How can a diode defend the coil during transistor cutoff?

[Femme_physics]

http://img252.imageshack.us/img252/6080/photo46k.jpg Once the transistor is at cut-off, it means Ib and Ic = 0. I don't see what the diode has to do with anything in terms of "protecting" another component.

Can anyone help me see that?

 

[truesearch]

If current is flowing through the coil of the relay and it is then switched off it is possible to get a high voltage induced across the coil (faradays laws).
The diode allows current to flow as a result of the induced emf which means that the current decreases slowly... giving a smaller induced emf...no danger to other components.

 

[Jony130]

The transistor is initially "open". so current is flow through the coil.
Once the transistor is at cut-off the coil tries keep current to flow in the same direction.
This means that the collector voltage goes positive relative to VCC.
So when we add a diode, the diode start goes into conduction when the switch is off. And this limits the inductor kick to Vd + Vcc = 12.7V

Current flowing through a coil creates a magnetic field which collapses suddenly when the current is switched off. The sudden collapse of the magnetic field induces a brief high voltage across the coil which is very likely to damage transistors and ICs. The protection diode allows the induced voltage to drive a brief current through the coil (and diode) so the magnetic field dies away quickly rather than instantly. This prevents the induced voltage becoming high enough to cause damage to transistors and ICs.

http://www.allaboutcircuits.com/vol_3/chpt_3/9.html

 

[Femme_physics]

I'm not sure I understand.

In your link:, it says

However, when the switch is opened, the coil's inductance responds to the decrease in current by inducing a voltage of reverse polarity, in an effort to maintain current at the same magnitude and in the same direction. This sudden reversal of voltage polarity across the coil forward-biases the diode, and the diode provides a current path for the inductor's current, so that its stored energy is dissipated slowly rather than suddenly in Figure above (c).

I don't get it, how can a diode can "suddenly" reverse it's bias, unless it's physically done? I thought that under so circumstance can voltage flow from the opposite side of the diode.

 

[DragonPetter]

The diode is not reversing its bias. The winding is biasing the diode with its inductance when the voltage spikes. When you learn about an inductor, you learn the voltage across it is:

V = L*di/dt

So, when you push the button and the transistor switches, the current goes from whatever it was to 0, and the faster this happens and the more current you had going through it, the voltage spike can get very big. The coil stores energy, and when the switch goes off, that energy has to go somewhere. Since di/dt is negative, the voltage on the lower end will be much higher than 12V, which can damage a transistor who has a maximum collector-emitter voltage rating. At the same time, the anode of the diode is actually at a higher voltage than 12V, so it is forward biased and begins conducting current, which is discharging the inductor safely, reducing its voltage spike before it gets big enough to hurt the transistor.

 

[Jony130]

Diode has nothing to do with reverse polarity.
The coil is responsible for reverse polarity.
When the switch is close, the upper end of the inductor gets set to a higher voltage than its lower end (collector). And current is flow through the coil form
Vcc ---> Coil ---> collector-emitter--->R2---> GND.
So when the switch opens, the input dc source gets disconnected from the inductor (transistor is at cut-off).
But the coil demands the current to keep flowing in the same direction as previously flowing. And this causes the lower end of the inductor to now be at a higher voltage than its upper end. So the voltage across the coil reverse his polarity. And diode can start to goes into conduction.


And I highly recommend you to read this pdf from page 22 "Understanding the Inductor".
http://www.elsevierdirect.com/samplechapters/9780750679701/9780750679701.PDF

 

[DragonPetter]

This explains it pretty well:

http://en.wikipedia.org/wiki/Flyback_diode

 

[FOIWATER]

Let me try, OP please read this:

The diode, normally does not allow current flow through it. It is across the COIL.

Imagine this, Current through that coil, produces a magnetic field.

Once power is REMOVED from that coil, the field collapses. As it collapses, it induces a voltage into the coil, this voltage causes current to flow back through the circuit due to lenz' law!

If the diode were not there to short the ends of the coil, current would back feed to the transistor.

