Emitter follower related question

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
Hello,

When biasing an emitter-follower with a current source in the place of a typical emitter resistor, can the equivalent emitter resistor be calculated by V-emitter / I-emitter?

I need this value in order to calculate input and output impedances.

Thanks!
 
AX tech editor
Joined 2002
Paid Member
Hello,

When biasing an emitter-follower with a current source in the place of a typical emitter resistor, can the equivalent emitter resistor be calculated by V-emitter / I-emitter?

I need this value in order to calculate input and output impedances.

Thanks!

The impedance from the current source can be calculated from delta-V/delta-I. That is much higher than Ve/Ie. So what you see from the base is that very high Isource impedance in parallel with the load, and most of the times it will be the load that dominates, meaning you can disregard the Isource.

The input impedance looking into the emitter is that Isource in parallel with whatever the impedance into the emitter itself is, and the latter will dominate, meaning you can disregard the Isource.

jd
 
Thanks for a quick answer janneman.

If I understand correctly, I can disregard the Isource impedance and simply use Rload in my calc.

How about for the base? I also have a current source in place of the base-to-ground resistor in the "voltage divider", how does that go?

ig
 
AX tech editor
Joined 2002
Paid Member
Thanks for a quick answer janneman.

If I understand correctly, I can disregard the Isource impedance and simply use Rload in my calc.

How about for the base? I also have a current source in place of the base-to-ground resistor in the "voltage divider", how does that go?

ig

Well in any case the Isource is (very) high impedance. So the impedance of the base divider is the Isource in parallel with the resistor from base to supply (remember that for ac and signal, the supply is considered ground).

But do you have a circuit diagram? I suspect what you are trying to do at the base will not be thermally stable and might need individual adjustment for each device.

jd
 
Here's what it looks like, don't look at component values, just topography.

ig
 

Attachments

  • Amp_V5.jpg
    Amp_V5.jpg
    29.8 KB · Views: 221
AX tech editor
Joined 2002
Paid Member
So for the input Z of the circuit, can I also "neglect" Isource and just use the +rail-to-base resistor || with Zin-base?

Thanks,

ig

Well, depends on the resistor values. If the Isource has 150k dynamic impedance and the other R is also 150k, clearly you can't ignore Isource: the parallel value is 75k.
If the R is 10k, the parallel value is about 9.3k. Are you familiar with calculating parallel impedances?

jd
 
Well, depends on the resistor values. If the Isource has 150k dynamic impedance and the other R is also 150k, clearly you can't ignore Isource: the parallel value is 75k.
If the R is 10k, the parallel value is about 9.3k. Are you familiar with calculating parallel impedances?

jd

Yeah, it's more how to translate a current source into an impedance that I was looking into here. I think I can see what you mean by "dynamic impedance" that should not be neglected if the lower Z it can take is not big enough to make the parallel impedance be close enough to the resistor above.

az
 
AX tech editor
Joined 2002
Paid Member
Yeah, it's more how to translate a current source into an impedance that I was looking into here. I think I can see what you mean by "dynamic impedance" that should not be neglected if the lower Z it can take is not big enough to make the parallel impedance be close enough to the resistor above.

az

An Isource can be considered as an impedance but you can't measure it by its voltage over its current. You measure it by varying the voltage a bit and see how the current varies, then the impedance (the dynamic impedance) is delta-V/delta-I. Ideally, the current doesn't vary with voltage and then the dynamic impedance is infinite...

jd
 
An Isource can be considered as an impedance but you can't measure it by its voltage over its current. You measure it by varying the voltage a bit and see how the current varies, then the impedance (the dynamic impedance) is delta-V/delta-I. Ideally, the current doesn't vary with voltage and then the dynamic impedance is infinite...

jd

Are we talking collector voltage and current here? I could simulate this by changing the value of the collector resistor, correct?

ig
 
AX tech editor
Joined 2002
Paid Member
Are we talking collector voltage and current here? I could simulate this by changing the value of the collector resistor, correct?

ig

If you sim your Q1 circuit from above, you can vary the voltage at the end of R2 & R3 (the supply) and measure the variation in Ic (you could measure VR4).

