How can a diode defend the coil during transistor cutoff?

[jimalex]

I work in industrial electrical and electronic machine manufacturing. we use these a lot; we call them flyb, ack diodes. Flyback is when the current through a coil is interrupted and all the energy stored in the coils of the solenoid is near-instantaneously delivered into the circuit literally as the switch is opened, in the form of an arc across the contacts of the switch that is opening. the flyback can be more than a dozen times the voltage of the one that was applied to the coil. when it is a switch in the primary of a transformer opening, the flyback on the secondary side is amplified bigtime; we call those a flyback generator, like an ignition coil in a car.

the diode in the drawing (shorting out a flyback generator) is going to give the flyback current a safe path to dissipate through, instead of blasting its way through the circuit. since the flyback is in the same direction as the applied voltage was, it will circulate through the diode until it is gone.

J

 

[Femme_physics]

If you hold P1, then release it and do not press on P2, what can you say is going to happen to the lamp's operation?

I'm not entirely sure, mostly because my big confusion that I will explain after quoting vk6kro

The reason a relay is used is to allow a circuit using only 12 volts to control a much more dangerous 220 volt AC supply. Also, most transistors will work on 12 volts so the choice of transistors for this circuit are easier than if it had to run on 220 volts AC (if it was rectified and filtered).

Well, we have 220 volts here with nothing in their way, so, if we look at them individually for a moment

http://img225.imageshack.us/img225/299/circuits.jpg
The first circuit supplies so much voltage (without any resistance in between it and the lamp!) that IMO that second circuit is pretty useless and downright ignored. (don't mind the incomplete line in the drawing I made)

That 7.3V figure is not derivable from the information in your schematic. We'd need to know the transistor's approximate current gain, β, to determine that figure.

Yes, I mentioned in brackets that was defined in the text of the question :) sorry.

If the switch was on the 220v side, well you would have to deal with 220 volts.

But the switch is on the 220 V side!

 

[vk6kro]

If you just wanted to turn a lamp on and off, you would get a 250 volt switch and switch it on and off. Of course.

But if you wanted a delay after pressing the switches, as in this case, or if you wanted to turn the lamp on and off using a computer or a low voltage electronic circuit, then you would need a relay driver circuit like this.

 

[Femme_physics]

If you just wanted to turn a lamp on and off, you would get a 250 volt switch and switch it on and off. Of course.

But the voltage on the lamp will always just be the AC voltage (in our case 220 V)

So basically the other circuit is just to allow for delay and computer control. But once the lamp turns on, the voltage stays at 220 V. Yes?

 

[vk6kro]

Yes, it is partly a safety measure too. The lamp turns on via a switch so it is very safe to do it this way.

The alternative would be to have a large high voltage FET or Triac and directly switch it with a low voltage circuit. This can be done, but it is risky for the low voltage circuitry if the FET or Triac failed.

As an example, you might like to turn a lamp on when it is night time. You could have a low voltage circuit to detect that it is dark and then a relay driver circuit to actually turn the light on.

 

[Femme_physics]

I see...

I want to see that I fully understand the original thread's question. So first of all I'm trying to understand how the currents work in this circuit for negative and positive...(steady state) how do I fare here? http://img341.imageshack.us/img341/115/currrr1.jpg

http://img14.imageshack.us/img14/5299/currrr2.jpg

 

[sophiecentaur]

The 220V part of the circuit is irellevant to all of this discussion - except in so far as it is the raison d'etre for the thing in the first place.
In both of your diagrams, the 'control' switch is in the same state so how could the two situations be different? (The transistor is On in both cases). That 'up arrow' in the second diagram implies current flowing from negative to positive through the diode. Why would it be like that? When the transistor turns off, the PD at the bottom of the coil increases ('trying' to maintain the original current through it) to a value which would be way above the 12V rail, The diode just provides a short circuit path to avoid this happening. (Look at the way the diode is connected - it will only conduct at the appropriate time if it is that way round- so the collector volts will increase until they are 12V+the diode drop and never higher)

PS, what's wrong with using the accepted symbols in a circuit schematic? That anglepoise lamp looks very nice bit I don't think you'd find it in a list of ISO symbols.

 

[NascentOxygen]

I want to see that I fully understand the original thread's question. So first of all I'm trying to understand how the currents work in this circuit for negative and positive

A relay is a solenoid-operated mechanical switch. Your transistor circuit switches current to the solenoid and the magnetic field causes the contacts of the solenoid-operated switch to close allowing 220V to power the lamp. When the 220VAC reverses polarity, the only current that changes direction is that in the 220V conductors to the lamp. Nothing on the transistor side changes polarity, nor does the current in the solenoid.

(I've used the word "solenoid" though it's actually a solenoid with a steel core, so it's an electromagnet.)

 

[Femme_physics]

Let me see if I get this straight. Are you telling me that the only point for the left sided circuit is to electrically close a switch?

