emitter follower resistor question

[caress]

when using an emitter follower like so at the end of a circuit to lower output impedance, i'm curious as to what effect the value of the emitter resistor has?  bear in mind that -V is ground.
i've just read a bunch of articles, but the math is slowing me down...

i tried a 10k in a circuit and it sounded ok, but i'm just wondering what difference it would make if i changed that value?

[PRR]

If you don't do math, you won't do circuit analysis.

Roughly: make the resistor smaller than your load. In guitar cord work, values from 47K to 4.7K are fine.

[alanlan]

It's a good question.  The first thing to say is that it doesn't really affect the AC output impedance very much, which is much lower than the value of the emitter resistor usually (due to the effect of negative feedback).  The value is selected to set the DC operating point i.e. the quiescent or DC operating current flowing through the transistor.  Why is this important?  It's important if you want the stage to have the ability to drive a given amount of AC current into a load.  This choice is, in turn, determined by what the stage is feeding.  So it depends on the circuit it is in.

To set the current, usually you would bias the base to a voltage roughly 0.6V higher than you want the emitter to sit at (because the emitter will sit 0.6V below the base as long as it is in the normal linear region of operation of the transistor (i.e. not cut off or saturated).  Then the quiescent DC operating current is the base voltage - 0.6V divided by the emitter resistor value.

i.e. (VB - 0.6 - (-Vsupply)) / RE

Vsupply would be 0V in a normal 9V batt application.

You would usually want the emitter (VE) to be sat at roughly half supply (4 to 5V for 9V application).

I hope this helps.

[guitarman89]

tweaking the value of the emitter resistor you can vary the bias of the bjt. In other words you can tweak (you must have properly biased the base voltage) input impedance and output impedance of the stage by tweaking that value. In terms of the impedance (AC analysis) the presence of a bipolar supply isn't critical. I'm agree with PRR, math is so important to understand ac analysis. I'm having Electronics I exam next tuesday and the experience of an AC analysis opens me a different world.
So try to read something about this in the bible of electronics: Sedra-Smith. It is a heavy monster book but it's very good for many not-understanding problems.
I hope this is helpful for you!

[Brymus]

tweaking the value of the emitter resistor you can vary the bias of the bjt. In other words you can tweak (you must have properly biased the base voltage) input impedance and output impedance of the stage by tweaking that value. In terms of the impedance (AC analysis) the presence of a bipolar supply isn't critical. I'm agree with PRR, math is so important to understand ac analysis. I'm having Electronics I exam next tuesday and the experience of an AC analysis opens me a different world.
So try to read something about this in the bible of electronics: Sedra-Smith. It is a heavy monster book but it's very good for many not-understanding problems.
I hope this is helpful for you!

@ guitar man 89
Sorry this is off topic...
I was listening to your band, and my wife wants to know what the first song on your bands my space page is ? (nulla)
Neither of us speak Italian,but she claims it is a cover and wants to know who the original band is what name is the song ?

BTW cool to see you guys rocking in Rome.

[guitarman89]

XDXD Thank you to have visit my myspace page!
It is an our song, we wrote it a few years ago. We don't think we have covered it XDXD. What song have you and your wife in your mind?

@caress and the other: sorry for the ot...

[caress]

I hope this helps.

it does. 
and PRR - i know, i know!

i'm directly coupling the base to a preceding stage's collector and the collector of the follower is linked directly to 9V.  this is on a single supply.
i tried some different values and they seemed to only have a negligible influence on the actual sound, but i'll take voltage readings and test it out driving another pedal.

[PRR]

> different values and they seemed to only have a negligible influence on the actual sound

As Alan says, you have to be able to drive the load. Since guitar-cord loads are usually E-Z, and 9V gives us some excess to play with, this is not a tough requirement.

Your power source has to be able to handle the buffer power demand.

220K EF reistor won't make hot guitar level into some guitar inputs.

100 ohms will run your battery flat quickly.

In between these extremes, no, it won't make any big difference, possibly "none".

> AC output impedance very much, which is much lower than the value of the emitter resistor usually

The emitter impedance is about 30mV/Ie. Taking Ie as 1mA, just 30 ohms. Taking 4V across the emitter resistor and 1mA, we have 4K. So the impedance could be 30||4K which is 30 for all practical purpose.

That's the small-signal impedance. If the load current approaches 1mA in the direction of transistor turning-OFF, then the 4K emitter resistor controls what happens. If we thought a 30 ohm output would "easily" drive a 300 ohm load, we are surprised. The standing 4V splits between 4K resistor and 300 ohm load to give 0.28V and then it clips.

