# Transimpedance amplifier

(Redirected from Current-to-voltage converter)
A transimpedance amplifier converts the input current to a voltage at the output.

In electronics, a transimpedance amplifier (TIA) is a current-to-voltage converter, most often implemented using an operational amplifier. The TIA can be used to amplify[1] the current output of Geiger–Müller tubes, photomultiplier tubes, accelerometers, photo detectors and other types of sensors to a usable voltage. Current-to-voltage converters are used with sensors that have a current response that is more linear than the voltage response. This is the case with photodiodes, where it is common for the current response to have better than 1% linearity over a wide range of light input.

The ratio between input current and output voltage is called transimpedance, i.e.:

${\displaystyle Z_{\text{trans}}={\frac {V_{\text{out}}}{I_{\text{in}}}}}$

Transimpedance amplifiers may be built using operational amplifiers or using bipolar or MOSFET transistors.

## Transimpedance amplifier with op-amp

Simplified transimpedance amplifier built using an operational amplifier

### DC operation

In the circuit shown above the photodiode 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

${\displaystyle -I_{\text{in}}={\frac {V_{\text{out}}}{R_{\text{f}}}},}$

so

${\displaystyle Z_{\text{trans}}={\frac {V_{\text{out}}}{I_{\text{in}}}}=-R_{\text{f}}.}$

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 FET input op-amps that have very low input offset voltages.[2]

Transimpedance amplifier with a reverse biased photodiode

An inverting TIA can also be used with the photodiode operating in the photoconductive mode, as shown in the figure. A positive voltage at the cathode of the photodiode applies a reverse bias. This reverse bias increases the width of the depletion region and lowers the junction capacitance, improving the high-frequency performance. The photoconductive configuration of a transimpedance photodiode amplifier is used where fast switching speed is required but high gain is not. The feedback capacitor Cf is usually required to improve stability.

The transimpedance amplifier presents a low impedance to the photodiode and isolates it from the output voltage of the operational amplifier. In its simplest form a transimpedance amplifier has just a large-valued feedback resistor, Rf. The gain of the amplifier is set by this resistor and has a value of −Rf (because the amplifier is in an inverting configuration). There are several different configurations of transimpedance amplifiers, each suited to a particular application. The one factor they all have in common is the requirement to convert the low-level current to a voltage.

The gain, bandwidth, as well as current and voltage offsets, change with different types of sensors, requiring different configurations of transimpedance amplifiers.[3]

### Bandwidth and stability

Incremental model showing sensor capacitance

The frequency response of a transimpedance amplifier is inversely proportional to the gain set by the feedback resistor. The sensors that 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

${\displaystyle \beta ={\frac {Z_{C_{\text{i}}}}{R_{\text{f}}+Z_{C_{\text{i}}}}}={\frac {1}{1+R_{\text{f}}C_{\text{i}}s}},}$

where ${\displaystyle Z_{C_{\text{i}}}}$ is the impedance of the capacitance Ci. This filter attenuates the feedback signal, which places a greater demand on the amplifier gain.

When the effect of this low-pass filter response is considered, the circuit's response equation becomes

${\displaystyle V_{\text{out}}=I_{\text{p}}{\frac {-R_{\text{f}}}{1+{\frac {1}{A_{\text{OL}}\beta }}}},}$

where ${\displaystyle A_{\text{OL}}}$ 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,

${\displaystyle V_{\text{out}}=-I_{\text{p}}R_{\text{f}}}$

as long as the loop gain,

${\displaystyle A_{\text{OL}}\beta ,}$

is much greater than unity.

Bode plot of uncompensated transimpedance amplifier.[5]

In the Bode plot of a transimpedance amplifier with no compensation, the flat curve with the peak, labeled I-to-V gain, is the frequency response of the transimpedance amplifier. The peaking of the gain curve is typical of uncompensated or poorly compensated transimpedance amplifiers. The curve labeled AOL is the open-loop response of the amplifier. The feedback factor, plotted as a reciprocal, is labeled 1/β. In Fig. 5 the 1/β curve and AOL form an equilateral triangle with the frequency axis. The two sides have equal but opposite slopes, since one is the result of a first-order pole, and the other of a first-order zero. Each slope has a magnitude of 20 dB/decade, corresponding to a phase shift of 90°. When the amplifier's 180° of phase inversion is added to this, the result is a full 360° at the fi intercept, indicated by the dashed vertical line. At that intercept, 1/β = AOL for a loop gain of AOLβ = 1. Oscillation will occur at the frequency fi because of the 360° phase shift, or positive feedback, and the unity gain.[6] To mitigate these effects, designers of transimpedance amplifiers add a small-value compensating capacitor (Cf in the figure above) in parallel with the feedback resistor. When this feedback capacitor is considered, the compensated feedback factor becomes

${\displaystyle \beta ={\frac {1+R_{\text{f}}C_{\text{f}}s}{1+R_{\text{f}}(C_{\text{i}}+C_{\text{f}})s}}.}$

The feedback capacitor produces a zero, or deflection in the response curve, at the frequency

${\displaystyle f_{C_{\text{f}}}={\frac {1}{2\pi R_{\text{f}}C_{\text{f}}}}.}$

This counteracts the pole produced by Ci at the frequency

${\displaystyle f_{\text{zf}}={\frac {1}{2\pi R_{\text{f}}(C_{\text{i}}+C_{\text{f}})}}.}$
Bode plot of compensated transimpedance amplifier[7]

In the Bode plot of a transimpedance amplifier that has a compensation capacitor in the feedback path, 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]

## Transimpedance amplifier using MOSFETs

A transimpedance amplifier may be built using a CMOS inverter and placing a feedback resistor between its input and output as shown in the image to the right. The input capacitance ${\displaystyle C_{\text{in}}}$ consists of the gate source capacitances of the MOSFETs and of the capacitance of the sensor.

Circuit diagram of a transimpedance amplifier built using MOSFET transistors.
AC equivalent circuit of a transimpedance amplifier built using MOSFET transistors.

To simplify the formulas, the transconcuctance ${\displaystyle g_{m}}$ is defined as the sum of the transconductances of the NMOS and PMOS transistors:

${\displaystyle g_{m}=g_{m,PMOS}+g_{m,NMOS}}$

And the output resistance ${\displaystyle r_{o}}$ is defined as the parallal combinaation of the output resistances of the NMOS and PMOS transistors:

${\displaystyle r_{o}=r_{o,PMOS}||r_{o,NMOS}={\frac {r_{o,PMOS}r_{o,NMOS}}{r_{o,PMOS}+r_{o,NMOS}}}}$

### Transimpedance

The small signal equivalence circuit of the transimpedance amplifier is shown in the image to the right. The transimpedance is given by:

${\displaystyle Z_{\text{trans}}={\frac {v_{\text{out}}}{i_{\text{in}}}}={\frac {\frac {r_{o}(1-g_{m}R_{F})}{1+g_{m}r_{o}}}{1+sC_{\text{in}}{\frac {R_{F}+r_{o}}{1+g_{m}r_{o}}}}}}$
${\displaystyle Z_{\text{trans}}={\frac {Z_{\text{trans,DC}}}{1+{\frac {s}{\omega _{p}}}}}}$
${\displaystyle Z_{\text{trans,DC}}={\frac {r_{o}(1-g_{m}R_{F})}{1+g_{m}r_{o}}}{\xrightarrow {g_{m}\rightarrow \infty }}-R_{F}}$
${\displaystyle \omega _{p}={\frac {1+g_{m}r_{o}}{C_{\text{in}}\left(R_{F}+r_{o}\right)}}}$

As in the case with the operational amplifier, the transimpedance converges to minus the value of the feedback resistor for low frequencies and high amplifier gain.

### Input impedance

The input impedance, i.e. the input voltage divided by the input current under the assumption that there is no output current, is given by:

${\displaystyle Z_{\text{in}}={\frac {v_{\text{in}}}{i_{\text{in}}}}={\frac {1}{sC_{\text{in}}+{\frac {1+g_{m}r_{o}}{R_{F}+r_{o}}}}}={\frac {\frac {R_{F}+r_{o}}{1+g_{m}r_{o}}}{1+sC_{\text{in}}{\frac {R_{F}+r_{o}}{1+g_{m}r_{o}}}}}}$

Notice that the input impedance is low because ${\displaystyle g_{m}r_{o}}$ is typically large.

## Transimpedance amplifier using bipolar transistor

Circuit diagram of a transimpedance amplifier built using an NPN bipolar transistor.
AC equivalent circuit of a transimpedance amplifier built using an NPN bipolar transistor.

To build a TIA using a bipolar transistor, a feedback resistor may be placed between input and output of a common emitter amplifier.

### Transimpedance

${\displaystyle Z_{\text{trans}}={\frac {r_{1}R_{2}(1-g_{m}R_{F})}{R_{F}+R_{2}+(1+g_{m}R_{2})r_{1}}}{\frac {1}{1+sC_{\text{in}}{\frac {R_{F}+R_{2}}{1+g_{m}R_{2}+{\frac {R_{2}+R_{F}}{r_{1}}}}}}}}$
${\displaystyle Z_{\text{trans}}={\frac {Z_{\text{trans,DC}}}{1+{\frac {s}{\omega _{p}}}}}}$

As in the circuit with MOSFETs, the transimpedance of the amplifier converges to ${\displaystyle -R_{F}}$ at low frequencies and for large transconductance of the transistor.

${\displaystyle Z_{\text{trans,DC}}={\frac {r_{1}R_{2}(1-g_{m}R_{F})}{R_{F}+R_{2}+(1+g_{m}R_{2})r_{1}}}{\xrightarrow {g_{m}\rightarrow \infty }}-R_{F}}$
${\displaystyle \omega _{p}={\frac {r_{1}(1+g_{m}R_{2})+R_{2}+R_{F}}{C_{\text{in}}\left(R_{F}+R_{2}\right)}}}$

### Input impedance

${\displaystyle Z_{\text{in}}={\frac {v_{\text{in}}}{i_{\text{in}}}}={\frac {1}{sC_{\text{in}}+{\frac {1}{r_{1}}}+{\frac {1+g_{m}R_{2}}{R_{F}+R2}}}}={\frac {\frac {r_{1}(R_{F}+R_{2})}{R_{F}+R_{2}+r_{1}(1+g_{m}R_{2})}}{1+s{\frac {R_{F}+R_{2}}{R_{F}+R_{2}+r_{1}(1+g_{m}R_{2})}}C_{\text{in}}}}}$

## 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

${\displaystyle {\sqrt {\overline {v_{n,out}^{2}}}}={\sqrt {4k_{\text{B}}TR_{f}\Delta f}}.}$

Though the output noise voltage increases proportionally to ${\displaystyle {\sqrt {R_{f}}}}$, the transimpedance increases linearly with ${\displaystyle R_{f}}$, resulting in an input-referred noise current

${\displaystyle {\sqrt {\overline {i_{n,in}^{2}}}}={\sqrt {\frac {4k_{\text{B}}T\Delta f}{R_{f}}}}.}$

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 gain-bandwidth product. The feedback resistance and therefore the sensitivity are thus limited by the required operating frequency of the transimpedance amplifier.

### Derivation for TIA with OpAmp

Schematic for output noise calculation of transimpedance amplifier with opamp and feedback resistor

The noise currenct of the feedback resistor equals ${\displaystyle {\overline {i_{n}^{2}}}={\frac {4k_{\text{B}}T\Delta f}{R_{f}}}}$. Because of virtual ground at the negative input of the amplifier ${\displaystyle {\overline {v_{n,out}^{2}}}={R_{f}}^{2}{\overline {i_{n}^{2}}}}$ holds.

We therefore get for the root mean square (RMS) noise output voltage ${\displaystyle {\sqrt {\overline {v_{n,out}^{2}}}}={\sqrt {4k_{\text{B}}TR_{f}\Delta f}}}$. A high feedback resistor is desirable because the transimpedance of the amplifier grows linearly with the resistance but the output noise only grows linearly with the squareroot of the feedback resistance.