Transimpedance amplifiers (TIAs) convert a current to a voltage. In many systems, this current comes from a photodiode. In order to test a TIA, it is best to drive the TIA with a current that emulates that of a photodiode. Current sources are high impedance, which means that the voltage can change as much as necessary to generate a given amount of current. For high-speed systems, there are few commercially available high-speed current sources; even most laboratories don’t have them. So, in this edition of Signal Chain Basics, I will show you how to use a 50 Ω high-speed signal generator and a bias tee to drive a current signal into a TIA.
The first step in emulating a signal is to define the components. For a photodiode, there are several ways to model the device characteristics. When simulating, I typically use an ideal current source with a parallel capacitor. The schematic snippet shown in Figure 1 is representative of a typical high speed photo diode model. Notice that the “diode” in the schematic is not connected because I am using the current source and capacitor as the diode model. The diode symbol is just a reminder that the actual circuit element is a diode.
SPICE Model for a Photodiode
In the laboratory, I do not have access to a current source capable of generating AC current at frequencies 1GHz and above. In order to test my TIA, I need either a light source and a photodiode, or another way to drive a known high-fidelity high-speed signal into my amplifier.
First, let’s look at what happens when driving the photodiode. The first stage of optical testing is to set up an optical signal source, which typically comprises a continuous wave (CW) laser and an amplitude modulator (or alternatively, a modulated laser). The modulation has an attribute called the extinction ratio (ER). The ER is defined as the on power divided by the off power. For example, if the laser on state is 1mW and the off state is 0.25mW, the ER = 10xlog(1/0.25) = 6dB.
Now that I have the light-source specifications, let’s move on to the diode to calculate the required currents. A typical photodiode will provide around 0.8A for every watt of light power. So going back to the example, the on laser state will provide 1mW x 0.8A/W = 0.8mA of current and the off state will provide 0.2mA of current. Now that I know what currents I need flowing into the TIA, I need to figure out how to generate those currents at the TIA input.
A typical TIA application will connect the current source directly into the amplifier input pin. This means that I will need to know the amplifier S11, or input impedance specification, in order to determine the input current for any given input-voltage stimulus. In some cases, the amplifier data sheet will have an S11 graph in one of the typical curves. Most of the time, however, the graph will have been made with a closed-loop circuit that includes a gain-set resistor or termination resistors, so it doesn’t have the information I need.
To calculate the input current for a TIA, I need to calculate the S11 for my circuit configuration. One easy way to do this is with a Texas Instruments TINA-TI simulation. Figure 2 shows the S11 for the LMH5401 fully differential amplifier in a TIA configuration.
Amplifier Input Impedance Graph and Schematic
Over the bulk of the frequency range, the S11 of the circuit in Figure 2 is 40 Ω . Given this input impedance, I can now calculate the input voltage necessary to generate a given input current. If I want 1mA of input current, I’ll need V=1mA x 40 Ω = 40mV. If I want 0.25mA of current, I’ll need V = 0.25mA x 40 Ω = 10mV.
Most high-speed signal generators are AC-coupled, which means that they will produce a ground-centered output signal. In order to have an input signal that varies between 10mV and 40mV, I’ll need a DC component of 25mV added to a 30mVpp signal to simulate my diode current. The tool that sums the DC and AC components is called a bias tee. Figure 3 shows an example schematic.
Bias Tee Schematic
Before you make your own bias tee, keep in mind that radio frequency (RF) bias tee circuit manufacturers will go to great lengths to keep the AC signal path matched to 50 Ω . You will need a 50 Ω , controlled impedance bias tee for frequencies over 100MHz.
An RF signal generator will have an output impedance of 50 Ω and is calibrated for a 50 Ω load. In this case, a 1Vpp signal applied to a 50 Ω load will be Vout = 2 x (40/(40+50)) = 0.89V. In other words, the load resistance changes the available voltage from the signal generator. So, the final settings for my photodiode emulation circuit are Vout (sig gen) = 33.7mVpp AC, plus 25mV DC, into a 50 Ω , controlled impedance bias tee circuit.
The DC voltage applied to the DC port of the bias tee may need adjusting based on the TIA’s default or quiescent voltage. For example, if the TIA is 0V when not driven, then the calculated voltage is fine. However, if the TIA input bias voltage is 1.1V, you will need to add the extra voltage to the quiescent voltage for a total of 1.125V.
Testing a TIA without a photodiode can be a challenge. Using a bias tee and a high-speed signal generator can produce good results if you know the amplifier input impedance and use that information to generate the necessary input currents.