Transistor Circuits Homework: Explaining Circuit Action as VA Changes

[Jimmy87]

Homework Statement

Please see circuit diagram attached which I am struggling to understand. The question is: explain in detail the action of the circuit as VA changes from –5.0V to +5.0V. I have looked through my textbook and I can see that when VA reaches +5V the base of the transistor switches it on but I have little understanding of what is going on. Why does the transistor not work when the base is at -5V? Also, when the base is at +5V I don't understand how current can flow up to the relay because the voltage at the top of the circuit is also at +5V?

Homework Equations

None

The Attempt at a Solution

Tried looking through relevant sections in textbook

 

[Curious3141]

Homework Statement

Please see circuit diagram attached which I am struggling to understand. The question is: explain in detail the action of the circuit as VA changes from –5.0V to +5.0V. I have looked through my textbook and I can see that when VA reaches +5V the base of the transistor switches it on but I have little understanding of what is going on. Why does the transistor not work when the base is at -5V? Also, when the base is at +5V I don't understand how current can flow up to the relay because the voltage at the top of the circuit is also at +5V?

Homework Equations

None

The Attempt at a Solution

Tried looking through relevant sections in textbook

It's been many years since I dabbled in electronics. So I'll give it a go because no one else has answered, but more experienced posters will likely chip in and correct my errors.

That's an NPN bipolar junction transistor. Your text should have a detailed chapter on how these work talking about semiconductor doping, the charge carriers (predominantly electrons in this case) and what happens at the base, collector and emitter. You should have something about diodes before this, and it's essential to understand how diodes work to get how transistors work. So review that.

Here the transistor is being used as a switch. The transistor "switches on" when the voltage between base and emitter goes above a certain threshold value. That value is usually +0.6V in a silicon transistor. The positive value indicates the base is at a higher potential than the emitter (which is held at ground, or 0V) because this is the switching condition for an NPN transistor. The reverse would be the case for a PNP transistor, when the base has to be at a lower potential than the emitter to switch on.

So at some point during the transition of $V_A$ from -5V to +5V (which happens quickly in a switching application), the transistor gets switched on. There will be a current flowing between base and emitter. This will be amplified into a much larger flow of current between collector and emitter, because transistors amplify current. Think of the base-emitter junction as an input diode. When it gets positive biased above +0.6V it acts to turn on an output diode (the collector-emitter junction) and allows a (greater) current to flow through the output than was fed into the input.

The transistor will not switch on when $V_A$ is at -5V because the base is clearly at a lower potential than the emitter. It will only switch on when $V_A$ is some positive value high enough that the base-emitter potential is at or above +0.6V.

Anyway, when the transistor is switched on, the output resistance (which was initially very high, infinite in the ideal case) suddenly goes very low (zero in the ideal case). This draws current from the positive voltage source (+5V) through the inductance, activating the relay.

The diode in parallel with the inductance is known as a kickback or protective diode, and its sole function is to protect the collector from a very high voltage that may develop across the inductance when the transistor switches off. This is because of a basic property of an inductor that causes them to resist changes in current. You can read more about this in your text. The diode is reverse-biased when the transistor is on but the moment it goes off and the inductor develops a high potential at the point connected to the collector, the diode is in forward bias and shorts out the inductor, protecting the collector of the transistor.

 

[Jimmy87]

Many thanks for your help! This makes more sense now. So when the transistor is activated does current flow from the top of the diagram to the bottom or does it flow from the bottom (i.e. where the transistor is) to the top?

 

[Curious3141]

Many thanks for your help! This makes more sense now. So when the transistor is activated does current flow from the top of the diagram to the bottom or does it flow from the bottom (i.e. where the transistor is) to the top?

I forgot to mention one thing in my previous post. "Saturation" - the technical term for what happens at the collector-emitter junction when a transistor is fully switched on.

Current always flows from a higher potential to a lower potential when there is a path between these. When the transistor is off, there is no path (because the output resistance is very high, it's like an "open circuit"). But when the transistor is on, a nice path is created between collector and emitter.

So which is at a higher potential (voltage)? Collector or emitter?

Sorry - it's past midnight here in Singapore. Have to turn in. Will check on your reply in the AM, unless someone's already handled it.

 

[Jimmy87]

Thanks again for the reply and information - very helpful. I think I was getting confused with the base being at +5V and the top of the circuit also being at +5V and therefore thinking that there is no potential difference. So when the transistor is activated, current will flow from the very top of the circuit (+5V) to 0V at the bottom, is that correct? So that collector (Y) will be at +5V and the emitter at 0V? Wow that is late! It's only 4pm here.

 

[lightgrav]

yes, it happens before V_A becomes 5V. When base becomes 0.7 V higher than the emitter (at V=0), the "PN junction" acts like a diode (the transistor symbol), and the transistor resistance becomes low. then, the 5V top pushes current thru the relay and thru the transistor emitter.

 

[Curious3141]

Thanks again for the reply and information - very helpful. I think I was getting confused with the base being at +5V and the top of the circuit also being at +5V and therefore thinking that there is no potential difference. So when the transistor is activated, current will flow from the very top of the circuit (+5V) to 0V at the bottom, is that correct? So that collector (Y) will be at +5V and the emitter at 0V? Wow that is late! It's only 4pm here.

