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Heathkit IP-18 Circuit Explanation Help

Discussion in 'General Electronics Discussion' started by Mmm, Sep 16, 2017.

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  1. Mmm

    Mmm

    4
    0
    Sep 11, 2017
    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.


    upload_2017-9-16_14-20-5.png


    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.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,505
    2,852
    Jan 21, 2010
    The jfet Q1 supplies a current that is sufficient to turn on the Darlington pair of Q3/Q4 on sufficient to supply the maximum voltage at the maximum current. All the rest of the control circuitry is dedicated to shunting some of this away to maintain voltage regulation, any to override the voltage regulation in order to provide a current limit.

    D1 and associated circuitry create a voltage reference which can be monitored and overridden externally.

    The reference voltage has its positive rail in common with the output of the power supply. Because Vbe of a bjt is relatively constant, varying the voltage to the base of Q2 wrt the output effectively forces a voltage across Vce of that transistor that is sufficient to cause the output voltage to track the voltage at the wiper of R3.

    The output current flows through R6 and R7. When the voltage across them exceeds about 0.6V Q5 begins to turn on. It causes current to be shunted away from the base of Q3 sufficient to maintain the voltage across R6/R7 at 0.7V (ish) or lower. Normally you would expect the base current to be shunted to ground, but in this case the output voltage is about 2V lower than the voltage at the base of Q3, so it works just as well. Know that when Q5 begins to reduce the base current of Q3, the power supply moves from having a regulated voltage output to a regulated current output.

    D3 has a dual role. Firstly it adds an additional 0.6V drop to counter the Vbe of Q2. Secondly it provides protection to the base-emitter junction of Q2 when the power supply is deep in current limiting. In this case, the voltage at the wiper of R3 may be sufficiently negative with respect to the output ground that it could exceed the breakdown voltage (about -7V).

    R5 exists for cost reasons. Lots of this circuit is designed with cost in mind. Half wave regulation is used for the reference supply because it's cheaper (and presumably sufficient). A centre tapped transformer was clearly chosen over a bridge rectifier. R5 takes the place of a far more expensive inductor or larger filter capacitors. It reduces the rate at which voltage rises across C2 to limit the required frequency response of the regulator (this reducing ripple on the output).

    C4 provides additional filtering of the reference power supply, but is switched out of circuit when an external reference is used, presumably to make response faster.

    D6 protects the base emitter junction(s) of Q4 and possibly Q3 where a voltage at the output exceeds the unregulated per supply by 7V or more. This can happen if you use the power supply to charge a capacitor to 20V and then turn off the power supply.

    R8 is a minimum load.

    Remove the "AC" jumper, place a voltage source across the reference input and use R3 to control the proportional of the reference voltage which appears at the output of the power supply. Be VERY aware that the reference voltage shares a common positive rail, not the more conventional negative rail.
     
    Last edited: Sep 27, 2017
    CDRIVE and duke37 like this.
  3. JRUBIN

    JRUBIN

    96
    15
    Jul 17, 2015
    As it would turn out I just restored an IP-20 about 2 months ago. Interesting that one vacuum tube is still used as a voltage regulator.
     
  4. AnalogKid

    AnalogKid

    2,650
    768
    Jun 10, 2015
    First - EXCELLENT job of posting a detailed, researched question with documentation. **** for adding reference designators.

    Generally speaking, there are two basic types of linear voltage regulator topologies. a) the core regulator circuit is a fixed gain amplifier with a variable input voltage coming from a voltage reference of some kind going through a pot. b) the core regulator is a variable gain non-inverting amplifier with a fixed input voltage coming from a voltage reference. Identifying which type you have comes down to what is driving the pot.

    In your case, the pot is driven by the output, so this is a type b. Again generally speaking in terms of standard opamp circuits, the part of the pot from the output to the wiper is the series feedback resistor, and the part of the pot from the wiper to GND is the shunt feedback resistor. Of course there can be other things like fixed resistors in series with either leg to set limits on the adjustment range.

    Q2 Vbe in series with D3 is the system voltage reference, approx. 1.2 V, and Q2 is the error amplifier. The whole second winding with D2 and R2 isn't "standard" these days, but it does allow the system to adjust and regulate all the way down to 0 V out. The output voltage rises until the R3 wiper is 1.2 V above GND, at which point Q2 starts to conduct, shunting current (from the Q1 current source) away from the output transistors, preventing the output voltage from increasing any more.

    ak
     
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