THE DIODE IS THERE TO CARRY COLLAPSING FIELD CURRENT OF THE COIL BACK TO THE OTHER END OF THE COIL, RATHER THAN LETTING IT FEED THROUGH THE CIRCUIT!

Have you ever opened a switch feeding a relay coil? It sparks almost every time. Put a diode in parallel with your coil? Doesn't happen

My source, I work as an automotive electrician on mining equipment, and I work with this daily, so I know it is correct. Once power is removed from a coil, a voltage is induced back into the coil, as now the field collapses and crosses it.

Please read and understand this as well... The voltage that is induced when DC is removed, is EXACTLY the same as inductive reactance in a AC circuit!

With DC the field is static until power is removed, then voltage is applied against the circuit, in an AC circuit, the field is expanding and collapsing all the time, therefore a constant voltage is applied against the circuit, since this voltage opposition does not dissipate energy as heat, rather store energy in a magnetic field, we call it reactance.

The diode is to dissipate the current that flows due to the voltage that is induced in the coil when power is removed or lost.

 

[FOIWATER]

in addition:

You can think of it like this, counter emf in a motor. All motors generate while motoring, because of one thing, relative motion between a conductor and a field during motoring. Once the power is shut off, you have the condition met for generation, relative motion (collapsing field) between a conductor (the coil) and a field (the field created by the current the 12vdc pushed through the circuit). Due to Lenz' law, and common sense really, since the field is now collapsing as opposes to expanding, the voltage will be opposite, IE, rather than acting like a load, the coil acts like a generator (as it would constantly if supplied with AC, the generated power we would attribute to inductive reactance, since it is DC, the voltage is only ever induced as the field changes, IE when power is removed and current collapses, so too does the field) The current the generated voltage from the coil pushes through the circuit could back feed, but since current flows in a loop from the source, connecting both ends of the coil across the diode allows a discharge path for that spike.

People, engineers, often say than coils produce spikes or oppose current changes etc.

The REAL reason, is field collapse. As current is removed from coils (IE, as current CHANGES, the field it produces CHANGES... now you have the conditions met for faradays induction law... the direction of voltage is simply opposite to applied voltage. hence the current spike.

Hopefully I can explain it good enough for you femme...

 

[FOIWATER]

So the diode does not defend the coil in transistor cutoff, rather the opposite.

I see these a lot in automotive relays, where power for the relay coil is provided by a PLC output card. The fear is that the coil will generate current back to the card, damaging it, in the event of power loss/power removal.

 

[Studiot]

You need to be aware that the coil always develops a reverse voltagewhen the current is changing, even when the transistor is conducting. The reverse emf doesn't suddenly appear when the transistor switches off.
This is masked when the transistor is conducting so we don't normally see it.
Since the largest current change is normally when the transistor current switches off abruptly this leads to the the largest reverse voltage, which if unchecked can be many times the supply and destroy the transistor.

This is also the principle of old fashioned contact breaker points ignition in automobiles.

The correct terminology is that the coil develops 'back emf' or 'inductive backswing' or 'backswing voltage'

Here are some good pictures

http://www.physics.brown.edu/physics/demopages/demo/em/demo/5j1023.htm

 

[truesearch]

When the current to a coil is switched on the back emf cannot be greater than the applied emf. It is given by e = LdI/dt and the rate of rise of current must be so that e = applied emf. so there is no problem at switch on.
At switch off the current stops instantaneously and that is when dI/dt is large and a large induced emf is possible unless some measure (the diode) is taken to allow the current to decrease 'slowly'

 

[FOIWATER]

You need to be aware that the coil always develops a reverse voltagewhen the current is changing, even when the transistor is conducting.

But the voltage is always higher to the coil, as you said, so current always flows to the coil. With DC this diode is only used when the Counter EMF of the coil exceeds that of the external circuit, so current flows outwards from the coil. The result of the Counter EMF otherwise is to limit current TO the coil, not to cause current to flow from it.

