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Bias Calculations (Bootstrapped?)

W

Warren

Jan 1, 1970
0
Upstream I was introduced to this 1968 PE article. In addition
to the phase shift osc used there, I found the bias
arrangement used in that project interesting.

Figure 5 and elsewhere in the same project, there is an an
interesting biasing arrangement between Q3 and Q4 (Fig 5). Or
Q1/Q2, which just adds a fuzz gain adjustment.

http://www.swtpc.com/mholley/PopularElectronics/Apr1968/PE_Apr
_1968_pg46.jpg

( http://tinyurl.com/64xybq2 )

Using Q3/Q4:

Basically it looks like the current is split between R19=10k
and R16=22k + R15=10k.

In parallel that amounts to Rp=10k||32k => 7.6k.

Supply is +60V.

Assuming the base of Q4 is approx 1/2 of supply, say 30V, then
it's emitter is at about 30-0.6 = 29.4V.

Then Iq4 = 29.4V / 7.6k => 3.9 mA

Due to voltage divider, Ir15 = 29.4v/32k => 0.92 mA

Vr15 => R15 * 0.92mA => 9.2V

Since the base current in R14 is very small,
then Vbe3 ~= Vr15 = 9.2v.

But if Vb3 = 9.2V then Vr18 = 9.2 - 0.6 => 8.6v

Iq3 => 8.6v / R18 = 8.6v / 10k => 0.86 mA

Vr17 = 47k * Iq3 => +40 volts

So my crude calculations are off: I had assumed the voltage at
Q4 to be about +30v, but when I do the rough calculations I
get +40v instead. The answer probably lies between.

I'd like to know if this is a form of "bootstrapping", since
the bias is determined by the 2nd stage.

Warren
 
P

Phil Allison

Jan 1, 1970
0
"Warren"
( http://tinyurl.com/64xybq2 )

Using Q3/Q4:

I'd like to know if this is a form of "bootstrapping", since
the bias is determined by the 2nd stage.

** Bootstrapping involves signals.

C12 removes any signal from the game leaving only the DC bias.


..... Phil
 
W

Warren

Jan 1, 1970
0
Phil Allison expounded in @mid.individual.net:
"Warren"


** Bootstrapping involves signals.

C12 removes any signal from the game leaving only the DC bias.
.... Phil

You're right.

My next question then is why use this particular biasing
arrangement? I.e. why bother with negative DC feedback using
R16 (22k)?

Wouldn't a voltage divider to bias the base of Q3 be stable
enough? The input & output is capacitively coupled.

Warren
 
P

Phil Allison

Jan 1, 1970
0
"Warren"

Ok folks- I figured it out for myself.
Now I won't forget the lesson. :)

This arrangement allows you to configure your input
impedance, regardless of any voltage divider
bias requirement you might otherwise have used.

** The most important info is not supplied in your links - ie where the 60
volts DC comes from. I suspect it comes from the ( unregulated ) supply rail
of a power amplifier.

Using a simple divider across the rails of that pre-amp to bias the base of
a transistor means the output point will be bouncing up and down like a
yo-yo in operation as the 60 volt rail bounces up and down. Be nice of the
designer has used a zener to stop that.

The feedback bias arrangement used is largely insensitive to variations in
the DC supply rail and also to variations in transistor beta.


..... Phil
 
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