What is it?
Typically you think of an amplifier as taking a voltage as an input and producing a larger amplitude voltage as an output. A transimpedance amplifier is similar except it takes current as an input and produces a voltage as its output. You can think of a transimpedance amplifier as a current to voltage converter.
This is particularly useful when interfacing to sensors, such as a photodiode, which produces a current output.
In the same way that feedback resistors can be used to implement inverting and non-inverting amplifiers using an op-amp, the addition of a single resistor allows you to make a transimpedance amplifier from an op-amp. The following figure depicts the circuit configuration.
An op-amp is a differential amplifier with very high gain (see Op-amp Basics for a primer on op-amps). This means that when the negative input is larger than the positive input, the output of the op-amp becomes a large negative voltage. With that in mind, the operation of the transimpedance amplifier is easy to sort out.
Assume that a small positive current is applied to the Iin input. This causes the op-amps inverting input (-) to be greater than the non-inverting input which is connected to ground. In response to this, the op-amps output begins to move towards a large negative output voltage. As it does, a current flows from the output of the op-amp via Rf to the inverting input of the op-amp. Note that this is a negative current flow due to the negative voltage output of the op-amp.
As the output of the op-amp continues to increase, a point is reached where the current through Rf matches Iin. At this point the summation of the currents at the op-amps inverting input yields 0 and op-amps output stops changing because a balance has been reached. Any change to Iin will unbalance the inputs to the op-amp and result in a change at the op-amps output. This balancing act occurs very quickly which means that the amplifies output is a voltage representation of the input current.
The gain of the amplifier can be found by realizing that Ohm’s Law describes the relationship between the current through Rf and voltage across it. The gain is shown in the following equation.
Transimpedance amplifiers are typically used to amplify very small currents. This means that a large gain, which leads to a large Rf, is needed. Aside from the inherent bandwidth limitations imposed in large gain configurations, another factor comes into play. A current source (such as photodiode) looks like an open circuit from which a current flows. The diodes capacitance, along with the input capacitance of the amplifier, act in conjunction with Rf to form a filter. For large values of Rf, this can substantially limit the frequency response. To get around this issue, it is common practice to add a capacitor in parallel with Rf as shown in the following figure.
The addition of Cf improves the frequency response by supplying a lower impedance path at high frequencies that more quickly moves charge to and from the input capacitance. The value of Cf is normally equal to or less than the capacitance at Iin. In situations where the amplifier is very close to the sensor for optimum performance, this means capacitors in the low to tens of picoFarads range are typically used.