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Transistor Circuit. Why darlingtons?

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patroclus said:
Yes, I suposed so.
As I said, the only thing I don't get now is that about leakage current. For the LED to have leakege current, it must be inversed biased, so it only happends when voltage in VM2 is above 4.1V or so. Then, the transistor should be off. But, how would this affect the transistor working area?

The resistor from base to emitter is usually a good idea, it guarantees the transistor switches off. As already mentioned, any current leakage from the base of the transistor down to ground will tend to switch it on, even if only slightly. Bear in mind that logic gates don't switch from 0V to 5V, they switch from low to high - the specification for which covers a wide range.

The suggestion of possible 'leakage' in the LED isn't reverse leakage, it's forward leakage before the LED turns on.

Also, I though that in PNP transistor, current came in through the emiter, and came out trhough base and collector..

It's probably best if you don't think of current in that way, there are actually two different thoughts on it any way - conventional current flow (from positive to negative) and electron flow (from negative to positive).

For PNP devices it's usually easiest to think of electron flow instead, that way everything works exactly the same - just a mirror image.

In this particular case though, don't think of it at all, think of it as a simple switch, take the base low to turn it on, and take it high to turn it off.
 
patroclus said:
Also, I though that in PNP transistor, current came in through the emiter, and came out trhough base and collector..
This is true when a PNP transistor is in its active or saturation region. The situation we are concerned with is the cutoff region.

By defintion, the current Icbo flows from collector to base as the "cbo" indicates (Icbo is negative in the PNP case).

The "o' indicates that the emitter is open which is not strictly correct in your circuit, but it makes little difference in practice.

Len
 
This is true when a PNP transistor is in its active or saturation region. The situation we are concerned with is the cutoff region.

Ah, so when transistor is off (and only off), there's this current Icbo, isn't it?
But why should this current turn on the transistor? (you mean into active region?) even if base voltage is high enough?

The suggestion of possible 'leakage' in the LED isn't reverse leakage, it's forward leakage before the LED turns on.

Forward leakage? You mean meanwhile LED is turning on?
again.. why should this small currents take the transistor into active region?
 
patroclus said:
This is true when a PNP transistor is in its active or saturation region. The situation we are concerned with is the cutoff region.

Ah, so when transistor is off (and only off), there's this current Icbo, isn't it?
But why should this current turn on the transistor? (you mean into active region?) even if base voltage is high enough?

The suggestion of possible 'leakage' in the LED isn't reverse leakage, it's forward leakage before the LED turns on.

Forward leakage? You mean meanwhile LED is turning on?
again.. why should this small currents take the transistor into active region?

Icbo can turn the transistor on if it is large enough. However, with silicon transistors, this is not very likely at room temperature. But it certainly did happen with Germanium transistors.

Sorry, I should not have called the LED current Ileak since the current is in the forward direction. It is the the forward current but will be very low since the voltage across the LED is small. I will edit the circuit I posted to correct this error.

However, the sum of Icbo and If could possibly turn Q3 on unless R3 is included.

Len
 
patroclus said:
Wow, great explanation. Thanks a lot! :)
the only thing I don't get is that about transistor and LED current leakeage. Isn't the current suppose to go OUT from the base, not in, in PNP? (Icbo?)

and.. maybe is a stupid question, I know, but would this circuit do the job?
I think the transistors cannot be saturated this way, but not sure.

This circuit will not work for another reason (additional to that posted by Nigel). The transistors in the original circuit inverted the signal since they were in common emitter configuration. Your proposal has the transistors in common collector mode which does not invert.

So the logic is wrong also.

Len
 
Yes, now I understand. It should work without R3 I suppose, but it's better to include it for safety. anyway.. why does R3 prevent the transistor turning on due to the Icbo and If currents??
 
patroclus said:
Yes, now I understand. It should work without R3 I suppose, but it's better to include it for safety. anyway.. why does R3 prevent the transistor turning on due to the Icbo and If currents??

R3 pulls the base positive, turning the PNP transistor off. If it's not there, any slight leakage towards ground will tend to bias the transistor on - by providing a larger 'leakage' through R3 to the positive rail it overcomes that possibility.
 
patroclus said:
Yes, now I understand. It should work without R3 I suppose, but it's better to include it for safety. anyway.. why does R3 prevent the transistor turning on due to the Icbo and If currents??

I expect it would work at room temperature without R3, but it would fail at higher temperatures.

What you need to understand is that a transistor is a voltage controlled current source, ie. the collector current depends on the base - emitter voltage. It is an expontial function and it varies with temperature.

Icbo approximately doubles for every 10 degree C rise in temperature for both silicon and germanium transistors.

Len
 
Great, I think I more or less understand all that has been comented here :) Many thanks!
But ljcox, you just said something that made me think...

the collector current depends on the base - emitter voltage. It is an expontial function and it varies with temperature.

I already knew that, but in practic.. doesn't colector current depend of base current, not really voltage? In active and saturation region, Vbe is always around 0,7V, isn't it?
 
patroclus said:
I already knew that, but in practic.. doesn't colector current depend of base current, not really voltage? In active and saturation region, Vbe is always around 0,7V, isn't it?

Personally I would consider a transistor current driven, but obviously current and voltage are very closely related anyway. You are quite right that around 0.7V is the Vbe for a silicon junction, but it's not an instant transition, it turns on gradually with a nice little curve.
 
and this R3 is needed in ALL transistor circuits to avoid leakages? Or just in PNP? also if I use Darlingtons?
I've seen many circuits and not many of them had this R3 resistor beteewn emiter and base.
 
patroclus said:
Great, I think I more or less understand all that has been comented here :) Many thanks!
But ljcox, you just said something that made me think...

the collector current depends on the base - emitter voltage. It is an expontial function and it varies with temperature.

I already knew that, but in practic.. doesn't colector current depend of base current, not really voltage? In active and saturation region, Vbe is always around 0,7V, isn't it?

The collector current is controlled by the charge in the base emitter region. The charge is proportional to Vbe.


Nigel, think about how a current mirror works.
Len
 
patroclus said:
Yes, now I understand. It should work without R3 I suppose, but it's better to include it for safety. anyway.. why does R3 prevent the transistor turning on due to the Icbo and If currents??

Attached is a quote that answers this question
Len
 

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