Computing stuff tied to the physical world

Power booster

In Hardware on Dec 6, 2012 at 00:01

The trouble with the Arbitrary Waveform Generator I use, is that it has a fairly limited output drive capability. I thought it was broken, and returned it to TTi, but they tested it and couldn’t find any problem. It’ll drive a 50 Ω load, but my habit of raising the signal to stay above 0V (for single-supply uses) probably pushed it too far via that extra DC offset.

I’d like to use a slow ramp as sort of a controllable power supply for JeeNodes and the AA Power Board to find out how they behave with varying input voltages. A simple sawtooth running from 0.5V to 4V would be very convenient – as long as it can drive 50 mA or so.

Here’s one way to do it:

Volt follower

This is an op-amp, connected in such a way that the output will follow exactly what the input is doing – hence the name buffer amplifier or “voltage follower”.

Quick summary of how it works – an op-amp always increases its output when “+” is above “-”, and vice versa. So whatever the output is right now, if you raise the “+” pin, the output will go up, until the “-” pin is at the same value.

It seems pointless, but the other property of an op-amp, is that the input impedance of its inputs is very high. In other words: it draws nearly no current. The input load is negligible.

The output current is determined by the limits of the op-amp. And the AD8532 from Analog Devices can drive up to 250 mA – pretty nice for a low-power supply, in fact!

Here’s the experimental setup (only one of the two op-amps is being used here):

DSC 4273

Here you can see that the input voltage is exactly the same as the output:

SCR17

(yellow = input signal, blue = output signal, a 500 KHz sine wave between 1V and 3V)

Well, almost…

As you can see, there’s a phase shift. It’s not really a big deal – keep in mind that the signal used here is a high-frequency wave, and that shift is in fact less than 0.1 µs. Irrelevant for a power supply with a slow ramp.

Tomorrow I’ll bombard you with scope shots, to illustrate how this op-amp based voltage follower behaves when gradually pushed beyond its capabilities. Nasty stuff…

Keep in mind that the point of this whole setup is to drive more current than the function generator can provide. As a test, I connected a 100 Ω resistor over the output, and sure enough nothing changes. The AD8532 will simply drive the 10..30 mA through the resistor and still maintain its output voltage.

The beauty of op-amps is that all this just works!

But there is a slight problem: the AD8532 can drive up to 250 mA, but it’s not short-circuit proof. If we ever draw over 250 mA, we’ll probably damage it. The solution is simple, once you think about how op-amps really work (from the datasheet):

Screen Shot 2012 11 24 at 20 26 21

The extra resistor limits the output current to the safe value, but the side-effect is that the more current you draw, the less “headroom” you end up with: if we draw 100 mA, then that resistor will have a 2V voltage drop, so the maximum output voltage will be 3V when the supply voltage is 5V.

If you look at my experimental setup above, you’ll see a 22 Ω resistor tied to each output.

That’s it. This simple setup should make it possible to explore how simple circuits work with varying supply voltages. A great way to simulate battery limits, I hope!

  1. Why not use a resetting PTC fuse instead? They have very low resistance depending on the type(in the range of 2 to 6 ohms at this level) and reliable tripping points. Most USB hubs contain these safety devices on each port set at 500mA. You do have to be careful in selecting them as there is a wide range between hold and trip (about double). Also short circuit current is not as fast as a normal fuse, it takes some time to trip, measured in tenths of seconds. I have seen them as low as 80mA hold, but there may be lower ratings.

  2. Hmm. The tiny mass of stressed silicon might well heat up much faster than the polyfuse. An alternative is to current limit the op amp chip supply.

  3. Yes – but what’s wrong with the above solution? With up to say 50 mA, it just means the power supply needs to stay ≈ 1.1V above the max output voltage. Which is why I’ll be using a 4x AA Eneloop, i.e. 5.2V.

  4. Well, yes if you limit the current to 50mA and have fully charged batteries you will be fine. I was offering a solution for the original scenario of 100mA and Vin of 5V. But the issue of the chip overheating before the fuse is a valid concern. A fuse with a smaller mass such as an SMD may work better than a lead type device.

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