Transimpedance amplifier
In
DC operation
In the circuit shown in figure 1 the photodiode (shown as a current source) is connected between ground and the inverting input of the op-amp. The other input of the op-amp is also connected to ground. This provides a low-impedance load for the photodiode, which keeps the photodiode voltage low. The photodiode is operating in photovoltaic mode with no external bias. The high gain of the op-amp keeps the photodiode current equal to the feedback current through Rf. The input offset voltage due to the photodiode is very low in this self-biased photovoltaic mode. This permits a large gain without any large output offset voltage. This configuration is used with photodiodes that are illuminated with low light levels and require a lot of gain.
The DC and low-frequency gain of a transimpedance amplifier is determined by the equation
so
If the gain is large, any input offset voltage at the non-inverting input of the op-amp will result in an output DC offset. An input bias current on the inverting terminal of the op-amp will similarly result in an output offset. To minimize these effects, transimpedance amplifiers are usually designed with field-effect transistor (FET) input op-amps that have very low input offset voltages.[3]
An inverting TIA can also be used with the photodiode operating in the
Bandwidth and stability
The frequency response of a transimpedance amplifier is inversely proportional to the gain set by the feedback resistor. The sensors which transimpedance amplifiers are used with usually have more capacitance than an op-amp can handle. The sensor can be modeled as a current source and a capacitor Ci.[4] This capacitance across the input terminals of the op-amp, which includes the internal capacitance of the op-amp, introduces a low-pass filter in the feedback path. The low-pass response of this filter can be characterized as the feedback factor:
When the effect of this low-pass filter response is considered, the circuit's response equation becomes:
where is the open-loop gain of the op-amp.
At low frequencies the feedback factor β has little effect on the amplifier response. The amplifier response will be close to the ideal:
as long as the loop gain : is much greater than unity.
In the
The feedback capacitor produces a zero, or deflection in the response curve, at the frequency
This counteracts the pole produced by Ci at the frequency
The Bode plot of a transimpedance amplifier that has a compensation capacitor in the feedback path is shown in Fig. 5, where the compensated feedback factor plotted as a reciprocal, 1/β, starts to roll off before fi, reducing the slope at the intercept. The loop gain is still unity, but the total phase shift is not a full 360°. One of the requirements for oscillation is eliminated with the addition of the compensation capacitor, and so the circuit has stability. This also reduces the gain peaking, producing a flatter overall response. There are several methods used to calculate the compensation capacitor's value. A compensation capacitor that has a too large value will reduce the bandwidth of the amplifier. If the capacitor is too small, then oscillation may occur.[8] One difficulty with this method of phase compensation is the resulting small value of the capacitor, and the iterative method often required to optimize the value. There is no explicit formula for calculating the capacitor value that works for all cases. A compensation method that uses a larger-value capacitor that is not as susceptible to parasitic capacitance effects can also be used.[9]
Noise considerations
In most practical cases, the dominant source of noise in a transimpedance amplifier is the feedback resistor. The output-referred voltage noise is directly the voltage noise over the feedback resistance. This Johnson–Nyquist noise has an RMS amplitude
Though the output noise voltage increases proportionally to , the transimpedance increases linearly with , resulting in an input-referred noise current
For a good noise performance, a high feedback resistance should thus be used. However, a larger feedback resistance increases the output voltage swing, and consequently a higher gain from the operational amplifier is needed, demanding an operational amplifier with a high
Derivation for TIA with op-amp
The noise current of the feedback resistor equals . Because of virtual ground at the negative input of the amplifier holds.
We therefore get for the root mean square (RMS) noise output voltage . A high feedback resistor is desirable because the transimpedance of the amplifier grows linearly with the resistance but the output noise only grows with the square root of the feedback resistance.
Discrete TIA design
It is also possible to construct a transimpedance amplifier with discrete components using a
See also
- Operational transconductance amplifier – converts differential voltage into current
- Optical communication
- PIN diode
Sources
- Graeme, J.G. (1996). Photodiode Amplifiers: OP AMP Solutions. Gain technology. McGraw-Hill Education. ISBN 978-0-07-024247-0. Retrieved 12 November 2020.
References
- ^ Electronic Principles Paul E. Gray, Campbell Searle, pg. 641
- ^ The Art of Electronics, Horowitz and Hill
- ISBN 978-1-249-07817-3. Retrieved 12 November 2020.
- ^ Graeme 1996, p. 39.
- ^ Graeme 1996, p. 40.
- ^ Graeme 1996, p. 41.
- ^ Graeme 1996, p. 43.
- ^ Pease, Bob. "Transimpedance amplifiers". StackPath. Retrieved 12 November 2020.
- ^ Graeme 1996, p. 49.
- PMID 23020394.