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I understand the need for flyback diodes when trying to change inductor current abruptly with a mechanical switch or when using a BJT/MOSFET as a switch.

Now, assume I (linearly) control the voltage across a coil with this op-amp circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

It seems to me that a flyback diode should not be needed at all with this circuit, even if we imagine that the op-amp has no internal diodes between its output terminal and its power terminals.

Let's say the circuit is steady at Vin = 4V so the coil current is 6A (from right to left). Let's then step Vin to 7V, which should decrease the stored energy in the coil. In my mind the op-amp should simply reduce current flowing through the lower transistor of its output stage at a rate which causes the EMF of the coil to result in 7V at the op-amp's output (when Vin is stepped the 6A through R1 mean that the right side of L1 is at 4V so the induced EMF across L1 would make it's left side 3V higher than the right side at that instant). This current change then ends when 3A are going through the inductor.

My question is: Is my understanding correct and theory dictates that a flyback diode is not needed for this circuit? If yes, is there anything missing from this simplified model which would require a flyback diode in real applications?

Note that this is more of a theoretical question aimed at improving my understanding and mental model of op-amps, so answers along the line of "just throw a diode in there to be safe" are unnecessary. This is a simplified version of my actual circuit which has dual-supplys and regulates current (using a current sense resistor between inductor and ground), so any explanation which allows extrapolation to more complex circuits is welcomed.

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I would still add diodes between the output of the opamp and both supply rails.

Such diodes are already present anyway inside the opamp (for ESD protection) but you do not want to damage these diodes as then you'd have to replace the opamp.

You're right that under normal working conditions the current through the coil is not interrupted (which is the cause for the high damaging voltage) but what about when you switch the circuit on, off or a glitch occurs on the supply ?

So I would not take the risk and just add reverse-biased diodes to protect the opamp's output. I'd use fast Schottky diodes which can handle 1 A forward current (this is just my guess).

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    \$\begingroup\$ +1 for start-up and shutdown protection. Designers often forget about that. \$\endgroup\$ – Trevor_G Jun 5 '17 at 12:29
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When you try to instantaneously change the current from 6 amps to 3 amps the left side of the inductor will react in a way to keep the current flowing at 6 amps (just for a short time until the stored energy is lost somewhere). So, its only option is to generate a voltage that is high enough to force current through the upper transistor in your push-pull stage inside the op-amp (or generate a spark).

Given that this top transistor is never actually intentionally turned on in either scenario, the inductor will pretty instantaneously generate a bigger and bigger voltage until that transistor breaks down or you get a spark somewhere.

It seems to me that a protection diode is needed.

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    \$\begingroup\$ In the mental model that I have, I'm not trying to change the current instantaneously, but rather the voltage. I agree that as soon as the op-amp starts limiting the current then the inductor's EMF will raise the voltage of the op-amp output net to try to force current through the op-amp. However, as soon as that net hits 7V my feeling is that the op-amp should "back off" and not reduce the current any further. (This then hapens an infinite number of times until the current has settled to 3A, per a typical RL diff.eq.) Am I missing something regarding the op-amp's behaviour in this scenario? \$\endgroup\$ – GummiV Jun 5 '17 at 16:31
  • \$\begingroup\$ @GummiV it falls down to ye olde formula V = L di/dt. if you are changing the current gradually then the voltage produced by the inductor that is attempting to counter that current change will be small. If you attempt to change current too quickly then that counter-voltage will be large enough to spark or destroy the top transistor. No, as soon as 7 volts is hit, things DO NOT become in equilibrium and settle because there will always be some energy left in the inductor but, if your driver is a hard-nut and can clamp at 7 volts then it will be acting as a reverse diode..... \$\endgroup\$ – Andy aka Jun 5 '17 at 18:25
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    \$\begingroup\$ .... Whether it does this is dependent on how good the driver is; usually people don't test/know this and use a diode. \$\endgroup\$ – Andy aka Jun 5 '17 at 18:26
  • \$\begingroup\$ Thanks. It feels like I'm still missing something. You mention "changing the current too quickly" when I feel like I'm changing the voltage and letting the current "settle" to a final value (in the similar manner as would happen when connecting an inductor to a voltage source, except here I'm decreasing current by dissipating the energy in the OA's transistor). That is, my thinking is that after the step change the op-amp will effectively regulate the di/dt part so that the op-amp output net remains at a fixed 7V until the current has changed from 6A to 3A (with the typical L/R time constant). \$\endgroup\$ – GummiV Jun 5 '17 at 19:08
  • \$\begingroup\$ If your driver can hack the ability to maintain the voltage then it will. You have to remember that I have to try and get inside your thoughts and sometimes answers can swing a bit back and forth but if the driver can handle sinking a current into the top transistor when it normally would only sink current then it should be ok. \$\endgroup\$ – Andy aka Jun 5 '17 at 19:29
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Often when you have a discrete push-pull circuit you don't need flyback diodes, but they will do no harm to put in, and if you can't analyze the (perhaps integrated circuit) amplifier output section of the amplifier to determine they are not necessary this is a good (and inexpensive) practice. In this case you can only sink current, so a diode to the +10 supply would be called for. A Schottky diode is much safer than a regular diode since little current will flow through internal junctions when the Schottky conducts in parallel. Current flowing through internal junctions could cause issues that are difficult to predict.

Note that there is still something to worry about- if there is such a diode (or the equivalent) and if the 10V supply is suddenly disconnected, the collapsing field may cause the supply voltage to the chip to rise to a high enough level to damage the chip (depending on bus capacitance etc.). So a Zener or a TVS capable of absorbing the energy in the coil may be a good idea- put it from +10 to ground, or from the amplifier output to ground.

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