# Simple Voltage Inverter without using IC

Say I need to create a negative supply rail of -5V (+-0.5V) from a DC input of +5V without the use of any ICs whatsoever. The components available to me are BJTs, Mosfets, diodes, resistors and caps. So far my basic understanding is this -

Make an oscillator circuit (maybe astable multivibrator) to convert the positive DC into a 0-5V square wave and then use a voltage doubler to get a DC voltage on the negative side.

The problem is that my output is at around -3.5V. Maybe I should amplify the square wave before passing it to the next stage? Or ditch this method altogether? Suggestions would be appreciated.

• You must edit your question (not add a a comment) to explain what supplies are available? I'm guessing you're creating a negative supply rail but you're not telling anyone what they have to work with. Voltage rails with current capabilities, please (x V @ y A). – TonyM May 23 '17 at 7:18
• No inductors? That would make things a LOT simpler. It's the diode drops that are killing your output. To eliminate that, you'd need synchronous rectification, using your MOSFETs. But the drive circuits quickly get complicated. – Dave Tweed May 23 '17 at 11:33
• If you want to build a real circuit then select real components in the simulator (eg. 2N2222, BAT54). – Bruce Abbott May 23 '17 at 14:07
• Even if you used ideal diodes (without forward drop) you'd get -5V only if there was no load on the output. Solutions are either having multiple stages (see sstobe's solution) or using an inductor (see Dave Tweed's comment). – Curd May 23 '17 at 14:22

Your solution is quite close. However you can take advantage of have 2 low ohmic output phases of your oscillator and build a charge pump. The schematic would be as the following,

simulate this circuit – Schematic created using CircuitLab

Where PhiA and PhiB connected to each phase of your relaxation oscillator.

On the charge phase of either pumping cap you loose 1 diode drop, 0.7 V. On the discharge phase you loose 1 Vcesat plus 1 diode drop of 0.7V.

So for a 1 stage charge pump as you have shown, you will get under no load approximately -(5-0.7-0.7-0.2) ~ -3.4 V.

For the two stage charge pump its approximately double that at -6.8 V.

• Charge pump is another name for voltage multiplier circuit right? In this case I should be using a voltage quadrupler rather than a doubler - is that what you're saying? – Md Ajwaad Zaman Quashef May 23 '17 at 15:49
• It operates on the same principles as a voltage multiplier, but achieves higher efficiencies than a multiplier which just take a single input phase. en.wikipedia.org/wiki/Charge_pump – sstobbe May 23 '17 at 15:54