# voltage regulator question

Discussion in 'Electronic Basics' started by tempus fugit, Dec 1, 2003.

1. ### Rich GriseGuest

True.

and adding a little more wouldn't
Not necessarily true, as you've noticed by now.

particularly when it was much smaller than the circuit's
Well, there's still X volts at the output of the regulator,
but what you did was create a voltage divider.

Also, the voltage drop varied
Use Ohm's law. The only way to change the load current with
a voltage-regulated source is to change the load resistance.
I haven't seen the circuit, but it sounds like you've got
16 ohms in series with whatever the load is, and the load
itself changes. So, if you're drawing 100 mA at 10 V, the
total load on the supply is 100 ohms. That's the sum of your
16 Ohm series resistor and the load itself, giving 84 ohms;
the load sees 8.4V. When you turn down the load so the total
is 10 mA, the total load is 1000 ohms, and the 16 is fixed,
so the adjustable part of the load is now presenting 984 ohms.
(please double-check my arithmetic. %-} ) So it sees 984 *
..01 = 9.84V, almost 1.4V difference.

HTH!
Rich

2. ### tempus fugitGuest

Thanks Rich

How? I've got the resistance in series with the output.

it sounds like you've got
I do have 16 ohms in series with the load, but the load doesn't change. All
I did was connect a 100 ohm resistor from the regulated out (after the 16
ohms) to ground. I did sweep the voltage output of the PS from 5-15v to see
what the effect on the regulation was.
Defaut has pretty much convinced me not to worry about the short circuit
protection, but I'd still like to figure out why this made such a
difference.

Thanks again

So, if you're drawing 100 mA at 10 V, the

3. ### tempus fugitGuest

Thanks again. The whole supply is fused at the AC input, so I should be OK
at the xformer - I was more worried about the regulator.

4. ### tempus fugitGuest

Sorry, that's what I meant.

5. ### Peter BennettGuest

As far as the voltage regulator is concerned, the 16 ohms is part of
the load. The regulator will keep the voltage at its own output
terminal constant. The 16 ohms and the load resistance are apparently
in series across the regulator output, so the current drawn by your
intended load also flows through the 16 ohms, and, according to Ohm's
Law, produces a voltage drop in that resistance. That voltage drop

6. ### defaultGuest

Like Peter wrote, the 16 ohms is part of your total series load
resistor. The regulator only cares about what its output pin voltage
is with respect to its "adjustment" terminal. If the numbers don't
calculate out - then you have a circuitry or part malfunction.

With the "16 ohms in series with your load resistor," (forgetting the
accuracy of that statement for the moment) and a load resistor of 100
ohms I calculate the following drops:

With the supply voltage set to 15 volts your 100 ohms will have 12.94
volts across it.

With the supply voltage set to 12 volts your 100 ohms will have 10.35
volts across it.

With the supply voltage set to 10 volts your 100 ohms will have 8.72
volts across it.

With the supply voltage set to 8 volts your 100 ohms will have 6.88
volts across it.

Assuming I didn't make a mistake, the laws of Ohm still prevail and
allowing for the ±5% tolerance in your meter, and ±5% tolerance in
your 16 and 100 ohm resistor, etc.. The actual reading you get should
fall within ±10% of those voltages at your 100 ohm resistor)

7. ### Rich GriseGuest

OK - that's what I meant by the "load" changing. The part of the
circuit _after_ the 16 oms in series is its "load." You've changed
the load from the point of view of the 16 Ohm. In a way, you could
trace current paths. Preferably on paper, with a pencil. Imagine
the water flow model - I've found that helps me a lot.

And maybe we should all be clear, in a situation like this,
what we're referring to as the "load." From the point of view
of the supply alone, the whole thing from the 16R out is the
load. From the point of view of the supply+16R circuit,
point of view of a circuit consisting of the PS, the 16R
and the 100R, the load on the 16-100 point is your DUT.

Or something like that - I think the chronic is kicking in. %-}

Cheers!
Rich

I did sweep the voltage output of the PS from 5-15v to see

8. ### tempus fugitGuest

ahhh. I think I see now. I don't think I measured the voltage at the output
pin of the regulator, but only at the output of the PS itself (which of
course is after the 16 ohms). So what you're saying is that the voltage I'm
reading is actually like measuring the voltage part way through into a
circuit? Hence the voltage drop.

That makes sense now. Simple thing - I shouldn't have overlooked it.

9. ### defaultGuest

I think you've got it.