Dual rail power

Discussion in 'General Electronics Chat' started by JasonL, Jul 28, 2013.

  1. JasonL

    Thread Starter Active Member

    Jul 1, 2011
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    I got this schematic from the web and I want to know how it works. I included calculations I've done in the attachments.

    I want to know what mode of operation the transistors are in. I'm assuming they're in forward active, but I don't think there is any current going through the collector for either of the transistors because I don't think there is no current going through the base. So, I'm a bit confused.

    I also want to know how this schematic is better than using a voltage divider to get a dual rail power supply.
     
  2. tubeguy

    Well-Known Member

    Nov 3, 2012
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    You've got the right idea, but in the real world, diode Vf and transistor Vbe aren't necessarily exactly a 0.7 volt drop. Diode drop may be higher than that and the transistor drop may be lower. The 1 k resistors will compensate for the difference. So, the transistors may be biased on just a bit.

    The transistor circuit may be a bit better than the resistor only divider.
    A better rail splitter is one which uses an op-amp, because the op-amp circuit uses feed back to compensate for changes in current draw.
     
    Last edited: Jul 28, 2013
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  3. tubeguy

    Well-Known Member

    Nov 3, 2012
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  4. LDC3

    Active Member

    Apr 27, 2013
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    The voltage divider allows only a little current to flow through the middle potential. Most of it is through -Vcc still. The first circuit will allow more current through the middle.
     
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  5. Sensacell

    Senior Member

    Jun 19, 2012
    1,515
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    The transistor circuit shown would not be very 'stiff' - it's output impedance is about 500 ohms, any appreciable load will cause the output voltage to deviate.

    You are on the right track, you basically want to create a voltage follower with a low output impedance that can source and sink current. The circuit also needs to be stable when driving a large capacitive loads, as you will inevitably have bypass capacitance in the powered circuit.
     
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