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

(Part 1 of this article provided the basic concepts and analysis. You can read it by clicking here)

Design Examples
National Semiconductor's LMV793 is the op amp used for the design example. This is a medium-speed part with a gain bandwidth product of 88 MHz. It is also a decompensated op amp, which gives the opportunity to investigate the operation of a decompensated op amp used in a transimpedance amplifier.

The input capacitance of the LMV793 is 15 pF. The PIN-HR040 photodiode from OSI Optoelectronics was selected for the detector. This part has bandwidth of about 300 MHz under the conditions being used in the circuit. This bandwidth prevents additional bandwidth limitation due to the photodiode. This diode also has a low capacitance, only 7 pF for the circuit operating conditions. Laser diodes were used for the light source. A laser diode has rise and fall times under 5 nsec, well under the rise and fall times of the op amp. A red laser diode was selected so that the beam of light can be seen, ensuring the safety of the setup.

A high gain will first be selected, setting RF to be 100 kΩ. Figure 7 shows the circuit used for the transimpedance amplifier.

Figure 7: Transimpedance amplifier
(Click to enlarge image)

When the photodiode is exposed to light, the current will flow into the inverting node of the op amp. The direction of the current flow is shown in the equivalent circuit in Figure 2. If the design goal is to have a minimum output voltage output of -20 mV, then the current flow from the photodiode must be 0.2 µA.

The expected current from the photodiode can be calculated from the expected optical power from the light source. For the PIN-HR040, the responsivity to light is given as about 0.38 A/W for the 630 nm wavelength of light from the laser diode being used. So for 0.2 µA of current the optical power on the photo diode is (Equation 14 ):

Solving for W (light power in Watts), we get (Equation 15 ):

If the light source has all of its energy on the active area of the photo diode, then this is the only calculation required. The active area of this photo diode is 0.77 mm2 and is a circular pattern.

Fiber optical cable may have an area larger than the photo diode and normally the optical power is measured in W/cm2. Expressing the area of the photo diode in cm2 , the result is 7.7×10-3 cm2 . For the same 0.2 µA output current from a light source measured in W/cm2 , the necessary power would be (Equation 16 ):

In Figure 7, the bypass capacitors for the power supplies are not shown. Each supply has a 6.8 µF tantalum capacitor for bypassing the lower frequencies. In addition, each supply has a 0.1 µF ceramic capacitor for bypassing the high frequencies. The ceramic bypass capacitor must be located as close as possible to the power pins of the op amp.

For the photodiode CJ = 7 pF. For the op amp CCM = 15 pF. Therefore CIN = 22 pF. Using the formula for CF gives CF = 0.53 pF. Note that not only is CF much smaller than CIN , but with high-gain configurations the value reaches the limits of standard available values. For this experiment, a 0.5 pF capacitor was available, which measured 0.64 pF, so the transimpedance amplifier had slight overcompensation. The bandwidth of the transimpedance amplifier is derived from the RF CF time constant, or about 2.5 MHz. Figure 8 and Figure 9 show the test results of this circuit.

Figure 8: 1 VP-P output
(Click to enlarge image)

Figure 9: 20 mVP-P output
(Click to enlarge image)

From the rise and fall time being around 108 nsec, the estimated bandwidth is 3.2 MHz, which is close to the 2.5 MHz calculated from the RF CF time constant. The overshoot does show a phase margin greater than the 45° phase margin that would come from the optimum value of CF . Waveform averaging was used for the Figure 9 photograph so the actual response of the amplifier for small signals can easily be seen.

Next let's look at a transimpedance amplifier with a lower gain. Referring to Figure 7, if RF is now reduced to 10 kΩ, then the gain of the amplifier is one-tenth of the first example. This lower RF resistor gives a wider bandwidth. However, the light output of the laser diode must be ten times brighter for the same output level from the transimpedance amplifier. The higher output level of the laser diode did create a thermal gradient in the optical output, Figure 10 .

Figure 10: 1 VP-P output
(Click to enlarge image)

The ideal value for CF is now 1.7 pF, so the standard value 1.8 pF will be used.

The location of the pole must be checked for stability concerns. The pole is located at 8.8 MHz. This makes the noise gain (1/F ) equal to 10; which is the minimum stable gain of the LMV793. Although this matches the requirements of an ideal two-pole system for stability, the lab results showed extensive ringing. Thus, this design requires overcompensation for acceptable stability.

There are other poles and zeros internal to the op amp that are located close to the second pole, making this setup far from the ideal two pole system, therefore creating the need for overcompensation. The final value for CF is 2.7 pF. Figure 10 and Figure 11 show the response of the transimpedance amplifier with RF = 10 kΩ and CF = 2.7 pF.

Figure 11: 20 mVP-P output
(Click to enlarge image)

The higher-bandwidth configuration has rise and fall times of about 33 nsec. This gives an estimated bandwidth of 10.6 MHz. The pole, when using the 2.7 pF capacitor for CF , is located at 5.9 MHz.

This article has shown that the transimpedance amplifier is the same as a differentiator amplifier for stability calculations. The only difference between the transimpedance amplifier and a differentiator amplifier is the photodiode, which is a current source attached to the input node of the op amp. The photodiode has no affect on the stability calculations for a differentiator amplifier, once the diode capacitance is added to the total input capacitance of the differentiator amplifier.

Two different gain configurations were used in the lab, confirming a good match between the theory and the circuits tested in the lab. The equation for calculating CF does apply for any differentiator amplifier design. Even though the transfer function for a differentiator amplifier and a transimpedance amplifier is different, the feedback factor used for stability calculations is identical.

About the author
David Westerman is a staff applications engineer for National Semiconductor Corp., Santa Clara, CA. He has worked in several different applications groups at National Semiconductor for the last 16 years. For the last three years, he has been in the amplifiers applications group, covering low-power and precision op amps. He holds a bachelor's degree from the University of California (Berkeley) and a master's degree from the Georgia Institute of Technology (Atlanta), both in electrical engineering. In his spare time, he enjoys square dancing, hiking, and photography.

2 comments on “Understand and apply the transimpedance amplifier (Part 2 of 2)

  1. Mohammed15
    December 1, 2013


    this is really a great explenation for TIA, But I didn't understand how did you calculate the TIA bandwidth? what equation did you used?

    I am trying to build one with 2000pf photodiod so it will be very helpful if I understand your example.



  2. 2Torr
    June 20, 2018

    6/20/18: Links to images are broken.

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