Mu-amp gain / output impedance

[Davelectro]

Hi all,

I've been wondering if it is possible to predict / calculate gain and output impedance for jfet mu-amps like the famous mini booster. As far as i know, these two things are hardly discussed. When it comes to jfet-based mu-amps people just say "their output impedance is quite high" or "they've got lots of gain" and that's it. So, could someone explain the subject to the rest of us?

Thanks in advance.

[R.G.]

Read GEOFEX.

See "Mod Your Mu-Amp" from 7/7/01  http://geofex.com/Article_Folders/modmuamp/modmuamp.htm
and Foolin' with FETs  http://www.geofex.com/article_folders/foolwfets/foolwfets.htm

Calculating the gain gets difficult. JFETs vary so much that it's nearly always simpler to build one or simulate it.

Technically, the point of the mu-amp circuit with triodes was to get a circuit with a gain equal to the tube mu. Mu in triodes is equal to the voltage gain of the tube when it's fed with a constant current on the plate. The gain then becomes the transconductance times the internal rp. The mu-amp circuit was an attempt to get the gain up to this.

In JFETs, the simplest thing to do is to either measure it on the devices you have, or to presume that a mu-amp circuit will actually get you up to that gain and then calcuate mu. For a JFET, you can calculate mu as mu =gm*rd, only they don't give you those numbers all the time, and they're ranges when you do get them. "gm" is the forward transconductance (output current change for an input voltage change) and rd is usually expressed as the inverse, 1/rd, and called the output admittance. So an estimate of mu for a JFET is mu = gm/Yfs.

And that's an estimate of the voltage gain for an UNLOADED mu-amp. The whole point of a mu-amp is that the amplifying device is working into a very high impedance. If you load it, the gain (which was gm*rd) is lowered by the real loading, and reduces to gm*Rd, which is the external load resistance. As a practical matter, if Rd is not much higher (i.e. 10x or more) than rd, the internal drain resistance (and that's not rds, the drain/source resistance) then you don't have a mu-amp. This pretty much says that real resistances are useless as loads on a mu-amp, and you need a semiconductor current source load.

The mu-amp *can't* do whatever gain it has into a real load. So you need to either buffer it with a high impedance input follower or self-buffer it by inserting the source resistor in the upper device that makes it into a SRPP amp.

With that as background, yes, you can write the equations for the gain of mu-amp made from JFETs, and calculate the gains and output impedance. In doing that, you have to take into account the variation in the specified values of the JFETs you're using, and if you want a generic answer, leave it in a form to plug in the actual JFET parameters, which vary. A lot. My eyes always used to glaze when I had to do an hour's worth of equation manipulation to get to a prediction. It's quite useful if you can get to a closed-form solution, but I didn't have the patience or fascination with algebra. S'probably why I didn't go into mathematics. Too impatient to dink with the equations for long enough. I always wanted to be designing something with explosive bolts in it instead. 8-)

Depending on the JFET, and whether you use a source resistor for the lower device or not, and whether that resistor is bypassed, you can get gains of several hundred from a mu-amp. You can't keep that gain if you load it with resistors. The gain drops back to gm*Rd. If you buffer the drain of the amplifying device, you can keep the gain. Which I interpret as saying that the output impedance of a mu-amp stage is of only academic interest, and might be the subject of some cheery evenings in front of the fire with a good pad and pencil, but I'd never use in a real design. In a real design, I'll buffer it every time.

But that's just me.

[Davelectro]

Thanks a lot for sharing your knowledge on this subject. Very useful post.

I guess I'll stick to SRPP.