True, we don't have 300 ohm loads on stage.

And in this case, the 0.28V clipping, while far short of the 2V-4V we might expect from a 9V supply, probably "is enough" for many guitar-cord signals.

In fact, if supply voltage and resistor drops are more than a part-volt, and signals are a part-volt or more, an emitter follower usually has "zero" small-signal impedance, and we design for large-signal limits (output drive verus power consumption).

Which is where "make the resistor smaller than your load" gets a workable answer without hard thinking. In a world of 47K-470K loads, 20K to 2K collector/emitter loads usualy work fine. 10K is certainly a good answer.

[alanlan]

I would recommend as a rule of thumb having the value of the emitter resistor about 1/10th of the load it feeds (assuming it is AC coupled into the next stage).  This isn't for source impedance reasons, it's because if the load is of the same order of magnitude as the emitter resistor, you start losing headroom on negative excursions i.e. the negative peaks of the signal output limit before they would do with a higher value load resistor.  It gets worse as the load resistance gets lower.

So, if your follow-on stage had an input resistance of say 100K, then use 10K at most for your emitter resistor.  Obviously, if the emitter resistor value is lower, the stage will consume more power.  This is one of the instances of there being more than meets the eye with transistor design.  There are always compromises to be made.

[PRR]

> emitter resistor about 1/10th of the load

A good guide. Maybe generous.

But... what IS that load? Has anybody compiled typical guitar-cord input impedances? Fender tube amp is 1Meg, but Ampeg had 3Meg, some early transistor amps ran 100K, some pedals go 10Meg and some go 70K maybe less.

And are we just talking guitar-cord work? Or do we sometimes use guitar-cord boxes in Line Level work? Line Inputs are often 22K and sometimes 10K.

Not just the boxes. 30 feet of cable gets down below 50K at the top of the guitar band.

So a useful design must start from "what is the LOAD?"

I dunno if Brian has a number for this. It looks like I picked a number "under 100K"; dunno why.

Also, as you point out: the "about 1/10th of the load" rule allows the negative swing to go 1/1.1 or 90% of the standing voltage. On a 9V supply with probable 4V across the resistor, that's 3.6V peak. Going into a guitar amp, we never need much over 1V to slam the amp senseless.

If we take "resistor similar to load" we only get half the standing voltage, but half of 4V is 2V which is still plenty for our purposes.

And we may have a power budget. Not so much on wart-power, but battery life is directly affected by power consumption. Being pessimistic with a 10K load, a 1K emitter resistor may suck 4mA, a 10K may suck 0.4mA. On battery, I like to see total box demand under 10mA, and less is best. 4mA is almost half the max target just in one stage.

Still in all, 9V pedals are not that critical. The suppy voltage is several times higher than max signal voltage, and the happy current is 100 times greater than our loads need.

At some point you have to go back to the Product Boss and ask: what's important? Output? Battery life? Cost? If "Darn the battery, slam the load!!", sure 1/10th load is a great plan. If the plan is to seal a watch battery inside, then some sharp-pencil figuring (and probable extra cost) is needed to get useful output with tolerable battery life.

After all that, 10K is looking like a very good dartboard answer.

[caress]

I dunno if Brian has a number for this. It looks like I picked a number "under 100K"; dunno why.

nope no number.  it's at the output so it's pretty unknown what it will drive... long cords, many other pedals, tube amp, mixer... who knows?
it was more of a specific, yet general question.   

And we may have a power budget. Not so much on wart-power, but battery life is directly affected by power consumption. Being pessimistic with a 10K load, a 1K emitter resistor may suck 4mA, a 10K may suck 0.4mA. On battery, I like to see total box demand under 10mA, and less is best. 4mA is almost half the max target just in one stage.

Still in all, 9V pedals are not that critical. The suppy voltage is several times higher than max signal voltage, and the happy current is 100 times greater than our loads need.

At some point you have to go back to the Product Boss and ask: what's important? Output? Battery life? Cost? If "Darn the battery, slam the load!!", sure 1/10th load is a great plan. If the plan is to seal a watch battery inside, then some sharp-pencil figuring (and probable extra cost) is needed to get useful output with tolerable battery life.

After all that, 10K is looking like a very good dartboard answer.

i'm not too concerned about batteries as i use power supplies for everything now, but it's certainly something to take into consideration.  having something work as intended while demanding less power is certainly not a bad thing...