Transformer repairs

---snip some --- Hi Ade No, Q1 is on the secondary side of T3, not the primary. IC1 generates the pulse that go through T3 and cause Q1 to turn on for the short pulses. I'll try to walk you through how this particular supply works. ---snip-- some more It is unlikely that you'd see anything more than a small spike on a scope The T1 seems to have about a 100:1 ratio or more. Let me try to describe how this supply works. This should help you fix it. First, T2 provides power to IC1. IC1 monitors the 12 volts out through R14, VR2 and R12 network. If it is more than 12 volts, IC1 will sorten the pulse on time to T3. If the voltage is too low it will increase the pulse width. The other components around IC1 provide time constants for the pulse frequency and filter of response time of the regulation. IC1 also has a current monitor that looks at R11. If the voltage across R11 is too high it will turn on a circuit to stop the pulses to T3. Now, lets look at the high voltage side. As Tony has stated, you can't see much on a scope because of the fact that everything is following the ac input. Where you'd want to connect the ground would blow a fuse, unless you ran differential or as I suggested, use an isolation transformer. First the input AC goes through the filter network of L1 and some capacitors. It then goes through a full wave rectifier, causing a DC voltage to be developed on C7. You should measure this voltage with an ungrounded meter. You should see about 300volts across C7. If not, something is open in the input circuits. If, as you say, you see pulses on T3's primary side, it must be in the bridge rectifier is open. The voltages measured relative to ground have little meaning. T3's secondary is connected to the emitter and base of Q1. The pulse will cause Q1 to turn on for a short time. Since this is a short pulse,T1's secondary will see a voltage spike as well. This will forward bias the leg of D6 from the transformer to L2, charging C14. When the pulse is gone on T3's output, Q1 turns off. This turns the top leg of D6 off. Because some energy is now stored in L2, the bottom leg of D6 will now conduct, further charging C14. It is this voltage that feeds back to IC1 to change the pulse width going to T3 and on to Q1. This completes the 12 volt part of the supply. The 5 volt output is completely supplied from the 12 volt line. If there is no 12 volts, there is no 5 volts. The 12 volts powers IC2 that sets the regulation of the 5 volt output. It is directly connected to the transistors that switch the DC from the 12 volt line to the 5 volt. IC2 turns on Q4 and Q3 for a short amount of time. This causes current to flow in L5 and charge C21. When IC2 turns off Q4 and Q3, L5 will still have some stored energy, causing D7 bto conduct, further charging C21. IC2 monitors the voltage and adjust the pulse width to keep the output at 5V, in the same way as IC1 did. Besides testing the voltage across C7, as I suggested earlier ( being very careful not to electricute one self, you might also unsolder Q1 and see if it is shorted. This supply is also unique in that the 5 volt part runs from the 12 volt side. It would be possible to connect a 12 volt bench supply to the 12 volt output leads ( with the mains not connected ) and see if the 5 volt output works OK. That is all I have for now. Dwight _________________________________________________________________ Change the world with e-mail. Join the i?m Initiative from Microsoft. http://im.live.com/Messenger/IM/Join/Default.aspx?source=EML_WL_ChangeWorld