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Heathkit IP-18 Circuit Explanation Help

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guitardenver1

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Just for fun, a friend sent this to me to help understand this circuit. But I can't seem to fully figure it out. Circuit is below, I added part designators to help discussion.


Capture.JPG


Here is what I think of the circuit, please correct anything that is wrong.

D1 makes a half wave rectifier and charges C5, to smooth it out to near DC, D2 regulates to 20V and R1 limits the current through D2. This 20V provides voltage to R3 voltage divider which sets the output voltage of the power supply.

R3 controls the current through the base of Q3, which in turn control the current through the darlington pair formed by Q3 and Q4, which controls the voltage of the power supply output by changing the voltage drop across Q4 Vce. But that would also control the current limit of the output.

My guess is that Q5 is the feedback in the system for current control. The voltage across R6 and R7 controls the current through Q5 base, which in turn controls the current through the darlington pair base. But that would also change the output voltage right?

Q1 JFET is in a constant current source configurations. I am not sure what is going on there.

D7 is there for protection if an outside voltage source of opposite polarity and D6 is there for protection against an outside voltage source of same polarity. Not sure to the current path in that scenario.

D4 and D5 create a full wave rectifier and C1 and C2 smooth it out to DC.

Can someone explain in detail the operations of the power supply. I'd love to learn it well.

Questions:
1) What is the purpose of D3?
2) What is the purpose of R5 and why was it placed in between C1 and C2 and not before or after?
3) Q1 JFET is in a current source configuration, what is the purpose of this?
4) Why is C4 there?
5) Why is the output of the power supply connected to the positive node of C4? Wouldn't that affect the current through Q2 base?
6) How does D6 protect the circuit?
7) Is R8 there for a minimum load requirement?
9) What is the purpose "External Programming" and how does it work.


Thanks for the help. I hope this circuit is simple enough to explain in detail but complicated enough to be fun.
 
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1. D3 provides a small voltage drop so the adjustment doesn't have a 'dead' area at one end.

2. It makes a crude extra filter to smooth the rectified AC a little better. It probably also has other functions - to limit current a little and to drop some of the excess voltage so the pass transistor doesn't have to dissipate as much heat.

3. It is a constant current source. It increases the gain of the feedback system by allowing the voltage across it to vary more than a resistor would allow.

4. Two reasons: firstly as the bottom part of the circuit is the reference supply, it keeps it's output clean and secondly, it take time to charge up through R2 so it give the circuit a 'soft start'. (output voltage rises gradually after switching on)

5. No, the voltage on Q2 base is derived from the output + voltage and the reference (-) voltage. C4 plays no part in that.

6. It protects the transistors from a reverse voltage situation that can occur when the power is turned off. Imaging for example that you used it to charge a battery then switched off, without D6 there would be more voltage on the emitter of Q4 than on it's collector. The diode ensures no more than about 0.7V can exist across the transistor by pulling the input up to close to the output voltage. In normal operation it doesn't conduct.

7. Yes, and also to ensure the voltage drops when R3 is reduced so it doesn't stay at whatever C3 is charged to if the load is light.

8. (whatever happened to 8?)

9. It just allows you to provide an alternative voltage selector (external potentiometer or power supply) instead of linking directly to the internal one.

My guess is that Q5 is the feedback in the system for current control. The voltage across R6 and R7 controls the current through Q5 base, which in turn controls the current through the darlington pair base. But that would also change the output voltage right?
Not exactly a feedback system, the voltage across R6 and R7 is proportional to the load current. The voltage depends on the value of R6. If it exceeds the B-E threshold of Q5 it makes it conduct and 'steal' the bias current provided from Q1 and reduces conductivity through Q3 and Q4. The voltage does indeed drop, it is supposed to so the current also drops.

Please note that purists will complain about this design because it has a number of deficiencies. It has no proper feedback to stabilize the output voltage and all the load current passes through a potentiometer. With little modification it could work far better. Given it's age (~1975 I would guess) it used easily available parts rather than ones best suited for the job.

Brian.
 

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