BJT Emitter Follower...with collector resistor???

Thread Starter

LordSte

Joined Aug 31, 2012
5
Hi, all. I've stumbled across a couple of schematics showing NPN Emitter Followers with a collector resistor. Usually it's rather low-valued such as 10Ω. Just wondering what this resistor does and why a circuit designer would include it. It's been bugging me for a while now, and I was hoping someone could clear this up for me. Thanks.
 

chuckey

Joined Jun 4, 2007
75
Either its part of a decoupling network (was there a Cap to earth from it?) or its to stop the transistor self destructing if the feed from the emitter is earthed accidently.
Frank
 

WBahn

Joined Mar 31, 2012
30,058
The most likely reason is to either limit the maximum collector current in the transistor or to reduce the power dissipation (probably the latter, primarily). Assuming the base current is small enough to ignore (for this discussion) so that we can assume that Ie~=Ic, when a current Ie is flowing in the emitter, the power that must be dissipated in the transistor is Ie*Vce. Vce is Vcc - Ve, so we have P=Ie(Vcc-Ve) and this can turn out to be a rather large value. By putting a resistor in the collector path, we drop some of that voltage across the resistor and the power dissipation is split between the resistor and the transistor.
 

Thread Starter

LordSte

Joined Aug 31, 2012
5
I'm seeing three distinct reasons:

1.) Power Supply Decoupling - This makes sense. I think I read in TI's Op Amp book that instead of using a ferrite bead you can use a low valued resistor instead. It's cheaper, but it affects the output swing. At least one of the circuits I've seen has a capacitor to ground right after the resistor (at the BJT collector node). So that's probably why they used it.

2.) Short Circuit Protection - If the emitter node is shorted to ground, too much current will flow through the BJT. It will dissipate too much power and get damaged/destroyed when it exceeds its maximum power dissipation capability. The collector resistor will attempt the prevent or curb this effect during an accidental short to ground.

3.) Power Dissipation - Since BJT parameters are sensitive to temperature, limiting the power dissipation during normal operation can be important in particular applications. Including a collector resistor shifts some of the burden of dissipation from the BJT and onto the resistor.

I think I have it right, but let me know if I missed something. Thanks guys.
 

Brownout

Joined Jan 10, 2012
2,390
None of the above. Typcially, the emitter resistor is simply to stablize quiescent current. It provides a little of negative feedback so that the transistor doesn't get into thermal runaway. It's not always required or used, and there are other reasons. That's the usual one, however.
 

WBahn

Joined Mar 31, 2012
30,058
None of the above. Typcially, the emitter resistor is simply to stablize quiescent current. It provides a little of negative feedback so that the transistor doesn't get into thermal runaway. It's not always required or used, and there are other reasons. That's the usual one, however.
He wasn't asking about the emitter resistor, but about the collector resistor.
 

Thread Starter

LordSte

Joined Aug 31, 2012
5
Please disregard my responses. Sorry.
No worries.
Actually, no, let's talk about the emitter resistor too. I was just playing around in LTSpice and do have questions on that subject as well. I have an NPN emitter follower driving a DC coupled 150Ω resistive load. I attached a picture of the schematic. I was increasing the value of R5 until I deleted it and the performance of the circuit increased (Vout was more ideal). Of course, this totally makes sense theoretically. Lower values of R5 in parallel with the 150Ω load would obviously lower the equivalent resistance. However, is there any practical reason to still include R5 such as negative feedback, thermal runaway, or something else like you described? Just wanna make sure, since I always have such a hard time finding these practical issues discussed in textbooks.
 

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WBahn

Joined Mar 31, 2012
30,058
The presence of R5 ensures that the circuit will see a maximum load of R5. If the real load is going to be variable, having this boundary established can be useful, particularly in more complicated circuits such as higher frequency and/or multi-stage amplifiers.
 

Thread Starter

LordSte

Joined Aug 31, 2012
5
If the load is always attached then yes, R5 is superfluous since it is swamped by the load.
And what happens if the load is not always attached? In other words, is it bad, for this particular circuit, to have the emitter node not connected to anything at all? Or is it a good idea to always throw in at least a high valued R5 to give the emitter follower some kind of load that's < ∞?
 

MKCheruvu

Joined Nov 20, 2012
30
If (R3+R4) > R5 and also if load is capcitively coupled , then R5 value becomes significant ,because it has to establish quicent operating current Ice(q) and also allow Collector leackage ICEO.
But here(R3+R4)<< R5 and also load is direct coupled R5 effect is negigible.
 

Brownout

Joined Jan 10, 2012
2,390
I agree with the above post except the load can be directly coupled and R5 remains significant. I've seen this in circuits where the load is another transistor, and R5 is needed to maintain a reasonable Q-point.
 

crutschow

Joined Mar 14, 2008
34,432
And what happens if the load is not always attached? In other words, is it bad, for this particular circuit, to have the emitter node not connected to anything at all? Or is it a good idea to always throw in at least a high valued R5 to give the emitter follower some kind of load that's < ∞?
If there is no emitter load than the transistor conducts no current. It doesn't hurt anything to operate without a load. But for it to operate as an emitter-follower it must have a DC resistive path to ground.
 
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