How can a diode defend the coil during transistor cutoff?

[vk6kro]

There is a current flowing in the coil, through the transistor to ground.

Now, the transistor switches off so the bottom end of the inductor generates a large positive pulse. Lenz's Law says this is an attempt to maintain the current as it was, but that is just a way of remembering it.

This pulse can rise to hundreds of volts and, unless the transistor is capable of handling this sort of voltage, it will inevitably fail.

The top end of the inductor is still at 12 volts because the power supply is unaffected, so placing a diode across the coil gives an easy path for this pulse to go.

 

[FOIWATER]

yeah i understand the diode, but the voltage across the coil ultimately determines the current through it, so how can that current produce a voltage that is/was any greater than the applied voltage.

I understand lenz' law, but to my best of my knowledge I don't see how that voltage could be greater than the applied voltage.

The large positive pulse you make reference to, is due to field collapse as you know. so what factors determine field collapse? well the amount of current that created that field, which was ultimately a result of the applied voltage in the first place. Since the relationships are linear, how can the voltage induced during field collapse be any larger than the applied voltage, this would suggest that the current generated by the coil is greater than the current that caused the field which collapses to create the counter emf, which isn't possible.

I can buy the premise that a voltage spike causes spike in cemf, but still not at no larger value than the initial spike which caused it.

If you can explain it to me in such a way my thick head gets it, I'd appreciate it, but I don't see it at all.

V = Ldi/dt.

For a constant inductance, the maximum change in current possible will still not yield a voltage greater than a supply voltage that caused that initial current to flow. That's what I am thinking.

 

[FOIWATER]

I mean it isn't like a case of a transformer, where there are two coils, and a larger voltage can be induced at a smaller current, there is only one coil, one number of turns, and one self inductance. The coil surely cannot produce a voltage higher than the voltage that caused current to flow which created the field to collapse to induce that voltage, less the coil be capable of supplying more power then it was supplied with, which is not possible.

Am I getting this wrong?

just looking at the equation you posted, the maximum voltage induced would occur at the time the instantaneous change in current was the greatest, which couldn't occur at a rate higher than the peak magnitude of the applied voltage?

I guess that might be where I am missing something. Can current be changing at a rate larger than the magnitude of the voltage across a coil.

But if so, wouldn't that mean the current output of the coil now exceeds the current that initially caused the field?

 

[vk6kro]

The voltage is a product of the inductance and the rate of change of current. It is not related to the supply voltage.

In this case, the current may be fairly small, but the switch-off time is extremely small, so a large voltage can be generated. Much larger than the supply voltage.

If you ever test a large inductor (like a transformer winding) with a simple ohmmeter, which has only a 1.5 volt battery driving it, keep your fingers off the inductor leads as you can get a shock of several hundred volts.
Enought to remind you not to do it again.

 

[FOIWATER]

okay, thanks.

but the current output capabilities are reduced as a feature correct

 

[vk6kro]

There is no current flowing once the transistor switches off, so the power is zero, but the voltage can rise to huge values.

This is the way that your car's ignition system generates a large voltage for the spark plugs.
With only a 12 volt supply, a voltage large enough to generate a quarter inch spark is generated.

Just saw your other post:

but the current output capabilities are reduced as a feature correct

If you did put a load on this, the current would only occur for a very short time, so the average current could be quite low, but the instantaneous current could still be high.

There is nothing magic happening here, though. The power in the spark does not exceed the power available in the coil. In fact, the high voltage is a result of power being maintained . As the current drops, the voltage goes up, keeping the power constant.

 

[FOIWATER]

i mean in the context of field collapse in general, not this particular application

 

[NascentOxygen]

I can buy the premise that a voltage spike causes spike in cemf, but still not at no larger value than the initial spike which caused it.

If you can explain it to me in such a way my thick head gets it, I'd appreciate it, but I don't see it at all.

V = Ldi/dt.

Imagine a DC current flowing through a solenoid or coil of an electromagnet. There is a switch in series with that coil, allowing you to stop (i.e., interrupt) the current in the solenoid. If the switch is magically fast, and moves its contactors wide apart in a fraction of a microsecond, then the current should drop from its operational value to zero in less than a microsecond. If this were to happen, and the current fall from say 2A to 0A in 10⁻⁶ seconds, then di/dt ≈ 2x10⁶ amperes/second and (inventing a value for L, say 0.5H) the voltage that would develop across the switch contacts would be 1,000,000 V. That's nothing to be sneezed at. And nowhere have I needed to explicitly mention the DC supply voltage, though it might be only 12V.

Now, it's bit fanciful to suggest that a mechanical switch could move so fast, but a transistor switch might be in that region.

 

[vk6kro]

i mean in the context of field collapse in general, not this particular application

Yes, it is similar to what happens in transformers. High voltage and low current or low voltage and high current can both mean the same power.

 

[Femme_physics]

Current won't flow from the emitter to the collector since the emitter is always at a lower potential than the collector in your circuit.

Remember the condition this diode is protecting against is when the NPN turns off - so no current is going to flow - because it is an abrupt cutoff of current that causes the inductor to spike in voltage.

You are right that the diode protects the transistor, just for the wrong reason. The reason it protects it is because transistors have a maximum voltage that can be applied between its terminals. The maximum allowable Vce, collector-emitter voltage, can be much lower than the voltage spike that the inductor will create, and if the diode wasn't there to clamp the voltage, you would have a high voltage applied between the collector and emitter that would damage the junction internally.

Thanks for this explanation. And thanks everyone else for the feedback.

 

[truesearch]

Before the days of readily available, cheap semiconductor diodes one way to protect circuits, relay or magnet coils and people from high induced voltages at switch off was to have a resistor (a light bulb was common) permanently connected across the coil.
This provided a path for current so that the induced emf was kept low.
The disadvantage was that, unlike a diode, current was always flowing through the protective component and this was wasted energy.