So I don't think for sure it is every time the current changes, although I agree that the current to the coil decreases each time the current changes, I don't lend myself to the idea that the diode has to do work at that time. Not until the Counter EMF exceeds the value of external circuit voltage...

But - I could very well be incorrect.

 

[Studiot]

Yes the diode protection only acts when the transistor switches off and the forward current stops. When there is current in the forward direction the diode is reverse biased and plays no part in the circuit.

I was just pointing out that a back EMF is always present across an inductor with changing current. That is what inductors do.

Yes when there is a drive voltage this is larger than the back EMF. That is why the normal current direction is forwards. This back EMF is used in some control circuitry to sense what is happening.

You do not need a transistor to effect the switching, as with my automobile points example.

 

[FOIWATER]

True I would say all those things as well

 

[Femme_physics]

This explains it pretty well:

http://en.wikipedia.org/wiki/Flyback_diode

That article help me best understand what's going on :)

Let me try, OP please read this:

The diode, normally does not allow current flow through it. It is across the COIL.

Imagine this, Current through that coil, produces a magnetic field.

Once power is REMOVED from that coil, the field collapses. As it collapses, it induces a voltage into the coil, this voltage causes current to flow back through the circuit due to lenz' law!

If the diode were not there to short the ends of the coil, current would back feed to the transistor.

THE DIODE IS THERE TO CARRY COLLAPSING FIELD CURRENT OF THE COIL BACK TO THE OTHER END OF THE COIL, RATHER THAN LETTING IT FEED THROUGH THE CIRCUIT!

Have you ever opened a switch feeding a relay coil? It sparks almost every time. Put a diode in parallel with your coil? Doesn't happen

My source, I work as an automotive electrician on mining equipment, and I work with this daily, so I know it is correct. Once power is removed from a coil, a voltage is induced back into the coil, as now the field collapses and crosses it.

Please read and understand this as well... The voltage that is induced when DC is removed, is EXACTLY the same as inductive reactance in a AC circuit!

With DC the field is static until power is removed, then voltage is applied against the circuit, in an AC circuit, the field is expanding and collapsing all the time, therefore a constant voltage is applied against the circuit, since this voltage opposition does not dissipate energy as heat, rather store energy in a magnetic field, we call it reactance.

The diode is to dissipate the current that flows due to the voltage that is induced in the coil when power is removed or lost.

Also a great explanation. Understood.

I appreciate the other replies, I'm all saturated though:)

 

[Femme_physics]

On an unrelated note, can anyone explain to me though how come we need 2 voltage sources here? Isn't 1 enough? We can just use a transformer on 1 voltage source to adjust a single voltage. Having 2, or 3 (I have a similar exercise with 3 such voltage sources!) seem to pointlessly overcomplicate the procedure..

 

[NascentOxygen]

You need to be aware that the coil always develops a reverse voltage when the current is [strike]changing[/strike] decreasing,

Coil has a forward voltage when its current is increasing.

 

[FOIWATER]

On an unrelated note, can anyone explain to me though how come we need 2 voltage sources here? Isn't 1 enough? We can just use a transformer on 1 voltage source to adjust a single voltage. Having 2, or 3 (I have a similar exercise with 3 such voltage sources!) seem to pointlessly overcomplicate the procedure..

It's more about safety I think, here the only voltage at the hands of the operator is 12 volts. If the switch was on the 220v side, well you would have to deal with 220 volts.

I don't know why they are doing it here seems to just be an example, I would agree with you.

 

[FOIWATER]

Coil has a forward voltage when its current is increasing.

I agree here.

 

[NascentOxygen]

On an unrelated note, can anyone explain to me though how come we need 2 voltage sources here? Isn't 1 enough?

Are you referring to the 12VDC and the 220VAC? The 12V is safe, and allows the use of miniature switches, cheap low voltage transistors and other components, and fine wire with a thin covering of voltage insulation. If a toddler chews on it, or you spill coffee on the switch, or you walk on it, there should be little danger of shock or electrocution. The relay (inside the dotted rectangle) interfaces the low voltage control circuit with the 110V to the bed lamp. 12VDC is safe, but it's inefficient for room lighting. 220VAC is efficient for room lighting and operating household appliances, but it poses a danger when brought close to the reach of people and animals.

