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Help with how this PSU works. (With revised design)

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Evangelopoulos Panagiotis:
[EDIT: Check out the revised design at the bottom]

Greetings everyone,

I found this linear power supply in a dumpster a while back and only now I decided to reverse engineer it.
I did this mainly because I wanted to understand how it works and later maybe build my own.
I created the schematic of the thing but I can't understand how it really works. Can someone provide me with an explanation on how it works?
I'm familiar with all its components but I can't piece them together right.

Moreover, there is a 10uF that goes from the positive to mains earth (not shown in the schematic). What's its use?

Finally, what changes should I make if I need it to supply more power? The poor TIP3055 gets quite hot at 4A even with that huge ass heatsink.

EDIT: I forgot to mention that I have already repaired it a few times. The TIP3055 seems to get burned really easily. I just fixed it again and I was about to test a DC-DC converter when a weird spike caused the thing to get destroyed again... Is it a good idea to parallel a couple of TIP3055s?

Thank you in advance for your time and effort,
E.P.
 

duak:
Greetings EP,

Yes, you will need more TIP3055 transistors in parallel.  In the best case, each can dissipate 90 W.  Under short circuit conditions they will have to dissipate 4 A x 45 V = 180 W.  Two is the minimum and I would consider 3 or 4.  That heat sink looks small for the possible power dissipation but it might be possible to mount four TIP3055s by using both sides. 

There should be a 1K0 resistor between the emitter of the BD243C and the right hand side of the 0R33 resistors.  This helps reduce collector to base leakage current at high temperatures.

Each TIP3055 will require a 0R33 resistor between its emitter and the ammeter terminal to help share current between them.  Each TIP3055 should also have a 10R resistor in series with its base to prevent oscillation.

The current sense voltage developed across each 0R33 resistor should connect to the - input of op-amp A with its own 4K7 resistor.  The 4K7 connected to the + input of the op-amp can be replaced with a 1K0 to equalize the resistances the op-amp sees.   The C ADJ variable resistor will have to be adjusted to maintain a safe maximum current.

Evangelopoulos Panagiotis:

--- Quote from: duak on February 02, 2020, 02:50:44 am ---Greetings EP,

Yes, you will need more TIP3055 transistors in parallel.  In the best case, each can dissipate 90 W.  Under short circuit conditions they will have to dissipate 4 A x 45 V = 180 W.  Two is the minimum and I would consider 3 or 4.  That heat sink looks small for the possible power dissipation but it might be possible to mount four TIP3055s by using both sides. 

There should be a 1K0 resistor between the emitter of the BD243C and the right hand side of the 0R33 resistors.  This helps reduce collector to base leakage current at high temperatures.

Each TIP3055 will require a 0R33 resistor between its emitter and the ammeter terminal to help share current between them.  Each TIP3055 should also have a 10R resistor in series with its base to prevent oscillation.

The current sense voltage developed across each 0R33 resistor should connect to the - input of op-amp A with its own 4K7 resistor.  The 4K7 connected to the + input of the op-amp can be replaced with a 1K0 to equalize the resistances the op-amp sees.   The C ADJ variable resistor will have to be adjusted to maintain a safe maximum current.

--- End quote ---

Thank you duak for the tips. I shall try them as soon as I get my hands on some more TIP3055s. I ran out of them... Hopefully sharing the load across 4 of them will make it a bit more durable. I'm guessing the extra ballast resistors of 0.33Ω should be rated at 5W as well right?

aheid:
I drew the most important lines for the control circuit in the attached image.

The core regulation is provided by the pass transistor, used as a variable resistor. In order for the NPN pass transistor to conduct, the base must have a positive voltage compared to the emitter. Since the emitter is hooked directly to the positive output of the supply, that means the control circuit must have access to voltage above the positive output rail.

This has been solved by using the secondary windings on the transformer to provide an independent set of power rails, and then by connecting the "ground" of the secondary rails to the positive primary rail (purple in drawing). The LM7808 then provides a stable 8V above the positive output for the control circuit.

The control circuit consists of two main elements:

1) voltage control (red tones in drawing)
2) current control (cyan tones in drawing)

They both use opamps to drive the pass transistor through a diode-OR configuration. The opamp which has the lowest output voltage will be in control. Thus the control opamps can ever only actively reduce the output voltage. This means that even if the voltage controller raises its output voltage, in an effort to increase the output voltage, the current controller can override this by keeping its output low. It also means that without either opamp "putting the breaks on", the pass transistor will be saturated, giving maximum output.

The current controller is relatively easy to figure out. It compares the voltage across a current sense resistor on the positive output with an adjustable reference voltage derived from the control circuit rails. If the current sense voltage (- input ) is higher than the reference voltage (+ input ), the opamp will pull its output down. This will reduce the voltage of the pass transistor base (in darlington configuration), increasing the resistance of the transistor and thus reducing the voltage on the positive output. This setup works because the control circuit ground is tied to the low side of the current sense resistor.

The voltage controller works similar except it compares the output voltage, but does so in a slightly convoluted way. The key here is that the negative output terminal has a negative voltage compared to the control circuit ground. The negative output voltage is mixed with the 8V rail (used as a voltage reference), such that the sum (red blob in drawing) is zero volts with respect to the control circuit ground (~= positive output terminal).

Note that the control circuit ground is buffered by the 220uF capacitor (far right), which means small changes in the positive output terminal will not affect the control circuit ground. Thus for small changes of the output voltage, the 8V and the control circuit ground will essentially stay fixed, thus only the feedback from the negative output terminal will affect the opamp.

If the negative output terminal is not negative enough (ie output voltage is too low), then the sum at the blob will be above zero volts, and thus input to the + input on the opamp will be higher than the - input and the opamp output voltage will go up. This "releases the breaks" of the voltage controller and, unless the current controller overrides it, will drive the pass transistor harder, reducing the resistance in the transistor, thus increasing output voltage. Similarly if the output voltage is too high, the negative output terminal will be too negative, causing the + input to be lower than the - input, driving the opamp output low.

At least that's my take on it, modulo typos and brainfarts :)

xavier60:
Or use TIP35C instead. Beware of fakes.
A problem with the manual tap changing is the very high dissipation when the output is shorted while in high range.
And also the hazard of having  secondary wiring near live wiring at the switch.

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