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saturated transistor and maximum current

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csdave

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This is a follow up of a parallel thread of mine, but general enough to deserve a new one.

What happens in the following circuit with varying Rb while Rc=0 and with varying Rc while Rb=0?

**broken link removed**

does the transistor break/overheat or are there semiconductor effects that limit the currents?
 

If you drive too much current through the device it'll overheat and fail. There is no "self-protect" inside a simple device like a transistor. This is why it's important that the circuit designer takes into account all possible scenarios of operation, and provides safeguards (like current-limiting resistors) to keep the device from operating at a point which will cause it to exceed it's operational region (die junction temp, in this case).

As for the "what happens if... resistors" questions, read up on one of the MANY tutorials about how BJT's work. Write down some of the equations, it's pretty straightforward. Or put it into a circuit simulator (LTSPICE is freeware) and sim it.

Start by looking at a few of these.
 

thank you for your reply.
maybe it helps if I am clearer and ask you to check my reasoning.

say Vcc=5, and let hfe be between 100 and 200, and max collector current be 100mA (values from datasheet)

If Rc=0, then device will work provided that Rb such that Ib<=0.5mA, that is Ic<=100mA.

If Rb=0 then there's no way to make the device work because the very large Vbe will kill it regardless of the value of Rc.

Is the above correct?

If you put it in common collector, thus remove Rc and add Re on emitter, then:

If Rb=0, device will work with Re>=43ohm so that emitter current<=100mA (or 101mA if that made a difference...)
If Re=0, then device will work for such that 200*Ib<=100mA, that is Ib<=0.5mA hence Rb>=2K.

Is this correct?

thanks!

PS: I'm going to install ltspice ;)
 

csdave said:
say Vcc=5, and let hfe be between 100 and 200, and max collector current be 100mA (values from datasheet)
Click to expand...
correct

csdave said:
If Rc=0, then device will work provided that Rb such that Ib<=0.5mA, that is Ic<=100mA.
Click to expand...
correct

csdave said:
If Rb=0 then there's no way to make the device work because the very large Vbe will kill it regardless of the value of Rc.
Click to expand...
correct

csdave said:
If you put it in common collector, thus remove Rc and add Re on emitter, then:

If Rb=0, device will work with Re>=43ohm so that emitter current<=100mA (or 101mA if that made a difference...)
Click to expand...
the emitter current when you use an emitter resistor is (Vb-Ve-0.7)/Re, the collector current will be Iemitter-Ibase.
so Re>=43ohm depends on the base voltage, assuming Vb=5v then you are correct

csdave said:
If Re=0, then device will work for such that 200*Ib<=100mA, that is Ib<=0.5mA hence Rb>=2K.
Click to expand...
when Re=0 and Rb=2k the base current is (Vb-Ve-0.7)/Rb , assuming 5v Vb you get about 2.15mA, not 0.5mA

Alex
 
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Be careful relying on base current multiplied by hfe to limit collector current, assuming that is what you are trying to do. hfe is very variable and cannot be assumed to be any specific value, simply a range. Also, watch out for power dissipation - that is an equally likely reason to kill the transistor.

Keith
 

keith1200rs said:
Be careful relying on base current multiplied by hfe to limit collector current, assuming that is what you are trying to do. hfe is very variable and cannot be assumed to be any specific value, simply a range. Also, watch out for power dissipation - that is an equally likely reason to kill the transistor.

Keith
Click to expand...

yep, this was mostly for understanding the behavior. What I am trying to do is driving a common collector while limiting current with a resistor on the emitter, see my thread:

https://www.edaboard.com/threads/207139/

---------- Post added at 20:59 ---------- Previous post was at 19:02 ----------

Indeed the question is whether I need a resistor on the base of the PNP transistor here:

**broken link removed**

---------- Post added at 21:03 ---------- Previous post was at 20:59 ----------
alexan_e said:
when Re=0 and Rb=2k the base current is (Vb-Ve-0.7)/Rb , assuming 5v Vb you get about 2.15mA, not 0.5mA
Click to expand...

