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Power Supply

  1. Mar 5, 2005 #1
    Hi guys! Need some help. Attached is a circuit that I designed. The Vin is 7.2V and the Vout (at the load) is 5V. Basically, thats exactly what I wanted!!

    Anyway - can someone please help me with the understanding of the circuit. I know the basics but I'm a bit stumped on the use of the op-amp (I think it's used for smoothening out the voltage or something) and the configuration of the transistors. I know that the BJTs are used for current limitation but the way they are connected here doesn't make much sense.

    Also can someone suggest another type of BJT to use coz I blew the 3904s the first time around!

    Thanks :smile:

    Attached Files:

  2. jcsd
  3. Mar 6, 2005 #2
    I assume the base of the top 3904 is connected to the collector of the other transistor. The op-amp is an error amplifier. Normally you see negative feedback in a linear op-amp circuit. It may appear to you that there is no negative feedback in the power supply but look closely. The zener diode sets a reference voltage. Then, the op-amp 'compares' the two inputs and sets the output in the appropriate direction until the two inputs are VERY close together in voltage. The top 3904 is the pass transistor. It is driven by the output of the op-amp through the series resistor. The other transistor is turned off because its base-emitter voltage has not been developed by enough current flowing through R3. So its collector appears as an open circuit. A fraction of the output voltage is fed back to the op-amp for 'comparison'. That is the basic feedback loop. When too much current is drawn by the load, R3 develops enough of a voltage drop to turn on the bottom 3904. The collector of the bottom 3904 now shunting current away from the top 3904 and it is not allowed to pass as much current. This is the over-current protection scheme.

    I use the term comparison in this post loosely. In no way is the op-amp being used as a comparator. I just use that term as an attempt to give you a clearer idea of what is going on.
  4. Mar 7, 2005 #3
    Thanks for that! I've attached the updated circuit here. NOTE: on the circuit, a 3.25 resistor is used - it is actually a 4.8 resistor.

    Anyway - any suggestions on a different transistor to use? I'm going to mount the circuit on PCB so I don't want components blowing all the time! MJE520s were suggested but they dropped the output boltage a bit too much. Any ideas?? :confused:

    Attached Files:

  5. Mar 7, 2005 #4
    Concerning the pass transistor: Determine the maximum wattage that the pass transistor will dissipate. If you want to be able to short the output terminals and not have anything blow up you will dissipate 7 volts x about .15 amps or about a watt. .15 amps is the most that the overcurrent protection will allow to pass. But, if you don't have a transistor that is rated at a minimum of 1 watt then you will blow it up if you short the supply output. Obviously you will need a transistor capable of passing .15 amps regardless of whether you want it to hold up under short circuit conditions or not. If you want to make the supply start shutting down at a lower current increase the value of the 4.8 ohm resistor.
  6. Mar 11, 2005 #5
    Thanks for that. The max wattage for the 3904 is 625mW and at the moment it's handling the circuit - only it is heating up under short circuit conditions. I think I'll keep it in though what about using a transistor with a heat sink - just for safety sake or will that be impractical cost wise (I'm trying to keep the circuit cost down to a minimum - after all this is engineering :smile: )
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