Yes, you can say the current will from from the top (+5V) through the inductance/relay and then through the collector and the emitter toward ground.

[strike]However, the actual voltage at the collector may not be a full 5V because there could be a potential drop across the inductance of the relay. An ideal inductance will not have any resistivity but a real-life inductance always has some resistivity. So the voltage at the collector may be some value less than 5V, but of course it'll be a lot higher than ground so current will still flow toward the emitter.[/strike]

Sorry, the above is misleading. Rude Man is right - when the transistor switches on, the collector-emitter resistance falls to a very low value, and therefore the voltage at the collector is almost zero. Nearly 5V is dropped across the inductance.

 

[rude man]

Thanks again for the reply and information - very helpful. I think I was getting confused with the base being at +5V and the top of the circuit also being at +5V and therefore thinking that there is no potential difference. So when the transistor is activated, current will flow from the very top of the circuit (+5V) to 0V at the bottom, is that correct? So that collector (Y) will be at +5V and the emitter at 0V? Wow that is late! It's only 4pm here.

No.

The collector will be close to 0V when base current flows. So the relay coil sees nearly 5V across it (current flowing thru the coil from the +5V power supply thru the transistor to ground.)

Now, if you want another challenge, explain the function of the diode connected across the coil!

 

[Jimmy87]

Thanks people this has really helped me a lot! I took your challenge on board rude man! I think the diode has to do with Lenz's Law and EM induction. When the coil is on there is a constant magnetic field. As soon as it is switched off the field falls instantly to zero which means there is a big change in the magnetic field which means a very large back emf is induced. I think the emf is in the same direction as it was before as Lenz's Law always tries to keep the magnetic field the same as it was before it was switched off. However, this emf is much larger due to abrupt change in the magnetic field. From the reading I have just done it looks like the diode diverts the current induced by the back emf around in a circle until it stops. But I'm not sure why this stops it going through and damaging the transistor which is what my book says the diode does? Is it because the diode has a lower resistance so the back emf will choose the route of the diode over the transistor?

 

[rude man]

Very good explanation, Jimmy! You could also say that the voltage across an inductor is L di/dt and of course di/dt is huge when the transistor suddenly shuts off.

Look at the way the diode clamps the collector voltage. If the cathode is at Vcc = +5V what will the anode be? & the anode is connected to the collector.

 

[Jimmy87]

Thanks for the info rude man! I'm still a bit confused with the diode. How does it stop current damaging the rest of the circuit? Also, what do you mean by the cathode and anode? and clamping?

 

[rude man]

Thanks for the info rude man! I'm still a bit confused with the diode. How does it stop current damaging the rest of the circuit? Also, what do you mean by the cathode and anode? and clamping?

A diode has an anode end and a cathode end. Current flows only from anode to cathode (in the direction of the implied arrow representing the diode). If the diode is carrying current the forward voltage (anode +, cathode -) is never more than about +1V for a Silicon diode.

So OK, the current thru the coil flowed thru the transistor & now (when the transistor is shut off) the current must continue, and so it gets diverted thru the diode.

So from what I just said, what is the max. voltage you can get at the collector? Remember, the idea is to prevent large voltages from appearing at the collector which would break down the transistor and quite possibly damage it.

 

[Jimmy87]

Thanks again for the info rude man. Oh ok that makes sense, sorry bit of a stupid question - I didn't realize they were called anode and cathode in diodes. I'm note sure about the max voltage at the collector. I don't quite understand - if you have a large voltage when the coil is switched off surely this will be much higher than 1V so how does the voltage get dissipated if the diode can only handle 1V? One other final question, the transistor in the diagram is supposed to be an NPN transistor according to my book. When the voltage at the base is negative the transistor is off - now is this because NPN transistors require a positive base to allow current to flow as its not very clear in my book?

 

[rude man]

Thanks again for the info rude man. Oh ok that makes sense, sorry bit of a stupid question - I didn't realize they were called anode and cathode in diodes. I'm note sure about the max voltage at the collector. I don't quite understand - if you have a large voltage when the coil is switched off surely this will be much higher than 1V so how does the voltage get dissipated if the diode can only handle 1V? One other final question, the transistor in the diagram is supposed to be an NPN transistor according to my book. When the voltage at the base is negative the transistor is off - now is this because NPN transistors require a positive base to allow current to flow as its not very clear in my book?

The diode will carry no more current than what the coil carried before the transistor was switched off. The diode forward voltage drop can't exceed about 1V. If it did it means it's carrying many amperes of current, way more than the coil current. It also means the diode will probably self-destruct since power = i times V and for a small diode like a 1N4148 the max allowed current is below 100 mA.

Don't think of an inductor necessarily producing a high voltage. Think of the current not being able to change instantaneously instead - ever! The diode "catches" the coil current when the transistor stops carrying it.

The diode is also called a "clamping diode" since it clamps the collector voltage to no more than about 0.7V above the power supply.

For the coil, di/dt = V/L with V about 0.7V. That tells you how fast the coil current is abating.

Yes, an NPN needs current flowing into the base to turn on.