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!

<p>I would be very careful with this circuit. It lacks many self-protection features of &quot;real&quot; switching power supplies.<br></p><p>It<br> is susceptible to &quot;shoot-through&quot;, where one MOSFET doesn't turn off <br>completely before the other turns on. They short power to ground, <br>potentially destroying themselves, or at least reducing the circuit's <br>efficiency. Maxim Semi has a nice definition: </p><p><a href="https://www.maximintegrated.com/en/glossary/definitions.mvp/term/Shoot-Through%20Current/gpk/1040" rel="nofollow">https://www.maximintegrated.com/en/glossary/defini...</a></p><p>There should be a dead-band between the two MOSFET gate <br>signals, when both transistors are off. The circuit as is may work for <br>some people, but variations in the MOSFET gate capacitance, threshold <br>voltage, and logic gate drive strength may cause it to fail [maybe dramatically] for <br>others. </p><p>The circuit also lacks a lock-out circuit to prevent <br>both transistors from both being turned on of the logic gates <br>misbehave. They wouldn't misbehave once the circuit is powered up and running, <br>but odd things can happen with turning the power supply on or off. The gate<br> outputs may all jump high or low, even if only briefly, and again short<br> power to ground.<br></p><p>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. ;-)</p><p>Caveat emptor.</p>
<p>Hi! If you take your time to read the circuit operation details <em>(step 3)</em> 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 <em>&quot;shoot through&quot;</em> and create the <em>&quot;dead band&quot;</em> you claim.</p>
<p>I think you should move this sentence to the start</p><p>&quot;555 based polarity inverter circuits can only provide a few milliamps of current output.&quot; 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.</p>

About This Instructable




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