The objective of this lab activity is to investigate simple Transresistance amplifier Input stage configurations.
A transresistance amplifier outputs a voltage proportional to its input current. The transresistance amplifier is often referred to as a transimpedance amplifier, especially by semiconductor manufacturers. The general description for a transresistance amplifier in network analysis is as a current controlled voltage source (CCVS) .
An inverting transresistance amplifier can be configured from a conventional operational amplifier and a single resistor. The resistor is connected between the output and the inverting input of the operational amplifier and the non-inverting input is connected to ground. The output voltage will then be proportional to the input current at the inverting input node, decreasing with increasing input current and vice versa.
The activity in this lab exercise investigates an alternate differential input structure which produces an inherently low input impedance (a current input) as opposed to the relatively high input impedance of the voltage differential pair that was investigated in BJT Activities and MOS. A complete transresistance amplifier will require the addition of possible further gain stages and an output driver stage such as what was investigated in the Activity on Amplifier Output Stages.
3 – 1 KΩ resistors
2 – 2.2 KΩ resistors
1 – 47 KΩ resistor
2 – 10 uF capacitors
2 – NPN transistors ( 2N3904 or SSM2212 )
2 – PNP transistors ( 2N3906 or SSM2220 )
The circuit and the connections to the ALM1000 hardware are as indicated in figure 1. NPN transistors Q1, Q2 and PNP transistors Q3 and Q4 should be selected from the available devices with the best matching of VBE. Transistors fabricated in the same package such as the SSM2212, SM2220 or the CA3046 tend to match much better than individual devices. Split I/O input AIN may optionally want to be connected to the junction of Q1 and Q3 emitters or the collectors of either Q1 or Q3 while investigating the operation of this circuit. The current input node at the junction of the emitters of Q1 and Q3 is nominally a low impedance so as to be driven from a current source. The AWG channel outputs of M1k are more like voltage sources. So the 1KΩ resistor RIN serves to convert the voltage output of AWG CH A to a current (IIN = VIN/1K),
Figure 1, Transresistance amplifier input stage with current drive
The channel A waveform generator (output on pin CH A), should be configured for a 1 KHz sine wave with 400 mV swing, Max value set to 2.7 and Min value set to 2.3. Channel B split I/O input (on pin BIN) should be connected to the combined output at the end of RL.
Observe the output at RL, which is the AC coupled sum of the signals at the collectors of Q1 and Q3. Measure the voltage gain from the AWG channel A output to RL and compare it to your calculated value. Observe the voltage amplitude of the signal seen at the current input node, at the emitters of Q1 and Q3. Based on the this amplitude calculate the input current amplitude (voltage across RIN divided by RIN ) and the effective input resistance of the amplifier. Compare these values to your calculated values.
Is the voltage at the emitters of Q1 and Q3 equal to the common mode level (2.5V)? How much different is it and why?
1 – 470 Ω resistor
Now we are going to reconfigure the input for voltage drive. Replace RIN with a 470 Ω resistor and connect the other end to the 2.5 V common mode level as shown in figure 2. Disconnect the emitters of Q2 and Q4 from the 2.5 V common mode voltage and connect them to the output of CH A.
Figure 2, Transresistance amplifier input stage with voltage drive
Observe the output at RL, which is the AC coupled sum of the signals at the collectors of Q1 and Q3. Measure the voltage gain from the CH A output to RL and compare it to your calculated value. Observe the voltage amplitude of the signal seen at the current input node, at the emitters of Q1 and Q3. Based on the this amplitude calculate the input current amplitude ( voltage across RIN divided by RIN ) and the effective input resistance of the amplifier. Compare these values to your calculated values.
To measure the current that is required from the input driver (CH A) in this voltage drive configuration, you can turn on the CH A current trace. How does this current compare to the current seen in across the 470 Ω RIN resistor?
How closely does the voltage at the emitters of Q1 and Q3 follow the voltage at emitters of Q2 and Q4? Explain any differences you observe.
For Further Reading:
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