It is a classic problem when designing with op amps: you need the output to go to zero, but — for most op amps — you can’t quite get down to the supply rail. If your power options are a positive voltage and ground, you can’t get down to zero without a special kind of op amp which might not meet your needs. The best thing to do is provide a negative supply to the chip. Don’t have one? [Peter Demchenko] can help. He uses a simple two-transistor multivibrator along with some diodes and capacitors to generate a minimal negative voltage for this purpose.
The circuit is simple and only produces a small negative voltage. He mentions that into a 910 ohm load, he sees about -0.3V. Not much, but enough to get that op amp down to zero with a reasonable load. Unlike other circuits he’s used in the past, this one is efficient. With a 5-volt input, it draws less than 1.5 mA.
His previous circuit used LEDs and photocells, which led to low noise but wasn’t the best for efficiency. However, it is still a clever idea.
There are many ways to get a negative voltage like this. If you have an RS232 interface anyway, you might be able to swipe some from the port or your own driver, although it might be noisier than you want.
We’ve written extensively about “the negative rail” and how to get it. Of course, you can always throw a module at the problem. What’s your favorite way of getting a negative supply?
When designing op-amp circuits, you’re often using a chip with 2, 4, 6 etc. amplifiers on the same die when you only need 1,3,5… so you end up with an extra. Instead of using many parts to make an oscillator, it might be easier to build a relaxation oscillator charge pump using the spare op-amp.
I count three fewer parts. Not that much better, but still.
https://electronics.stackexchange.com/questions/210233/any-downside-to-using-a-charge-pump-for-an-op-amp-negative-supply
By winding a small transformer on a toroid, one can reduce the parts count to five: one transistor, one resistor one transformer, plus one diode and one capacitor.
That kind of oscillator is commonly known on the net as “joule thief”, and used to light LEDs using almost depleted batteries (hence the name), and they can easily converted to generate negative pulses that can charge a bigger capacitor through a diode.
Speaking of batteries, I’ve always wondered how efficient the Joule Thief circuit actually is. Obviously it’s useful if you want to run off of double AA batteries that dip below your white LED’s forward voltage at about half empty.
But if you have a third cell in series, you’ll have about 20% left by the point the LED starts to shut down. Is it better to pump that last 20% from the bottom of the battery, or do you lose more than that in the operation of the circuit?
The Joule Thief’s exact efficency depends on the ratio between the output voltage and the diode’s forward voltage, but “less than 50%” is a decent first guess. Don’t run a fully charged battery down using it, certainly.
That class of circuit is best used when you have no other choice.. energy harvesting, low-voltage solar, etc.. and the total power is so low the absolute loss is small even when the percentage loss is terrible. It can be the bootstrap that generates initial power for a more efficient boost circuit that can’t start on its own.
Isn’t that a chicken-egg problem? Using an opamp to generate the negative voltage needed by an opamp on the same die…
Yeah, I’m not sure bootstrapping an opamp by itself would work. The diodes will make sure V- is initially at 0.6V or lower, so the oscillator will start up fine. But it appears that in the negative half-cycle, all current that you draw from the V- output capacitor, goes through the opamp to V- and back to the capacitor.
You’re right. Switching the capacitor to V- would simply dump the charge across D1 and leave V- at positive 0.6 Volts. You need to add a transistor buffer in between that switches to ground instead.
And that loses the 3 component advantage.
It doesn’t need the negative voltage to be below zero to start oscillating.
Ahh I used to use the good old ICL7660 for making -ve voltages or as a voltage doubler, later the tech went into the MAX232 and made a legend in the RS-232 era.
Today, just use an MCU digital output or push-pull drive to a capacitor and two diodes, then a filter cap.
Same. I still have half a tube of 7660 and am wiring up something that needs one now. There are a couple newer variations that are better. Digikey has the Microchip version for about $1. The newer ones run better at the low voltages used in a lot of op-amp circuits today, like +-3V or less. Microchip TC1044 goes as low as -1.5V 20mA, check TPS6040 1.6 to 5.5 V 60mA 95 cents.
There are others up to 100mA that are as simple but I can’t recall.
I just found the MAX232 trim k recently and made up some small PCBs that straddle the edge of a breadboard — making prototyping with op-amps simpler.
I guess the advantage of the MAX232 is that you get the (approx.) +/- 9V from just 5V (and you get 5V too of course).
https://engineersneedart.com/blog/opamphelper/opamphelper.html
(And by ‘trim k’ I mean ‘trick’, ha ha.)
Wrong answers only? I prefer to get a bunch of (indirectly heated) triodes and use them as diodes or electrically-heated thermionic converters if you prefer. There’s a weak but steady and measurable DC voltage difference that forms because some electrons do still “boil” off a hot cathode and “condense” on the cool anode even though you haven’t specifically applied a positive voltage to attract them. If you ground the cathode, and electrons leave it and enter the the anode (or the grid, but you could tie that to the anode to get it out of the way or to whatever’s the most positive rail you have convenient to try and get more current), then there has been a current flow and the anode must be more negative than ground in that case.
What if the lights are off?
Then we’ll have to watch the TV by candlelight?
(Obligatory 555 comment)
(Obligatory Satan’s 666 response)
Unprompted 777
There’s also from TI (and possibly other companies) the LM7705 which is designed to make a -0.23V rail specifically for this purpose. The datasheet mentions that with the low voltage, it doesn’t tend to exceed 5.5V ratings for 5V opamps, and even if something swings negative, it won’t inject current through the ESD diodes to ground.
That’s the much better option. Aside from experimenting, you’d want a simple and reliable component, that does not overcomplicate your overall circuit.
Maybe a single cell battery. One component. Maybe a holder..
anywhere theres a pushpull oscillating signal, theres potential for a charge pump. And a ton of designs have such signals, or, at least, the ability to create such with a spare gate or gpio.
like so: https://www.edn.com/modded-charge-pump-extracts-power-from-digital-signal/
lately, i like to use an sn74lvc2g04 and a few tiny passives to generate additional rails with a few mm^2 of board area. you could also drive a tiny flyback or push-pull transformer this way too.
If you have a microcontroller that you can just output a pwm on, then you just need 1 diode and 2? capacitors?
Yes. It is that simple, although usually you might also want a regulator to reduce the switching noise.
High frequency switching noise will go right through most LDO regulators. An LC filter is a better choice to get rid of switching noise.
True, though the main point of the regulator is to get rid of the oscillator frequency ripple. Both is better.
Went down this rabbit hole a couple of years ago. I started with a simple 555 approach, but efficiency was poor and the output was noisy. Moved on to play with the push pull architecture and things got better, but the complexity got way out of hand.
One thing stood out, my home wound torroid transformers don’t work nearly as well as the commercial transformers.
Tried the Linear Tech LT1533 parts, and they work well. Not cheap, and not a real low parts count, but if you want a real bipolar supply of modest current, this is a winner. They will regulate the output voltages, and the high frequency noise is very low. Efficiency is not super high, but not too bad.
I think the reason he’s only seeing -0.3V output is that the circuit as drawn only works because of the leakage current through the diodes. A proper charge pump would need two more diodes and 2 coupling capacitors. Where they go is left as an exercise for the reader.