[Exam problem for an instrumentation course]

Can anyone help me understand how this circuit works? I'm not too familiar with power circuits... This somewhat reminds of a BJT class B amplifier (but the diodes are not connected to the base), not sure if they are related or not. I also don't quite understand what the OpAmp adds to the circuit.

The problem asked for a sketch of the DC voltage transfer characteristic, but any elucidation related to the function of this circuit will help.

Thank you in advance.

Power-driver circuit

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    \$\begingroup\$ Do you know what the DC voltage transfer characteristic would be if it was only an op-amp? (Where the output of the op-amp would be your Vo and it would go to the same place as in your schematic, to the negative input) \$\endgroup\$ – Harry Svensson Jun 22 '19 at 12:45
  • \$\begingroup\$ "The problem asked for a sketch of the DC voltage transfer characteristic ..." Was that the op-amp transfer characteristic or the complete circuit? What exactly was asked for? \$\endgroup\$ – Transistor Jun 22 '19 at 13:00
  • \$\begingroup\$ assuming rail-rail opamp output,and assuming no currents thru the transistors, then the output will move to 0.5v or 0.6v of either rail. \$\endgroup\$ – analogsystemsrf Jun 22 '19 at 17:32

If you assume that the opamp is a rail-to-rail device you can then say from inspection:

  1. The configuration has unity gain.
  2. At zero load current the output limits at approximately +/- 4.3V (one V(be) drop from each supply).
  3. The configuration will clip at approximately +/-70mA and the output voltage will reduce linearly (the 10 Ohm resistor) above that current. This is calculated from the current flowing through the 10 Ohms resistor that would cause the transistor to be fully saturated (Collector just a few mV above Emitter). It is only an approximation of course without detailed transistor specs.
  4. The diodes serve no apparent purpose at all, positioned as shown on an output. To cause conduction you'd have to raise the output terminal above/below the supply voltages. If the output voltage is within the supply rails then the diodes can never conduct. They were possibly added to the circuit simply as obfuscation.

With device characteristics you could further calculate the limits.

| improve this answer | |
  • \$\begingroup\$ Thank you for your response. I think I have a better sense of what the circuit is doing now. But the diode placement still confuses me... And I'm sorry if I'm lacking some basic knowledge, but how did you estimate that current? One of the other assignments was to estimate the maximum output current across the transistors and I'm a bit lost on that too \$\endgroup\$ – João Miguel Jun 22 '19 at 14:47
  • \$\begingroup\$ @JoãoMiguel Added to the answer. \$\endgroup\$ – Jack Creasey Jun 22 '19 at 15:04
  • \$\begingroup\$ @JoãoMiguel The diodes would serve a purpose if whatever your output is connected to (the load) is inductive and/or if the output somehow goes too far (~0.7V=forward voltage of a typical diode) below 0V=ground or above 5V=VDD. See it as a safety precaution, such as the one who designed it thinking "I don't know what my users will do with this, better protect against it anyways". \$\endgroup\$ – Harry Svensson Jun 22 '19 at 16:54
  • \$\begingroup\$ @HarrySvensson I agree ...but they don't change the transfer characteristics of the configuration. \$\endgroup\$ – Jack Creasey Jun 22 '19 at 16:55
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    \$\begingroup\$ @HarrySvensson This is the difference between discontinuance of current (a switch) and continuous current (an amplifier) driving an inductive load. We don't have protection diodes like these on the output of speakers (even with highly inductive/capacitive filers in them) as an example. So not silly territory at all, but perhaps not intuitive. There are just endless examples of servo drive mechanisms with continuous conduction, all inductive loads but not producing spikes because of the amplifiers driving them. But I readily admit ...it's complicated. Thanks for your comments. \$\endgroup\$ – Jack Creasey Jun 22 '19 at 18:01

All op-amps "try" to set 0V between its + and - terminals, so connecting the output directly for feedback should make it a voltage follower or buffer (amplifier with gain 1, Vo=Vi). The push-pull part made by the two BJTs is just for power output.

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  • \$\begingroup\$ Thank you for your response. Yes, it makes sense that it is just a buffer. So the objective is to maximize current output with the push-pull stage, while keeping the voltage gain unity? The purpose of the diodes is to prevent crossover distortion? \$\endgroup\$ – João Miguel Jun 22 '19 at 13:48
  • \$\begingroup\$ @JoãoMiguel if I'm not mistaken, it is. \$\endgroup\$ – Iaka Noe Jun 22 '19 at 13:55
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    \$\begingroup\$ The diodes do not have any effect at all on the transfer characteristics, they are a simple clamp. \$\endgroup\$ – Jack Creasey Jun 22 '19 at 14:01

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