 

[sophiecentaur]

Let me see if I get this straight. Are you telling me that the only point for the left sided circuit is to electrically close a switch?

Yes.

Imagine you wanted to take a bath and you needed to turn an electric heater on and off. Would you rather have a 12V circuit in your actual hand or a 230V circuit, whilst you were 90% immersed in water?
Sometimes relays are reelay useful. Don't knock 'em.

 

[Femme_physics]

I see. And returning to the flyback diode. To my understanding, it defends the transistor not the coil itself. Right? Is that because the NPN transistor doesn't like too much current flowing FROM the emitter to the collector?

 

[DragonPetter]

I see. And returning to the flyback diode. To my understanding, it defends the transistor not the coil itself. Right? Is that because the NPN transistor doesn't like too much current flowing FROM the emitter to the collector?

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.

 

[FOIWATER]

I'm not convinced the diode is to protect only the transistor.. although I agree it could be for that.

The application speaks to me because of the push buttons. There is no seal-in contact, meaning field collapse of the coil occurs when you release the button, which means the button contacts would arc upon release without the diode.

PS: I know it's low voltage, people say it would be unlikely to arc, but the button will definitely cause an arc to drag even at that low voltage, I personally have seen it happen quite frequently at 24 volts where coils are involved.

 

[FOIWATER]

I see. And returning to the flyback diode. To my understanding, it defends the transistor not the coil itself. Right? Is that because the NPN transistor doesn't like too much current flowing FROM the emitter to the collector?

But it's definitely not to defend the coil you have it right there.

Although - I have seen a flyback application that does in fact sort of protect the coil.. that is a huge power diode across a large DC motor. If the system shuts down unexpectedly during operation, the field collapse could induce a large enough current to flash the motor, so the current is dissipated in the diode.

Although - It always seemed odd to me, since the normal operating voltage must be higher than the collapsing field voltage.

Come to think of it, maybe the diode was so that the current doesn't somehow damage the SCR's feeding the motor in the event a contactor failed to open...

 

[DragonPetter]

I'm not convinced the diode is to protect only the transistor.. although I agree it could be for that.

The application speaks to me because of the push buttons. There is no seal-in contact, meaning field collapse of the coil occurs when you release the button, which means the button contacts would arc upon release without the diode.

PS: I know it's low voltage, people say it would be unlikely to arc, but the button will definitely cause an arc to drag even at that low voltage, I personally have seen it happen quite frequently at 24 volts where coils are involved.

Well, the one push button is connected between a regulated 12V and a huge capacitor that is either at ground, 12V, or somewhere in between depending on how much the cap is charged. The other has a path from the coil through the collector-base junction of the transistor, but that would mean the transistor would be damaged first. Also, the 200uF cap would help absorb some of the energy if it made it through the BJT for some reason.

 

[FOIWATER]

I see what you mean yeah

 

[truesearch]

Please do not have a 230V heater in your bathroom ! Even if the control is only 12V

 

[NascentOxygen]

I'm not convinced the diode is to protect only the transistor.. although I agree it could be for that.

The application speaks to me because of the push buttons. There is no seal-in contact, meaning field collapse of the coil occurs when you release the button, which means the button contacts would arc upon release without the diode.

You have a good point, and yes the switch in series with R1 may in some circumstance be subject to the coil's back emf and arcing could feasibly be a factor if a low voltage switch were used here. Though arcing at that switch may not be paid much heed in this case as it can only follow the transistor being destroyed (by reverse C-B punchthrough).

 

[NascentOxygen]

But it's definitely not to defend the coil you have it right there.

Although - I have seen a flyback application that does in fact sort of protect the coil.. that is a huge power diode across a large DC motor. If the system shuts down unexpectedly during operation, the field collapse could induce a large enough current to flash the motor, so the current is dissipated in the diode.[/color]

https://www.physicsforums.com/showpost.php?p=3916773&postcount=25

Although - It always seemed odd to me, since the normal operating voltage must be higher than the collapsing field voltage.

The voltage induced by the collapsing field = L.di/dt and so is unrelated to the magnitude of the DC supply, meaning self-induced spikes can be many times greater.

Come to think of it, maybe the diode was so that the current doesn't somehow damage the SCR's feeding the motor in the event a contactor failed to open...

Conceivably a high voltage spike could punch a hole in insulation between motor windings, or other insulation in the motor, since it is quite possibly a high energy spike in the situation where current levels are high.

 

[FOIWATER]

makes sense.

yeah again i meant through the diode.

actually though in the application it was a bank of resistors connected to pulse xfrms and capacitors. when the system shut down, the cap discharged into the transformers which gates scrs which then put the grid across the motor..

i don't get how spikes could induce a back emf larger than the applied voltage, because the flux, and hence current is directly related to the amount of counter emf. ultimately only a spike in supply voltage would cause a spike in current, causing the emf no?

the inductance is constant, the current change by time is related to the source voltage changing...

 

[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.