Are flyback diodes needed for this op-amp voltage controlled inductor?

Question

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:

^{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.

Answer 1

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).

Answer 2

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.

Answer 3

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.