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Voltage/Current Converter OpAmp Circuits
9 questions By Tony R. Kuphaldt
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Question 4 of 9
Explain how the operational amplifier maintains a constant current through the load:
Write an equation solving for the regulated load current, given any relevant variables shown in the schematic diagram (R1, VZ, Vsupply, AV(OL), etc.). Also, describe what would have to be changed in this circuit in order to set the regulated current at a different value.
Reveal answerIload = VZ R1Follow-up question: is the transistor sourcing current to the load, or sinking current from it?
Challenge question #1: modify the given equation to more precisely predict load current, taking the β of the transistor into account.
Challenge question #2: modify the location of the load in this circuit so that the given equation does precisely predict load current, rather than closely approximate load current.
Notes:This is a good example of how operational amplifiers may greatly improve the functions of discrete-component circuits. In this case, the opamp performs the function of a current mirror circuit, and does so with greater precision and reliability than a simple current mirror ever could.
It should be noted that the equation provided in the answer does not directly predict the current through the load, rather it predicts current through resistor R2. This is equal to load current only if the transistor’s base current is zero, which of course it cannot be. The real equation for predicting load current will be a bit more complex than what is given in the answer, and I leave it for your students to derive.
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Question 5 of 9
At first glance, the feedback appears to be wrong in this current-regulating circuit. Note how the feedback signal goes to the operational amplifier’s non-inverting ( ) input, rather than the inverting input as one would normally expect for negative feedback:
Explain how this op-amp really does provide negative feedback, which of course is necessary for stable current regulation, as positive feedback would be completely unstable.
Reveal answerIf current increases, the feedback voltage (as measured with reference to ground) will decrease, driving the op-amp’s output in the negative direction. This tends to turn the transistor off, properly correcting for the excessive current condition.
Notes:The purpose of this question is to get students to realize negative feedback does not necessarily have to go into the inverting input. What makes the feedback “negative” is its self-correcting nature: the op-amp output drives in the direction opposite a perturbation in the measured signal in order to achieve stability at a control point.
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Question 6 of 9
Predict how the operation of this current regulator circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):
- Resistor R1 fails open:
- Zener diode D1 fails shorted:
- Resistor R2 fails open:
- Zener diode D1 fails open:
- Load fails shorted:
- Wire between opamp output and transistor base breaks open:
For each of these conditions, explain why the resulting effects will occur.
Reveal answer- Resistor R1 fails open: Load current falls to zero.
- Zener diode D1 fails shorted: Load current falls to zero.
- Resistor R2 fails open: Load current falls to zero.
- Zener diode D1 fails open: Load current increases.
- Load fails shorted: Load current remains the same.
- Wire between opamp output and transistor base breaks open: Load current falls to zero.
Follow-up question: which of the two opamp power terminals (Vsupply or Ground) carries more current during normal operation, and why?
Notes:The purpose of this question is to approach the domain of circuit troubleshooting from a perspective of knowing what the fault is, rather than only knowing what the symptoms are. Although this is not necessarily a realistic perspective, it helps students build the foundational knowledge necessary to diagnose a faulted circuit from empirical data. Questions such as this should be followed (eventually) by other questions asking students to identify likely faults based on measurements.