Besides, the more voltage sources there are, the more interesting it is for students to analyze. :tongue2:

BTW, I think your title could be "How does the diode protect the transistor during transistor cutoff?" https://www.physicsforums.com/images/icons/icon6.gif

 

[FOIWATER]

I mentioned that in a previous post as well - the diode definitely protects the transistor.

In another application, it could be used to protect anything electronic, anything that could be sensitive to the feedback produced by the coil during shut-down

 

[Studiot]

Good catch Nascent Oxygen I was not clear.

By 'reverse' I meant that the coil developed voltage opposes the change (Lenz law as already mentioned)

 

[Studiot]

On an unrelated note, can anyone explain to me though how come we need 2 voltage sources here? Isn't 1 enough? We can just use a transformer on 1 voltage source to adjust a single voltage. Having 2, or 3 (I have a similar exercise with 3 such voltage sources!) seem to pointlessly overcomplicate the procedure

Yes it is more safe but I think the safety explanation is incomplete.

Your circuit diagram shows the pretty conventional practice of taking a connection in the circuit to a suitable voltage supply, without detailing where that supply comes from.

So the 12 volts DC could be derived from the mains via a suitable circuit or from a battery or from somewhere else.
The point of the 12 volts DC is that the transistor switch is designed to work on 12 volts DC and use a low voltage transistor. Transistors capable of operating directly at mains voltages are relatively few and far between and much more expensive.

 

[NascentOxygen]

By 'reverse' I meant that the developed voltage opposes the change (Lenz law as already mentioned)

I think that is an unnecessary and unhelpful complication, though. I'd simply justify inductor voltage on the basis that v=L di/dt is the characteristic of an inductor.

FOIWATER, this needs a bit of polish: "The diode is to dissipate the current that flows..."

 

[NascentOxygen]

Transistors capable of operating directly at mains voltages are relatively few and far between and much more expensive.

They can be found in every old TV set (transistorized picture tube type); cheap enough when mass produced.

 

[FOIWATER]

Provide a discharge path for

 

[Femme_physics]

So let me see if I understand how this circuit works..

When button P1 is off no current flows through the circuit, and as long as the switch parallel to the transformer is off no current flows through the lamp. When P1 is pressed and the switch closes, current begins to charge the capacitor. Once the capacitor charges to 7.3 V the transistor turns on (-> according to the description) then the 220 AC voltage along with the 12 volts provide the voltage to light up the lamp.
P2 short-circuits the transformer circuit unit.Did I get it right?

 

[vk6kro]

Mostly OK except that there is no transformer in this circuit.

The thing with an inductor in the dotted box is a relay.

When current flows through the inductor, it attracts a piece of soft iron attached to a switch, turning the switch on.
When the switch is on, it allows current to flow through the lamp from the 220 volt mains supply.

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).

This circuit's main function seems to be to produce a slight delay after pushing switch P1 before the lamp comes on.

 

[NascentOxygen]

So let me see if I understand how this circuit works..

When button P1 is off no current flows through the circuit, and as long as the switch parallel to the transformer is off no current flows through the lamp. When P1 is pressed and the switch closes, current begins to charge the capacitor. Once the capacitor charges to 7.3 V the transistor turns on (-> according to the description) then the 220 AC voltage along with the 12 volts provide the voltage to light up the lamp.
P2 short-circuits the transformer circuit unit.


Did I get it right?

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.
Switches labelled Pn are probably push-buttons that connect only while held down, then when released the contacts open by spring operation. So the lamp stays on while you keep your finger on P1. But if you release the pressure on P1 then no further current can flow via R1 into the transistor base.

https://www.physicsforums.com/images/icons/icon2.gif 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?

 

[jimalex]

I work in industrial electrical and electronic machine manufacturing. we use these alot; 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.