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Let's say I have a transistor with an antiparallel diode in a circuit with inductance. Now suppose that the antiparallel diode conducts an intertial current of the inductance \$i_L\$.

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During this time period a control signal to the transitor is send e.g. from the MCU. I would say that the transistor won't turn on due to the fact that it is reverse biased via the voltage drop across the antiparallel diode. Is my idea correct? Thanks in advance for clarification.

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  • \$\begingroup\$ Please simulate it. Depending on voltage, current will commutate from the diode to the transistor. If your transistor has lower voltage drop than the diode, you will still have carriers within the diode which needs to be recombined before the current has fully commutated to the transistor, so it won't happen immediately. \$\endgroup\$
    – winny
    Commented May 3, 2023 at 12:28

2 Answers 2

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Using the direction of \$i_L\$ as shown in your diagram then, the upper transistor's emitter will be raised approximately 0.7 volts above the positive rail.

Then, for the upper transistor conduct (in reverse mode), the base needs to be several hundred millivolts above the positive rail and, the roles of collector and emitter are swapped.

The transistor will of course be conducting some of \$i_L\$ towards the positive rail but, once that current has depleted, it will conduct in the normal direction.

If the drive voltage to the base of the upper transistor cannot rise above the positive rail then the diode in parallel with the upper transistor will continue to conduct \$i_L\$ until it is depleted.

It's a different story for the lower transistor of course.

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I would say that the transistor won't turn on due to the fact that it is reverse biased via the voltage drop across the antiparallel diode.

Collector and emitter are interchangeable to an extent. The current gain is very low when conducting backwards (e.g. C->E in a PNP), but the saturation voltage is also typically lower than in "normal" direction of conduction (e.g. C->E in an NPN).

So, for a transistor to conduct between collector and emitter, what matters is whether the base-to-emitter junction is biased into conduction, with the understanding that the emitter is just whichever of the two junctions starts to conduct first.

In the case you asked about, as in any other case actually, the transistor's conduction will depend on what the base is doing relative to the functional emitter, not the nominal emitter. That of course assumes that the transistor hasn't broken down due to exceeding maximum absolute ratings.

In an NPN, the functional emitter is the one with lowest potential relative to base. In a PNP, the functional emitter is the one with highest potential relative to base. The functional emitter can switch places as the transistor's biasing changes.

During this time period a control signal to the transitor is send e.g. from the MCU.

That doesn't tell us what sort of amplification/buffering/level translation exists between the source of the control signal and the base of the transistor. So we don't really know the voltage range on the base relative to the functional emitter - in the topic circuit, the upper terminal of the upper NPN transistor. So we can't say anything about whether the transistor will turn on or not.

More specifically, there's nothing that implies that the base can't be driven above the positive rail. That doesn't even take anything fancy - a base drive pulse transformer is all it may take.

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