You can also supply R2 & R3 from different supplies, and keep R3 supply (that biases the cs) constant and vary the supply to R2 and see how taht influences the current. It will give you some idea of the impedance.

jd
 
I simulated a stand-alone current source loaded with a resistor.

I did it the way I described, by changing the load resistor, and the way you described, by varying supply voltage. In both cases, deltaV / deltaI were close enough, as my measuring ranges were a bit different.

I guess the thing when performing this is to use a delta that stays into the transistor's linear range, as things get out of whack at cutoff and saturation. Any recommended range of deltaV ?

az
 
Last edited:
AX tech editor
Joined 2002
Paid Member
I simulated a stand-alone current source loaded with a resistor.

I did it the way I described, by changing the load resistor, and the way you described, by varying supply voltage. In both cases, deltaV / deltaI were close enough, as my measuring ranges were a bit different.

I guess the thing when performing this is to use a delta that stays into the transistor's linear range, as things get out of whack at cutoff and saturation. Any recommended range of deltaV ?

az

For measurement or sim you just select a delta-V within the linear region because that is where the circuit is supposed to work.

jd
 
The variation is small. For a current source like Q1, the "effective impedance" looking into the collector is typically in the order of 1000 times the value of the emitter resister - i.e. in this case several tens of kilohms - probably about the same as R1 and also about the same as Q2's input impedance (assuming a current gain of a few thousand for the output stage).

Input impedance of the amp as a whole is a parallel combination of R1, R2, Q1's collector, and Q2's base. R2 will totally dominate and I'd expect an input impedance of about 1K4 overall.

I'd consider Q1's contribution to be negligible, but if it bugs you, you could always use a better current source like one of those in the attached gif. I'd vote for the one on the right.
 

Attachments

  • sources.GIF
    sources.GIF
    5.5 KB · Views: 108
The variation is small. For a current source like Q1, the "effective impedance" looking into the collector is typically in the order of 1000 times the value of the emitter resister - i.e. in this case several tens of kilohms - probably about the same as R1 and also about the same as Q2's input impedance (assuming a current gain of a few thousand for the output stage).

Input impedance of the amp as a whole is a parallel combination of R1, R2, Q1's collector, and Q2's base. R2 will totally dominate and I'd expect an input impedance of about 1K4 overall.

I'd consider Q1's contribution to be negligible, but if it bugs you, you could always use a better current source like one of those in the attached gif. I'd vote for the one on the right.

That second current source is certainly more elegant than using diodes. I'll redesign around that for sure.

Thanks for the advice and help guys!

ig
 
Hi IG

Sorry if this is a bit late but ....
The original circuit is going to have quite a bit of hum, mostly injected by the 1k5 bias resistor to the base of Q2, and also a little bit via Q2, Q4 and Q5 even if Q2's base voltage is clean.

An easy way to get super-clean results is just to use a positive-earth arrangement (see attached simplified schematic).

This way, there's no signal current flowing through the power supply, either (for those who care).

The capacitor I added to the bias network is kind of an optional extra - will help with the hum too, but mostly I'm thinking you might need something like that to tame the switch-on thump.

The monster output caps are nice but if you're not careful, your speakers are going to be doing some serious stretching exercises every time you switch on (not sure about switching off - you can have fun modeling that)

The other attached circuit is just how to hum-proof a current source with a zener. Tip for when you have free choice of zener voltage: those around 6V are the stiffest i.e. lowest delta V / delta I.

Right, time for me to shut up and let you get on with it :)
 

Attachments

  • ckt.GIF
    ckt.GIF
    3 KB · Views: 72
  • src.GIF
    src.GIF
    1.8 KB · Views: 76
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.