[kanamin]

Quick question for anybody who wants to help.

I'm currently using tangent's opamp power supply to power an AD827-based preamp, the opamp controlling the virtual ground is an opa2111. This is hooked up to 2 9v batteries. The circuit is the first image attached.

I don't want to keep replacing batteries (it doesn't even need to be portable), so I thought I might make a power supply for it.

also attached is the recommended circuit from the lm317 datasheet. Instead of R1 and R2, could I connect a 100K pot as shown in the third pic? or would Iadj be too low for the regulator?

I'll be hooking this up to a 24v dc unregulated wall-wart.

 

[FallenAngel]

It should be possible, just don't expect it to be very usable. For half of the adjustment range, you'll get 1V and you won't get above 10V until you get to the last 90% of the adjustment range at which point any tiny twist of the adjuster will send the voltage a few volts in either direction.

 

[kanamin]

darn, that's what I thought

what about a 10k in the same place? 5k?

 

[ericj]

why don't you just have a look at tanget's STEPs schematic?

 

[tangent]

Quote:

Originally Posted by kanamin /img/forum/go_quote.gif could I connect a 100K pot as shown in the third pic?

No.

Quote:

would Iadj be too low for the regulator?

Yes, except right near one extreme of the pot's adjustment range, where the resistance is <= 240 ohms.

The right way is shown as R2 in the second pic. I recommend adding a fixed resistor in series with it to set a reasonable floor, below which you can't adjust. It also lets you use a smaller pot value, for more precise adjustment. You can see this in my TREAD schematic, which is simpler than the STEPS, recommended by Eric. I think it's closer to what you're after.

Quote:

Originally Posted by FallenAngel /img/forum/go_quote.gif For half of the adjustment range, you'll get 1V and you won't get above 10V until you get to the last 90% of the adjustment range

You're thinking of a log pot. There are linear 100K pots.

If you put a 100K linear in place of R2 in the second schematic, you'd get the full unregulated voltage of the power supply, less probably some tiny drop across the pass transistor in the regulator, for most of the adjustment range. Only when resistance drops low enough that the regulator can begin regulating will the pot do anything. At that point, every degree of turn will drop voltage by the same amount, unlike with a log pot.

 

[DKJones96]

Quote:

Originally Posted by FallenAngel /img/forum/go_quote.gif It should be possible, just don't expect it to be very usable. For half of the adjustment range, you'll get 1V and you won't get above 10V until you get to the last 90% of the adjustment range at which point any tiny twist of the adjuster will send the voltage a few volts in either direction.

That's assuming it is a log taper, and if it is a log taper he can wire it backwards so the first portion of the adjustment is the part that will send it a volt or two in either direction and the last part is precise. I wired my audio pot backwards once and it acted like that.

 

[FallenAngel]

Actually I was thinking it was linear but that would adjust R1 and R2 at the same time where R1 + R2 = 100K.

With that ratio and V = R1/R2 + 1, you would have 2V at 50k/50k, 10V at 90k/10k (90% of the adjustment), 21V at 95K/5K (95% of adjustment), etc.

I'm likely missing something, but that was the idea.

 

[tangent]

What you're missing is that the resistance between OUT and ADJ has to be low enough to give 5-10 mA, given that the voltage between these two is always 1.25 V. So, 240 ohms or less.

 

[FallenAngel]

Really? Wow, I've been using whatever is in my parts bin all the time and some are definitely in the in the 1K+ area.

 

[tangent]

If the resistance is too high, the regulator can sort of work, just not regulate as well as it should. Also, this is a worst-case current requirement, for extremes of output voltage, temperature, and manufacturing variability.

We're not talking about ADJ to ground here, just OUT to ADJ. ADJ to ground can be huge.

 

[kanamin]

I just realized

I DID buy a 10K pot XD. So forgetful.

I'll do the right thing and put a 240 ohm in between adj and out, hook up the 10K pot with its wiper shorted to one end and another maybe 1k or so in series, the two together making up R2. So the 1.25 in the equation is the difference between adj and out?

 

[tangent]

Quote:

Originally Posted by kanamin /img/forum/go_quote.gif I'll do the right thing and put a 240 ohm in between adj and out

I prefer 120 ohms, myself, as that covers the absolute worst case. It means the idle current of your regulator is > 10 mA, even when the downstream circuit is off if there is a power switch between the regulator and that other circuit, but if that bothers you, you can move the power switch to between the wall wart and the regulator.

I sometimes use 100 ohms, when that makes it easier to pick a value for R2.

Quote:

hook up the 10K pot with its wiper shorted to one end and another maybe 1k or so in series

Yes, though I'd put another resistor in parallel with the pot, or pick a smaller pot. 10K is big enough to allow adjustment almost up to 50 V! Not that you could get away with that, as the regulator would die first, but still.

Your 24 V unregulated wall wart will probably put out about 30 V under light loads. That means the highest output voltage you can hope for is about 28 V. Call it 26 V just to be safe. You only need 2.5 K or so to do that, with no series resistor. With 1K in series -- ~12 V floor -- you only need 1.5 K or so. Try 2K or 2.2K in parallel.

Quote:

So the 1.25 in the equation is the difference between adj and out?

Yes.

 

[kanamin]

I... soldered before your post.

Well, I just desoldered and soldered again your values, gonna test in a bit.

It's one of those multi-turn pots so I'm not expecting it to be overly sensitive to adjustment.

 

[kanamin]

adjusted pot for 25V, multimeter read 25V, all is good

now for casework D: