# Understand and apply the transimpedance amplifier (Part 1 of 2)

Most engineers know that to design a transimpedance amplifier circuit, they just need a large-enough resistor to convert the input current to a reasonable output voltage range. To stabilize this circuit, a large enough capacitor must be placed in parallel with the feedback resistor. This article will show how to calculate the value for the feedback capacitor to ensure that the design has the largest possible bandwidth, and will still be stable.

Feedback factor calculation

Calculating the feedback factor for an op amp that is set up for current-to-voltage conversion may be a bit of a mystery to many engineers. By deriving the transfer function for a transimpedance amplifier and using a voltage amplifier op amp, the conversion will be easy to understand. Here we use the LMV793 op amp as an example for a transimpedance amplifier design. A basic transimpedance amplifier configuration is shown in Figure 1.

Figure 1: Basic transimpedance amplifier configuration

The figure shows the complete transimpedance amplifier; only the power supply decoupling capacitors are not shown. In most cases, the selection of the photodiode will allow the designer to use the same supply for VBIAS and +V. Using split supplies keeps the inverting input of the op amp at virtual ground. To derive the feedback factor, it is necessary to examine the equivalent circuit of the photodiode, Figure 2.

Figure 2: Photodiode equivalent circuit

The diode is an ideal diode in the equivalent circuit. Since the photodiode must be back-biased for proper operation, the ideal diode is not included in the feedback factor calculations. CJ is the capacitance that occurs from the depletion region of the diode and it is included in the photodiode specifications. IPH is the current that occurs from the photodiode action. The impedance of the current source is the series resistance of the photo diode, shown as RSH . The series resistance is at least 10MΩ and typically much higher, usually over 100MΩ.

Since RSH is such a high value, the current source will be considered ideal when deriving the feedback factor. The contact resistance and the resistance of the undepleted silicon are represented by RS . Normally RS is under 100Ω and can be ignored. Even though RS is being ignored in this analysis, this resistance can exceed 1kΩ in certain diodes, making it large enough to affect the transimpedance circuit feedback factor.

The op amp also has an input resistance. Figure 3 shows the circuit used in the feedback factor calculations. CCM is the input common mode capacitance of the op amp.

Figure 3: Equivalent AC circuit

For the circuit analysis, +V is an AC ground. From this circuit it is easy to derive the transfer function of the transimpedance amplifier. The op amp is assumed to be ideal; therefore, the inverting input is at virtual ground. CJ and CCM have no effect on the transfer function of this circuit since both nodes are at ground. The transfer function is (Equation 1):

Or (Equation 2):

This equation is quite familiar to designers of transimpedance amplifiers. For low frequencies, VOUT =-RF , and for very high frequencies VOUT =-1/sCF. This circuit inserts a pole at fP = ½πRF CF and this pole stabilizes the circuit. (The effect of this pole on the circuit is explained later.)

To derive the feedback factor, the circuit in Figure 3 is modified as shown in Figure 4.

Figure 4: Circuit for feedback-factor derivation

As noted at the beginning of this article, the op amp is a voltage amplifier, not a current amplifier. The current source in Figure 4 has infinite impedance; therefore, it has no effect on the feedback factor! The two input capacitors have been combined by setting CIN =CJ +CCM . Therefore, with no effect from the current source, the circuit in Figure 4 is just a differentiator circuit with the input capacitance to the circuit (CIN ) grounded. Even though a transimpedance amplifier is a current-to-voltage converter, the feedback factor is still based on a voltage-amplifier configuration, the differentiator circuit.

The feedback factor is simply what is fed back to the input from the output of the op amp. This is calculated by assuming that the node at the input to the op amp is not connected to the input of the op amp, then calculating the ratio of the input voltage to the output voltage, VIN /VOUT . Figure 5 shows the circuit used to calculate the feedback factor.

Figure 5: Feedback-factor derivation

The parallel impedance of CF and RF is (Equation 3):

This is also used in the transfer function. Now the calculation becomes a voltage divider network giving the equation (Equation 4):

With some algebraic manipulations the equation for the feedback factor, F, becomes (Equation 5):

The feedback factor for a transimpedance amplifier is identical to the feedback factor of a differentiator circuit. The only difference is that CIN of a differentiator circuit is added to the inverting input of the op amp, but for a transimpedance amplifier CIN is just the sum of the capacitance of the photo diode and the input capacitance of the op amp. Note, that for low frequencies, F = 1.

Optimum value for CF

The noise gain for an op amp circuit is 1/F. In a differentiator circuit, CIN will insert a zero in 1/F, thus making a differentiator circuit inherently unstable. Since a transimpedance amplifier is a differentiator circuit, it is inherently unstable. CF must be added to make the circuit stable. CF limits the bandwidth of the circuit. To insure stability a designer can add a feedback capacitor that is rather large, thus overcompensating the circuit. However, finding the optimum value for CF will maximize the bandwidth of the circuit and avoid using an op amp with a wider bandwidth than necessary. Figure 6 shows the relationship of 1/F and the open loop gain for an optimum design.

Figure 6: Open loop gain, A, and noise gain, 1/F

Figure 6 shows the 1/F curve with the optimum value for C. The pole of 1/F is located on the A curve. Therefore the optimum value for CF is where A=1/F, or A*F=1. Without CF there would be only a zero for the 1/F curve, giving close to a 180° phase shift at the intersection of A and 1/F. Adding the pole fP gives a lead-lag compensation with a phase shift of only 135° at the intersection of the two curves; giving the transimpedance amplifier a phase margin of 45°. If using a decompensated amplifier the intersection of 1/F and A must occur before the second pole of the op amp.

From the equation for F the pole for 1/F is (Equation 6):

And the zero for 1/F is (Equation 7):

The addition of a zero at fZ changes the slope of 1/F from 0 dB to +20 dB. For stability the slope of 1/F must be changed back to 0 dB and this is done by the pole created by the addition of CF.

Figure 6 also shows the overcompensated case, where the value of CF is larger than its optimum value. By making CF larger, the pole and zero for 1/F will occur at lower frequencies. In addition, CIN becomes a smaller portion of the denominator for the value of the zero; therefore, the pole and zero both drop in frequency and come closer together as the value of CF is increased for over compensation. Overcompensation may be necessary when using de compensated op amps and the intersection of 1/F and A is located near the second pole of A. The equation for A is simply the gain bandwidth product which is (Equation 8):

In this equation, fGBW is the frequency where the open loop gain is 0 dB (1V/V) for unity gain stable op amps. For decompensated parts, if the -20 dB slope was extended to the 0 dB line, the frequency at this point would be the value for fGBW . Using the equation A*F = 1, we now solve for CF (Equation 9):

Because A intersects 1/F at the pole of 1/F, we have the following two relationships (Equations 10a and 10b):

Using these two formulas to solve the A*F equation, the magnitude is found by taking the square root of the sum of the squares (RSS) of the real and imaginary parts (s = jω). This results in (Equation 11):

The above result simplifies into the equation to use to solve for CF (Equation 12):

This formula is difficult to solve for CF . It is best to set up an Excel spreadsheet with all the values and use iteration solve for CF . In almost all applications, the assumption that CIN >> CF will apply.

With this assumption the above formula can be simplified to easily calculate the value for CF as follows (Equation 13):

This completes the theory and calculations for a transimpedance amplifier. The above formula is the one to use for calculating the optimum value for CF . If the value calculated for CF gives too much ringing in the circuit, then some overcompensation can be used to reduce the ringing. Overcompensation will, of course, reduce the bandwidth of the transimpedance amplifier.