0
\$\begingroup\$

Please, can someone explain me how the transistors here reduce the voltage from +40 VDC to +15V in the positive side (upper part of circuit)

circuit schematic

These are the voltages observed across different components of the power supply card.

O/P Positive side : + 15.40
O/P Negative side : - 13.79

R1   0.525 VDC           R2     0.445 VDC  
C1   44.6                C2    44.8 
R3   14.40               R6    18.16 
R4    9.57               R5    12.05 
C3   29.40               C4    26.94 
ZD1   5.08               ZD2    5.20 
R9   10.28               R10    8.48 
R11   9.70               R14    7.89 
R12   5.65               R13    5.82  
R7    4.71               R8     0.536  

There is a striking difference between voltage drops across R7 & R8 which are current limiting resistors. Is R7 responcible for TR1 heating up ?

And which is causing difference of O/P voltages on both sides ?

Power Supply Card

\$\endgroup\$
  • \$\begingroup\$ I would suggest there is a mistake in your diagram. You are showing a junction on the base of TR2 shorting our base and collector. It is not going to work as drawn. It would help if you used the circuit editor provided in the tools to put your diagram together rather than photos of diagrams then your intention is made absolutely clear. \$\endgroup\$ – RoyC Sep 13 '17 at 8:53
  • 1
    \$\begingroup\$ It is a (n inefficient) linear regulator. What is your actual question? RoyC, I assumed that was meant to be a mirror of the negative side, with an error. And Siva Prasad... Where in the world are you coming up with 150 Amps? You have nothing but small signal transistors there. \$\endgroup\$ – R Drast Sep 13 '17 at 8:56
  • \$\begingroup\$ Sorry, that is not actually junction...it is R9 resistor between collector and positive supply...please compare it with negative side of the circuit. Both halves are complimemtary circuits. Thank you \$\endgroup\$ – Siva Prasad Sep 13 '17 at 8:57
  • \$\begingroup\$ It is the actual circuit diagram provided in the OEM manual. Yesterday I trsted the card in the test jig, the output on the positive side is +15.45 and on the negative side is -13.87. But TR1 on the positive side is getting heavily heating up, i could not even touch it. And TR2 and zener are also comparitively warmer than that of Negative side ones. TR4 was just normal and not heating up. I did not understand why \$\endgroup\$ – Siva Prasad Sep 13 '17 at 9:00
  • \$\begingroup\$ @R Drast Actually we have a Float Charger 51.5V & 150A. It has a firing cards for SCRs and Amplifier card for output voltage correction (Regulation). There is power supply card which gives 15-0-15(-) output for the Amplifier card. This float charger is made by Forbes & Forbes and inducted here around 1972. We are still using this equipment for supplying 50V DC to communication eqpt. \$\endgroup\$ – Siva Prasad Sep 13 '17 at 9:09
0
\$\begingroup\$

enter image description here

Figure 1. The region of interest.

  • On power-up all transistors are off. \$ V_c \$ = 0.
  • As the input voltage rises \$ V_a \$ will rise and TR1 will begin to turn on. \$ V_b \$ will rise 0.7 of a volt "behind" \$ V_a \$ because TR1 is wired as an emitter-follower.
  • Similarly \$ V_c \$, the output voltage, will follow 0.7 V behind \$ V_b \$. Note that R7 limits the current in TR1 to a safe value.
  • As the output voltage rises R9 and ZD1 will form a voltage regulator holding TR2's emitter at the voltage determined by ZD1. Note that there is no junction at 'e' on the schematic. It is a cross-over.
  • As the output voltage rises further \$ V_f \$ rises and TR2 starts to turn on. This steals the bias for TR1 and holds \$ V_a \$ slightly above the Zener voltage - maybe a volt or so.

The output voltage is now stable. If it tends to rise TR2 turns on more pulling it down. If the voltage falls due to increasing load TR2 turns off a little allowing more bias into TR1 turning on the big fellow some more to raise the output voltage.

\$\endgroup\$
  • \$\begingroup\$ thank you for taking time & answering.... Can you please explain me further, how the voltage across C1 cap which is >+40VDC is reduced to +15 at the output, which part of that circuit is working to reduce the voltage? \$\endgroup\$ – Siva Prasad Sep 13 '17 at 11:05
  • \$\begingroup\$ I did. Do you understand what an emitter follower is? \$\endgroup\$ – Transistor Sep 13 '17 at 11:09
  • \$\begingroup\$ When I tested the card in the Test Jig,. TR1 is heating up heavily while TR4 is not. And TR2 and ZD1 are also warmer than TR3 and ZD2 (on negative side of the circuit). And the output on postive side was +15.45 VDC and on negative side it was -13.87 VDC which is less than the required 15V. We checked all the resistors and other components, they are all serviceble and correct rating. But still the problem is there. \$\endgroup\$ – Siva Prasad Sep 13 '17 at 11:11
  • \$\begingroup\$ Emitter follower is Common Collector Configuration where we take output from Collector and emitter. \$\endgroup\$ – Siva Prasad Sep 13 '17 at 11:13
  • \$\begingroup\$ You are forgetting to tell us what the load current is on both outputs during the measurements. What happens when the load is disconnected? \$\endgroup\$ – Transistor Sep 13 '17 at 11:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.