250VOLT - 320 POWER SUPPLY FOR VACUUM TUBE EQUIPMENT TESTS


THE POWER SUPPLY UNDERGOING TESTS




CONTACT : COMMENTS AND QUESTIONS
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CURRENT LIMITED AND BUFFERED POWER SUPPLY

Discussion of Requirements Design and Performance



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   INTRODUCTION

  When applying voltage for the first time to a new piece of equipment employing vacuum tubes, it is convenient to have a current limited power supply.
Power supplies, including regulated supplies, using vacuum tube rectifiers and pass tubes are usually not current limited and depend entirely on fuses for protection.
This can be a source of great annoyance during debugging.
To minimise damage, the current limit should rapidly decrease should there be a direct short to earth ,say, by an electrolytic capacitor connected with the wrong polarity.
This can be accomplished by connecting incandescent lamps in series with the power supply output.
The resistance of the lamps rapidly increases and limits the fault current below the current limit.
The voltage of the regulated supply can be varied from 250 to 320 volts for testing.



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   SIMPLIFIED CIRCUIT

  The simplified circuit of the regulated supply is shown below.

    SIMPLIFIED CIRCUIT DESCRIPTION

  VOLTAGE REGULATION
  Q7 and Q8 provide loop amplification for the voltage control loop.
(30- 0.6) volts appears across R11 (27K) and injects Ia (1.089 Ma) into the base of Q8.
The current through the variable resistance R12 is V/R12 and must be equal to Ia at balance.
Therefore V = 1.089x 10-3 x R12.
The voltage of the output rail is 14.4 + V, and so the output voltage is controlled by the settimg of R12.
The dimensions of R12 are chosen to give an output voltage variable between 250 and 320 volts.
Vout = 14.4 + V = 14.4 + 1.089x10-3 x R
R = 218K Vout = 322V :: R = 282K Vout = 251V
The error amplifier controls the gate voltage of the output MOSFET.
This acts as the current generator Q2 supplying the load current.
The load is shunted by the output capacitor C1.

  LOOP STABILISATION
  There are two dominant poles in the feedback loop.
[1] The 100uF capacitor C1 and the load resistance form a loop stabilising pole on the negative real axis near the origin.
Unfortunately, this cannot be relied on to provide complete stabilisation for the following reasons:-
(1) The capacitor C1 will cease being a capacitor at some high frequency and go inductive.
(2) Most vacuum tube equipments have a large bypass capacitor across the B+ input, and this forms a series LCR circuit across the output capacitor of the regulated supply.
The major component of the inductance L is wiring between the units, and this is variable and not under control.
[2] A pole on the negative real axis formed by the the output resistance of the loop amplifier R9 (3.3K) and the input capacity of the FET Q3. This is somewhat reduced by the current sensing resistors in the emitter , but still quite large as shown in the graph opposite. A shelf transfer function produced by R8 and C2 makes the loop response virtually independent of the load and capacitor C1.
The shelf attenuation starts at 1KHz and ends at 33.9KHz.
The network reduces the output resistance of the amplifier to 97 ohms at high frequencies, and so moves the pole formed by the FET input capacity out along the negative real axis where it has little influence.

  CONSTANT CURRENT LOAD
  The pass transistor Q3 is unidirectional and can only increase the voltage on the output.
Normally, when negative voltage changes are required on the output, the current in Q3 decreases and the load current discharges the output capacitor C1.
During initial setup of output voltage there is no load and an internal load must be included to accomodate negative changes on the output.
A constant current load is desirable to give the same rate of change independent of the output voltage setting.
The value is a compromise between load dissipation and speed of response.
Q1 and Q2 provide a constant current load.
The system will settle when the voltage across R2 (56ohms) equals the base voltage of Q2 - 0.6volts.
The constant load current is then (0.6/56) = 10.7Ma.
The speed of response dV/dt = I/C = 10.7x10-3/(100 x 10-6) = 100.7V/sec. Acceptable.
The constant current load is returned to the power supply side of the current sensing resistor R1 so it does not register on the output current meter.

  CURRENT LIMITING
  All regulated power supplies used in test work should be current limited - both to protect the supply and the equipment under test.
The current limit with a short circuited output should be much lower than the current limit with full output voltage.
SHORT CIRCUIT CURRENT LIMIT
The output current develops a voltage across R3 and turns on Q6 which pulls the gate of the pass MOSFET in the negative direction so limiting the short circcuit current.
CURRENT LIMIT AT FULL VOLTAGE
R6 injects 1.089Ma into the base of Q5 turning it hard on and shorting the base drive to Q4 so it is off.
When the base voltage of Q5 rises to 205 volts the current through R7 equals 1.089Ma and Q5 is turned off.
The current through R5 then turns Q4 hard on placing R4 in parallel with the shunt R3 and so increasing the current limit.



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   FULL CIRCUIT

Full Circuit Diagram:-

 Note:
The five circuit blocks can be printed out and joined.

  NOTES

    TRANSFORMER AND RECTIFIER
   The C core transformers were salvaged from old TV sets , so a voltage doubler rectifier had to be used to get the required 400V DC output. Any transformer - rectifier combination which gives an output of approximately 400 volts may be used.
Thr 48 ohm resistor R1 limits the peak current on switching. It also reduces the dissipation in the rectifier and transformer.

    CURRENT LIMITING
   The voltage required to activate the current limiting transistor Q6 is 4.7 volts ----- 4 volts across the LED diodes and 0.7 volts on the base of the transistor.
The short circuited output current through R20 R21 R22 R23 (100 ohms ) in parallel (25) ohms must produce 4.7 volts.
The total short circuit output current is therefore 4.7/25 = 0.188 amps.
For output voltages greater than 205 volts the diode resistance networks D6 R16 D7 R17 D8 R18 D9 R19 are parelled with the current sensing resistors R20 R21 R22 R23.
Allowing for the diode drop, an extra current of 4 x (4/39) = 0.41 amps is added to the current limit giving a total of 0.6amps.


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  BOTTOM VIEW


  EXPANDED CIRCUIT


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