It seems as if it is working from the VAS to the output stage but lets do some sanity
checks. Q305 and Q306 form a current source their purpose is to pull a constant
current at the collector of Q305. The current is simply the voltage you measure across
R 313 put into I = V/R. That determines the emitter current of Q305, since the base
currents are low we can and are ignoring current into the base of Q306.
Now, we also know the collector current of Q305 since it is a high beta transistor
Ie ~= Ic so it is just the Ie that you calculated.
Let's calculate the drop on the 4.7K resistor, from V - IR where I is the current source
value that we just calculated. Take that voltage and subtract it from the positive supply
and that should be the voltage on the collector of Q304. We are ignoring any current
into the bases of Q307 and 308 because base currents are small for high beta transistors.
You can substitute a transistor but it must have a Vce rating of about 110V or the rail to rail
voltage. Also, compute the power dissipated at idle as Icurrent source times the rail (single)
voltage. Ic minimally would be twice the current source, but 10X is probably safer in
overload. Often I like to see a 500 mA part there but when the output is highly buffered as
it is here you can get away with less.
I'd want to be sure that nothing is broken around the diff pair so that we can be sure the
new transistor will not blow. Any electrolytic caps around the diff pair, input and feedback
to ground should be tested and really, at this age, replaced. If you put a resistor equal in
value to R303 from the base of Q302 to ground and lift one end of R314 both diff pair
transistors will have the same DC path resistance to gound and they should be balanced.
Compute the current through R307, by measuring the voltage across it. Then compute
it through R304, half of the previous value should pass through R304 if the diff pair is
balanced and working properly. Q304 should still be removed for this test.
The balance will depend on the matching of the diff pair transistors, 20% is close enough.
checks. Q305 and Q306 form a current source their purpose is to pull a constant
current at the collector of Q305. The current is simply the voltage you measure across
R 313 put into I = V/R. That determines the emitter current of Q305, since the base
currents are low we can and are ignoring current into the base of Q306.
Now, we also know the collector current of Q305 since it is a high beta transistor
Ie ~= Ic so it is just the Ie that you calculated.
Let's calculate the drop on the 4.7K resistor, from V - IR where I is the current source
value that we just calculated. Take that voltage and subtract it from the positive supply
and that should be the voltage on the collector of Q304. We are ignoring any current
into the bases of Q307 and 308 because base currents are small for high beta transistors.
You can substitute a transistor but it must have a Vce rating of about 110V or the rail to rail
voltage. Also, compute the power dissipated at idle as Icurrent source times the rail (single)
voltage. Ic minimally would be twice the current source, but 10X is probably safer in
overload. Often I like to see a 500 mA part there but when the output is highly buffered as
it is here you can get away with less.
I'd want to be sure that nothing is broken around the diff pair so that we can be sure the
new transistor will not blow. Any electrolytic caps around the diff pair, input and feedback
to ground should be tested and really, at this age, replaced. If you put a resistor equal in
value to R303 from the base of Q302 to ground and lift one end of R314 both diff pair
transistors will have the same DC path resistance to gound and they should be balanced.
Compute the current through R307, by measuring the voltage across it. Then compute
it through R304, half of the previous value should pass through R304 if the diff pair is
balanced and working properly. Q304 should still be removed for this test.
The balance will depend on the matching of the diff pair transistors, 20% is close enough.