How can a diode defend the coil during transistor cutoff?

by Femme_physics
Tags: coil, cutoff, defend, diode, transistor
 PF Gold P: 2,551 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?
 P: 349 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.
P: 409
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.

PF Gold
P: 2,551
How can a diode defend the coil during transistor cutoff?

I'm not sure I understand.

 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.
 P: 834 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.
 P: 409 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. http://www.youtube.com/watch?v=LXGtE3X2k7Y And I highly recommend you to read this pdf from page 22 "Understanding the Inductor". http://www.elsevierdirect.com/sample...0750679701.PDF
 P: 834 This explains it pretty well: http://en.wikipedia.org/wiki/Flyback_diode
 PF Gold P: 372 Let me try, OP plz 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.
 PF Gold P: 372 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...
 PF Gold P: 372 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.
 P: 5,462 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...emo/5j1023.htm
 P: 349 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'
 PF Gold P: 372 [QUOTE=Studiot;3913550]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.
 P: 5,462 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.
 PF Gold P: 372 True I would say all those things as well
PF Gold
P: 2,551
 Quote by DragonPetter 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 plz 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:)
 PF Gold P: 2,551 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..
HW Helper
Thanks
P: 5,246
 Quote by Studiot You need to be aware that the coil always develops a reverse voltage when the current is changing decreasing,
Coil has a forward voltage when its current is increasing.

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