What does this cap do in the AMZ Mini boost

[rosssurf]

I was wondering what effect the 3.3 non polarized cap in the AMZ booster does and how other values like 2.2 and 4.7 might affect the sound or performance of this circuit. Thanks you

http://www.muzique.com/amz/mini.htm

[peterg]

The second last paragraph should help:
http://www.muzique.com/tech/mini-a.htm

[rosssurf]

thank you peterg. Sorry I missed that.

[GibsonGM]

Still didn't say *what* it does, tho... (?)  it's bypassing AC, that's for sure, but to what end - to stabilize the next Q's drain?

[PRR]

We are asking the top transistor to be a CONSTANT current source.

To do this, the Gate-Source voltage must be held constant for audio, but allowed to shift to suit the transistor and the supply voltage.

To keep the voltage very-very constant, the cap reactance must be much-much less than the R2||R3 network. While we might use 0.02uFd to mostly-pass audio into 500K, we need more to keep losses very low (and current nearly constant). Hence a value near 1uFd.

Caps as big as 1uFd tend to be large/costly in Film. A several-uFd Electrolytic is cheaper. However the leakage in cheap or old or very-large electros may be a problem.

There is no need for it to be NP. A polar electro can go with the + side to the Q1-S pin, the - side to Q1-G.

To reduce cap cost you can adopt the values in the "National appnote". 10Meg gate resistors and 0.1uFd cap. All modern (post 1963!) JFETs can take 10Meg gate resistors; however not all experimenters stock 10Meg.

[GibsonGM]

Thanks Paul!  It was a little confounding....

[R.G.]

From what I've read about the SRPP stage that the mu-amp is a variation of, that upper device is not a constant current source - it's a very curiously driven push-pull stage.

See http://geofex.com/Article_Folders/modmuamp/modmuamp.htm for some looks at the circuit and how it can be made more tolerant of things like loading.

[amptramp]

With an SRPP (shunt regulated push pull) stage, the gate of the upper transistor is returned to the drain of the bottom transistor and the source of the upper transistor is connected to the drain of the lower transistor by a resistor which should be the reciprocal of the transconductance of the upper stage to get equal swing from the push and pull stages.  A load resistance is required to consume the difference in current between upper and lower stages.  An infinite load will result in generation of a square wave.

The upper transistor in the Muzique article is AC-coupled from gate to source making it a current source.  It is fairly linear and can use any load down to the AC voltage divided by the current.

[R.G.]

As you wish... 

The mu amp is a special case of the SRPP, depending on how you look at it. Or the SRPP is a special case of the mu amp.   

The cap is there to force the AC signal voltage on the upper gate to be the same as the AC signal voltage voltage on the lower drain. How the source interacts with the drain varies with the impedance between them.  In the messing I did with the circuit, running the resistor down does indeed force the output impedance up - which I thought was only useful in some limited circumstances. By changing the milliohms of wire resistance between the source and drain upwards, you get useful and interesting changes in the output impedance.

But it's all in how you look at it. 

[PRR]

> curiously driven push-pull stage.

When load current flows in a resistor not present in the linked plans (Jack's and the National appnote).

As shown, the upper FET shows its full pentode-like plate impedance to the lower FET (plus the two 1meg resistors).

Point to note: the AC gain is ~~500, but the DC gain is ~~1, so the active node biases-up essentially where you put it. A stage where DC gain is huge will usually slam one rail or the other. But this is also why you need a relatively huge capacitor: so your bass falls more near ~~500 than near ~~1.

One thing never mentioned in casual treatments of the Mu-follower: you don't get sky-high gain unless the external load is nearly-infinite. For any real-world purpose, you really need three devices: the gain-amp, the upper load, and a buffer. If you instead deploy three devices as cascade common cathode/source amplifiers, the total gain into a practical load is far higher than the Mu-follower: Mu about 500, 3-stage about 8,000 (20^3).