Abbie said:
The fact that the minus input of the op-amp is
connected to the ground confused me at first, but then I realised that since
ground is also psu red o.p. means that there is a negative feedback from
the o.p. of the op-amp, through all the transistors, and back to the minus
i.p. of the op-amp. Therefore from op-amp basic principles both minus
and plus inputs are equal, and are equal to gnd, i.e. 0V.
That is one of the basic simplifying assumptions of linear opamp
circuits.
This means that the voltage at VDIAL (top) is 0V. The voltage at Zener 431
is 2.5V. Hence the voltage across the 910 ohms resistor is always 2.5V
(assume the variable resistor below the 910 one is not there).
Now, constant voltage, so we can know the current through the 910 resistor.
This current must go through VDIAL, so the voltage of the negative rail will
be VDIAL times this current. That's it. By varying VDIAL we vary our
negative rail.
Right. The voltage divider between the negative output and the +2.5
volt reference must always produce 0 volts. You adjust the negative
output with respect to this internal zero volts by changing one of the
divider resistors.
This circuit was difficult for me to understand because all the circuits I have
seen up to now regulate the positive rail, whereas here we are regulating
the neative rail.
It's the same thing, but it is like looking at something upside down.
Yes. you have to be flexible.
I don't think so, but tell me what you mean or why.
An opamp has very high gain, and if the feedback loop has enough gain
at the frequency where the net phase shift turns negative feedback
into positive feedback is higher than 1, the loop will oscillate.
The capacitor across VDIAL provides phase lead (increasing gain with
increasing frequency), so it cannot be rolling the gain off at high
frequencies to prevent oscillations. The opamp has an internal
capacitor that rolls of the gain proportional to frequency that gets
the gain below one at a lower frequency than the one that has 180
degrees of phase shift internal to the opamp, but it does not account
for the extra gain and phase shift of the three transistor amplifier
chain hooked onto the output of the opamp. I would have expected some
high frequency roll off or other phase compensation network somewhere
in the loop to reduce the high frequency loop gain.
By the way, this chain has a lot more phase shift than it could. The
turn on bias for the transistor chain is provided by the 10k pull up
resistor above the zener diode, and this is stolen by either the
voltage regulating opamp or the current regulation opamp (whichever
pulls down more) to turn off the transistors.
The first current gain stage lowers this signal source impedance from
10k to 1k, which is about right for a single stage of emitter
follower, but the second emitter follower should drive a base to
emitter resistance (across the pair of 2N3055) of something more like
100 ohms, not 1k, to speed up the turn off of those big transistors
when the upstream transistors turn off.