# Simplest possible 5V to +/- 10-15V converter (for op-amps)

Im looking for simplest possible solution to power op-amps from USB or phone charger.

Requirements:

• +5V input voltage
• ±10V to ±15V output voltage
• at least 10mA output current on both rails (+/-)
• very low cost
• low part count
• simple and accessible parts (no specialized ICs)
• low noise and ripple would be nice (small and cheap linear regulators at output allowed)

Is this possible without a transformer, when I need no separation and I don't care about GND shift relatively to USB GND?

• If you could manage with +/- 10V then you could cheat with a MAX232 chip... Not sure what current you can get though. – Majenko Dec 31 '14 at 15:01
• I'm OK with MAX232. I just don't want to use relatively expensive inverter ICs, I want to use parts that are easy to buy in bad/small/local electronics shops :) – Kamil Dec 31 '14 at 15:14
• I assume you've looked at DC-DC converters... ~$5 each for what you want. What is a very low cost? – George Herold Dec 31 '14 at 15:15 • @GeorgeHerold I was hoping that it can be done with some transistors, resistors, coils and capacitors which can be found in my workshop. I can buy 2VA 230V/15+15V transformer for 4.50USD, few diodes, capacitor and linear regulators and build very simple 15+15V/60mA power supply. I want something really cheap. – Kamil Dec 31 '14 at 15:21 • I suppose it's possible to create your own switch mode boost converter. Eg: vin to inductor to diode to out (+caps). The node between the inductor and diode is PWM switched (with a fet) to ground. PWM is from, say, 555. This will produce a larger vout which can be linear regulated. Um. Ok - here's an example: learn.adafruit.com/diy-boost-calc. The switched node can be tapped to also produce the inverted voltage via a capacitor-diode-diode charge pump. It's all very manual though... – carveone Dec 31 '14 at 19:57 ## 4 Answers (comments got a bit long!) It is generally possible to create your own switch mode boost converter with varying levels of functionality. Any boost converter switches an inductor to ground with a FET to produce a higher voltage waveform. This can be rectified and smoothed to produce the higher voltage. All chip type boost converters take this output and feed it back so they can adjust the PWM switching waveform to keep the output stable. The following circuit is take from an Adafruit note and shows the concept: The adafruit note isn't concerned with absolute output stability, only a range of output voltages at a pretty static output current and input voltage. It has no feedback so is only approximate. However, with a small range of output currents, a linear regulator can take up the slack here. Inversion can be done by utilising the higher voltage PWM waveform, present at the inductor-fet-diode node and feeding it to a capacitor-diode charge pump. The wave at the node is put through a capacitor. The other side of the capacitor is diode clamped to ground, thus shifting the waveform negative. This image, pulled somewhat randomly from google images, shows the idea: In this case, this circuit is generating gnd to +Vcc transitions which are then inverted by the capacitor and two diodes. The output is smoothed and then zener regulated. The combined concept (ie: the PWM input to the FET is "someone else's problem!) might look like this: For opamp use, it would be recommended to filter the switching waveform through a PI inductor filter before feeding it to the linear regulators. Edit: I asked a question about this inversion tactic on stackexchange a short while ago which may also be relevant: Inverting charge pump with boost converter loading characteristics Take a look at the LT1930. On page 8 of that datasheet is a 5V to +/-15V converter using a single inductor. I've built one and it works quite well. Not exactly "built from stuff already in my workshop", but you can buy the necessary bits and pieces for about$10 from Digikey and it uses about 1.5sqcm of PCB space.

• That's a boost converter with the inverting charge pump tacked on after the inductor. I wonder what D2 and C4 help with in this instance. In my case I found that if I didn't pull current from the +ve line, the -ve line would just collapse to Vin as soon as I drew current. Not unexpected of couse due to the feedback line being solely from the +ve rail! Unless you like to reinvent things, I do agree with this style of solution. Even though I find Linear a bit pricey here (from Farnell anyway!). – carveone Jan 1 '15 at 13:40
• Looks great and not that expensive as you wrote. – Kamil Jan 1 '15 at 18:46
• Good find! That's pretty neat chip. – Cloud Jan 1 '15 at 21:48

The MAX743 chips (datasheet) will do the job, and are low cost, at about $13.00 per chip at the moment on Digikey. If you want cheaper than that, you'll still end up spending$5.00 per chip anyways, and will have to buy a pair of separate DC-DC switching ICs plus external circuitry anyway. This chip is specifically designed to convert +5VDC to ±12/15VDC, so it's an all-in-one solution, plus the external circuitry needed to run it. As an added bonus, they come in SPDIP 16 pin packages, so if you're a hobbyist without your own CNC or toner-transfer/etching capabilities, you can test it out on a solderless breadboard.

Also, consider the Project Management Triangle: Fast, good, cheap: Pick any two.

• +5V input voltage
• Requirements met
• ±10V to ±15V output voltage
• Requirements met
• at least 10mA output current on both rails (+/-)
• Requirements met
• 100mA generated per rail in 15V mode, 125mA in 12V mode
• very low cost
• Requirements met
• For what you're getting and what the normal is, this is considered low cost, unless you want to manually design your own switching dual-output symmetric supply from discrete components
• low part count
• Requirements met
• One chip
• simple and accessible parts (no specialized ICs)
• Defined "specialized"; It's a common chip from a company that won't be going out of business any time soon
• low noise and riple would be nice (small and cheap linear regulators at output allowed)
• Requirements met
• Good line regulation (0.05 %/%) and load regulation (1%)
• Well... Looks like perfect solution, but that price will triple my device cost. For that price I can make or buy power supply with two rails. – Kamil Dec 31 '14 at 19:43
• @Kamil with probably one-tenth of the PCB real estate used. You get the option to create your own system from scratch. I assume the time spent by the design engineer is finite and costs money. – Cloud Jan 1 '15 at 3:23
• @Kamil with extra risk, design time, time-to-completion/lead-time. I'm sceptical it will increase your device cost based on PCB fab times alone. – Cloud Jan 1 '15 at 3:32
• But I need 2 pieces for 2 rails... Right? – Kamil Jan 1 '15 at 18:47
• @Kamil No, just a single chip is needed (ie: Dual output: one is step-up/boost (+5VDC --> +12VDC), and the other is inverting (+5VDC --> -12VDC). Also, other comments have posted similar solutions which are also viable. Good luck! – Cloud Jan 1 '15 at 21:45

Check out the LT3463 - very small integrated solution