DC-DC Polarity Inverter




Introduction: DC-DC Polarity Inverter

About: Music: my profession for over 40 years... Electronics: my beloved hobby always.

This circuit generates a negative polarity voltage from a positive polarity one. This is useful for OP-amp circuits and low power audio amps where you need simultaneous +V and -V supplies from a single voltage source as a battery. Maximum input voltage is 18V, and output is up to 10W at 1 Amp. 555 based polarity inverter circuits can only provide a few milliamps of current output.

Step 1: 10 Ohm / 2W Resistor Burning From 10Volts 1Amp Output (10W) of Inverter Circuit

Step 2: Circuit Diagram

Main blocks of circuit are an oscillator stage (10KHz), a complementary power FET driver, and a capacitive charge pump. FET gates are preceded by a delay circuit to avoid simultaneous ON state of both FETs. Time constant is provided by a 470K resistor and input stray capacitance of 4011 gates (around 6pF). A 1N4148 diode cuts off FET with no delay. Schottky diodes in the charge pump provide fast switching and low forward voltage drop.

Step 3: Circuit Operation Details

Step 4: Charge Pump Operation

Step 5: Charge Pump Animation

Step 6: Output Ripple

Output ripple amplitude varies with load current and has a frequency of 10KHz

Step 7: FET Pinout

Step 8: Parts List

Step 9: For More Details, Watch the Video!

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    4 Discussions


    2 years ago

    Hi. If I want to move outside audible spectrum and increase pump frequency four times higher, to around 40kHz. What beside the multivibrator speed would change? Would the protection delays still work? Should the 500uF caps be reduced? Would it work? Thanks.


    3 years ago

    I would be very careful with this circuit. It lacks many self-protection features of "real" switching power supplies.

    is susceptible to "shoot-through", where one MOSFET doesn't turn off
    completely before the other turns on. They short power to ground,
    potentially destroying themselves, or at least reducing the circuit's
    efficiency. Maxim Semi has a nice definition:


    There should be a dead-band between the two MOSFET gate
    signals, when both transistors are off. The circuit as is may work for
    some people, but variations in the MOSFET gate capacitance, threshold
    voltage, and logic gate drive strength may cause it to fail [maybe dramatically] for

    The circuit also lacks a lock-out circuit to prevent
    both transistors from both being turned on of the logic gates
    misbehave. They wouldn't misbehave once the circuit is powered up and running,
    but odd things can happen with turning the power supply on or off. The gate
    outputs may all jump high or low, even if only briefly, and again short
    power to ground.

    I'm sure a more experienced power supply designer will have additional warnings. Commercial switching power supplies and ICs have extensive protection circuitry. They're not included just for fun. ;-)

    Caveat emptor.


    Reply 3 years ago

    Hi! If you take your time to read the circuit operation details (step 3) you will see R3 and R4 (470K) together with the input capacitance of 4011 gates (2 x 6pF), delay the turn-on of the FETs to prevent "shoot through" and create the "dead band" you claim.


    3 years ago

    I think you should move this sentence to the start

    "555 based polarity inverter circuits can only provide a few milliamps of current output." It confused me and left me initially with the impression yours could only produce a few milliamps . Thanks I will give this a try as my digital oscilliscope needs 12V, 5 V and -12V . This will cover it from one source.