Current Flowing Indicator.

[Mark Hennessy]

Mis-understood again. I thought that by 'Near the rail' you meant that the voltages on the inputs couldn't be too near to the supply voltage. Presumably this is why +/- supplies are best for them?

Either rail. I first mentioned that in post #4.

When it comes to understanding and using op-amps, free yourself from the text-book notion of equally split +/- supplies. As I said, op-amps don't know about earth - they don't have an earth pin, and have no idea where an earth might be. Earth is a relative concept.

Equally split supplies are convenient for many applications, but not for others. Unequal splits or single rails are just as workable. For example, I have a signal generator that supplies op-amps from all sorts of voltages - one is +/-15V, another sits between ground and 24V, another sits between 0V and -24V. In that PSU I built a while back, the op-amps in the positive regulator are powered from +22V and -5V, while the op-amps in the negative section run from +5V and -22V. You see, it's all relative to what you need to do, and where the signals sit.

In your case, the inputs to your op-amp travel over a range of 1.25V up to the 18V (or so) that your supply can provide.

Because both inputs are pretty close together as they travel over this range, this range is called the "common-mode" signal. You might have seen this term before, but perhaps not understood it. The key thing is the op-amp responds to the difference between the two inputs, but (hopefully!) ignores any common-mode signal.

Having just looked at the datasheet, it appears that the typical common-mode input voltage range is +/-13V when powered from +/-15V. Actually, the minimum guaranteed is +/-12V

In other words, it should work to within 2V of each supply rail, and definitely will work to within 3V of the rails. If your op-amp is powered from between ground and 24V, then the common-mode input range is 2V to 22V. Or 3V to 21V if you're feeling pessimistic

So, for reliable operation below about 2-3V, you need to make pin 4 slightly negative. This could be done using a capacitor, in a similar way to your voltage doubler capacitor. Then, just use a Zener diode (perhaps 3V or thereabouts) to ensure that the negative rail is sufficiently below ground. But not so far that the safe PSU rail voltage limit (36V) is exceeded.

Personally, if you were playing this trick, then I'd also add a -1.25V reference (two diodes in series) that would allow you to adjust the supply right down to 0V. Just return your variable resistor to this rather than directly to ground.

For the sake of 2 or 3 capacitors, a few diodes, a resistor or two, and a Zener, I'd be tempted. You'd learn a lot more. As if you haven't learnt enough already.

But, before going to that effort, try the battery trick I suggested previously. Although I'm 99% sure I'm right, don't trust me!

I have studied the data-sheets but the required information isn't on them in a way I can understand.

Datasheets aren't easy to understand when you're not really across how the device itself works. But you'll get there...

If you don't already have a copy, I'd recommend The Art of Electronics by Horowitz and Hill. It's a bit of a weighty tomb, but is the reference - and, most importantly - is incredibly readable and accessible. It's quite unlike most textbooks. Look out for s/h copies, as new it's 60 quid! The first edition will do, and should be much cheaper, but the second edition is the one I have. It was half that price when I bought mine in 1992...

I had a play with another 741 on a breadboard and that was plain weird: I'd only connected the power and a 1kΩ resistor and LED to the output and it wouldn't go out with nothing on the inputs or even with them shorted.

Yes, but try to understand that the inputs have an incredibly high input impedance, and will have been picking up noise and all sorts - perhaps leakage through the breadboard.

According to the datasheet, the 741 op-amp has an open-loop DC voltage gain of 106dB. To work this out in voltage terms without using a calculator, start by saying that 106dB is 20+20+20+20+20+6dB. Knowing that a gain of +20dB is a voltage gain of x10, and likewise, a gain of +6dB is a voltage gain of x2, we can work this out:

20 + 20 + 20 + 20 + 20 + 6dB

is

10 x 10 x 10 x 10 x 10 x 2 = x200,000

So, to drive the output over a range of 20V, you only need 100uV. That's noise. Given how high the input impedance is, that's going to easily find its way in.

I'm starting to think and EM80 would be easier!

Hardly. Magic eyes need more of a voltage swing than the LED, and the one I know about (EM84) requires -20V w.r.t. to the cathode to close up the bars. So you'd still need the op-amp, and as well as the inputs going to +15V or thereabouts, the output would need to swing negative by nearly as much. Not to mention generating the HT and heater volts!

Now, clearly I get that that wasn't serious, but it doesn't hurt to point out that you're nearer to understanding this that you think. It's taken me a lifetime to get here - wherever "here" might be - so don't expect all this to make sense immediately. But, it really is very understandable, and you'll be able to do so much more with the knowledge - op-amps are a far more useful building block than valves or transistors. And actually easier to understand!

Cheers,

Mark