedance amplifier (except in an edge case like \$R_{\text{L}} = 0\$) is a good thing.

Null
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  • So, the op-amp output will provide some current to the RL right? In that case, how much current will it provide and what determines it? –  Jun 25 '21 at 12:46
  • Suppose, the voltage at the Load R is defined by the current flowing through it right? So, if Ir and output of op-amp also provides some current, wouldn't the voltage across the load resistor change? Wouldn't it defeat the purpose of the entire circuit if the op-amp output also provides some current? –  Jun 25 '21 at 12:47
  • @Newbie The op amp output will provide whatever current to RL is necessary to set its output such that its inputs are at the same voltage. Vout which sets its inputs to the same voltage is -Iin*RF, so the op amp will source or sink current from its output to make that true (which also sets the voltage across RL). – Null Jun 25 '21 at 12:51
  • @Newbie I added the calculation of the RL current in terms of IR. – Null Jun 25 '21 at 13:01
  • Can this I to V converter be a non-inverting type also? –  Jun 25 '21 at 13:11
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    @Newbie No, you can't just switch the op amp's input terminals because then you'd have positive instead of negative feedback. If you need a non-inverting transfer function you might be able to switch the polarity of the input current, or you can add an inverting amplifier at the output of the transimpedance amplifier (the double inversion results in an overall non-inverting transfer function). – Null Jun 25 '21 at 13:14
  • For positive transfer function, just use a voltage follower from the input (use the resistor R as the current-to-voltage converter). In this case however, the input impedance will be R instead of R/Aol – tobalt Jun 25 '21 at 13:18
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\$R_L\$ plays no active role in that circuit. Its value does not matter to a large extent.

The voltage \$V_{out}\$ is varied to satisfy the feedback equilibration, which means IN- should reach 0V. The only connection from IN- to OUT is through \$R\$. Voltage across both Resistors will be the same, but the current through \$R_L\$ is basically parasitic and the opamp will source it no matter its value (within its capabilities)

tobalt
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