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## How can a diode defend the coil during transistor cutoff?

 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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 Quote by 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..
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.

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 Quote by NascentOxygen Coil has a forward voltage when its current is increasing.
I agree here.

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 Quote by Femme_physics 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.

BTW, I think your title could be "How does the diode protect the transistor during transistor cutoff?"

 Recognitions: Gold Member 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
 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)

 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.

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 Quote by Studiot 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..."

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 Quote by Studiot 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.

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 Recognitions: Gold Member 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?
 Recognitions: Science Advisor 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.

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 Quote by 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?
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.

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 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

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 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

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!

 Recognitions: Science Advisor 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.

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 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?