Right, I should have said 10k ;).
 

You wouldn't normally bother with a resistor on the base of an emitter follower. The variability of the hfe makes base current a poor means of trying to limit collector current.

Keith
 

thanks keith!
So does my resistor on the emitter limit both collector current and base current?

thanks

Davide
 

the emitter current when you use an emitter resistor is (Vb-Ve-0.7)/Re, the collector current is always lower than that, it will be Icollector = Iemitter - Ibase, Ibase depends on the hfe

Alex
 

and is it ever possible that hfe becomes smaller than the minimum values indicated in the data sheets?
 

I assume that the datasheet minimum is the minimum you can find in components but there are many different brands of the same transistor, maybe the minimums are not the same for all of them.

When you have an emitter resistor you don't care about that (assuming that your source can provide the base current), the base current will be as much as needed to give the collector current set by the emitter resistor.

Alex
 

csdave said:
thanks keith!
So does my resistor on the emitter limit both collector current and base current?

thanks

Davide
Click to expand...


The emitter resistor is all that is required. As alexan_e has said, the base current will be as much as is needed to satisfy the collector current. You have a feedback look although it is not obvious. If a base voltage change would result in too much current it doesn't actually happen because the emitter also moves therefore keeping the base-emitter voltage low.

Keith.
 

alexan_e said:
When you have an emitter resistor you don't care about that (assuming that your source can provide the base current), the base current will be as much as needed to give the collector current set by the emitter resistor.
Click to expand...

That's why I was asking about hfe. I mean, in this case my source is actually a 74HCT00 (not LS as in the picture) and can provide up to 25mA, therefore an hfe smaller than 3 or 4 would indeed require too much current. What would happen in that case? burn the source or simply discharge the capacitor more slowly?

I guess this won't happen as the datasheet lists a minumum value of 100 :p
 

As you say, minimum hfe is 100 (although there may be circumstances where it could be lower) so hypothesising about what will happen if hfe is 3 is a bit pointless.

Keith
 

I agree,
I am simply wondering whether it is the available base current that determines the collector current, or whether the collector current required by the load drives the base current.
 

I think it is the same as trying to use a load that needs a current higher than the source can provide, the source voltage drops.
In the transistor case when the base tries to pull more current than the amount available then the base voltage will drop (because the driver voltage will drop) and that will change the current set by the emitter resistor (which is (Vb-Ve-0.7)/Re ) and this two factor will balance to a value that can be maintained .
I think it would be best to avoid a case like that because it will probably stress the driver stage to a current level higher than the recommended one in the datasheet.

Alex
 

Hi csdave,
when you study the working principle of a BJT you will see that it is a device with an output current controlled by the base-emitter voltage. And because the input node has an input resistance other than infinite, there will be a corresponding input current (base current). Because of the pn junction properties this current has a more or less fixed relation to the output current (this leads to the current gain hfe).

---------- Post added at 13:18 ---------- Previous post was at 13:11 ----------

Hi Alex,

can you please justify/verify the emitter current equation as given by you: (Vb-Ve-0.7)/Re . Thank you.
LvW
 

I was trying to understand what was wrong with the equation but you are correct, Ve is a wrong way to describe what I meant because I meant the voltage at Re (the end at the opposite side of the emitter).
For example when the base is connected to 5v and the Re connected to 0v then it is (5v-0v-0.7)/Re

Thank you
Alex
 

Firstly Tr has 3 mode of operation--cutoff, normal, saturation.
When Vbe<(0.6-0.7V) Tr is in cutoff thats mean no current will flow C-E path, Ic=0.
When Vce is nearly (0.2-0.3V) Tr is in saturation, that is no matter what Vbe has no effect on Ic.

Now Ic-max=(Vcc-Vce)/Rc
I think Ic-max= 100mA according data-sheet.

When Vbe>0.7V & Vce is nearly (0.6-0.7V)Tr is in normal mode.Now
Vcc-0.7V
Ib=--------- & Ic=hfe*Ib
Rb
in this mode Ic depends on Ib only.

Thank you.
 
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