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  1. Help Under standing Schematic PSU

    Hi Islam, Thanks for letting me know I helped, that is thanks enough. Makes it worth the effort.
  2. Help Under standing Schematic PSU

    Hi Islam, I'm a new user here (just signed up to answer your question), so I hope this helps. There seems to be several things that need to be understood about the overall topology of this PSU to get to answering your questions The power supply has two main parts: 1) The control section with the GHI outputs from the transformer, generates +12, -6, and -15. The ground of the control section (center point, more or less) is connected to the Positive side of the output. See ground below C4, and on the far right of the page (has the number 15 near by). (I think these numbers scattered through the schematic are test point numbers). So all of the control circuitry (anything using +12, -6 and -15) Is referenced to this ground. What is a little surprising, is that means that as the power supply output goes up/down, the control ground is doing the same. Also on this signal, is one end of the each of the six Load-Balance/Current-Sense resistors (6 x 0.36 Ohms). 2) The second part is the power section, which includes the lower half of the transformer (ABCDEF), the relays that select between the 4 voltages you listed (15-20-30-50), the bridge rectifier D7..10. The Negative side of the bridge is directly connected to the negative side of the output. (the line at the bottom of the page with label 14.) The Positive side of this bridge rectifier goes to the 6 collectors of the 2N3055s, then the 6 resistors mentioned above, then to the Positive side of the output, which as described above is also the ground of the control electronics. Since the maximum current is 3 Amps, and it will be shared evenly between the 6 resistors (with very minor differences depending on the exact characteristics of the six 2N3055s), the maximum current for each resistor is 0.5 Amp. So the max voltage across these resistors is 0.18 Volts , which gives about 90 mW , so these resistors are grossly over-sized at 5W, and will run cold (unless they are mounted on the heatsink). This also means the voltage on the 6 emitters is 0.00 to 0.18, which is averaged by the six 100 Ohm resistors and then sensed by U2 which implements the current limit circuit. U3 implements voltage control, follow pin 2 to the right which goes to the ground symbol, which we have established is power supply's positive output, and pin 3 which is connected to a voltage divider (including the front panel control) which connects to the negative side of the power supply, marked with the label 14. So now to get to your question. Regardless of what the output voltage is, the six emitters are no more than 0.18 volts above the ground of the control circuits, which tracks the positive output of the supply. The bases of these transistors will be 0.6 to maybe 0.75 volts above that. So the base voltage is always in range of the control circuit driving it. For the same reason, the 2.5V reference is an offset from the control section ground, and the power section's positive output. Yes, the six resistors also carry the base current, but this is minimal, compared to the collector current, due to the gain of the transistors. At DC, 25 deg C, and collector current of 0.5 Amps, the gain is over 100. I just did a Science (see picture below) to make sure I wasn't lying: With 15 volts on the collector, I got the following results: Vbe Ib Ic Volts mA mA 0.5 .05 3 0.56 .15 22 0.6 .36 82 0.61 .5 122 0.63 .7 190 0.65 1.0 280 0.67 1.26 370 0.7 2.2 600 (Note, Vbe is using the voltmeter in the PSU, which is not very accurate, and the transistor heated up on the last few measurements, and needed to be cooled) Anyway, the measurements I made show a gain on my test 2N3055 of about 300 when the collector current is around 0.5 amps which is the high end of what is expected in this power supply, and would contribute about 0.3% of the emitter current. Thinking a bit further, it doesn't really matter since it is sensed along with the collector current, and it all goes out through the positive output of the supply. Cheers