voltage regulator question

tempus fugit wrote:
"> Well that would certainly do it. Putting 16 ohms in series with the
output drops voltage.

The water pipe analogy: pipe (resistor) is too small for the flow
(current) and pressure builds up on the inlet side and drops on the
outlet side (voltage drop across the resistor).

OK, but I figured that there would already be a certain resistance between
the output and ground (say, 100 ohms)
Click to expand...
True.

and adding a little more wouldn't
change things much,
Click to expand...
Not necessarily true, as you've noticed by now. :)

particularly when it was much smaller than the circuit's
resistance (i.e., If it was 100 ohms at 10v, the current draw would be 100
mA, so it would be a little less at 116 ohms).
Click to expand...
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
with the current draw - it was pretty major when drawing 100mA, but only a
few 100 mv whn drawing 30 or 40 mA. Why would that be?

Use Ohm's law. The only way to change the load current with
Click to expand...
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


If I'm not mistaken, the LM317 adjustable three terminal regulator has
built in over current protection at something like 1.5 amps (will
protect itself and the transformer if the transformer can output 1.5
amps safely. Voltage drops as current tries to increase beyond its
design value. (load resistance decreases)

That's good news. I'll check the datasheet and see if it does.


You don't say that you need the current to be adjustable just want
short protection - the LM317 already does that.

No, I don't really need adjustable current, just wanted something to prevent
me from having to replace the regulator IC every time I accidentally touched
the hot out to ground.
Thanks again
Click to expand...
 
Thanks Rich
Well, there's still X volts at the output of the regulator,
but what you did was create a voltage divider.
Click to expand...

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


it sounds like you've got
16 ohms in series with whatever the load is, and the load
itself changes.
I do have 16 ohms in series with the load, but the load doesn't change. All
Click to expand...
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
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


If I'm not mistaken, the LM317 adjustable three terminal regulator has
built in over current protection at something like 1.5 amps (will
protect itself and the transformer if the transformer can output 1.5
amps safely. Voltage drops as current tries to increase beyond its
design value. (load resistance decreases)

That's good news. I'll check the datasheet and see if it does.


You don't say that you need the current to be adjustable just want
short protection - the LM317 already does that.

No, I don't really need adjustable current, just wanted something to
prevent
me from having to replace the regulator IC every time I accidentally
touched
the hot out to ground.
Thanks again
Click to expand...
 
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.



"default" <R75/5@defaulter.net> wrote in message
news:s321tvsjrtg879g931stso6o2chn4aoqa2@4ax.com...
On Thu, 4 Dec 2003 21:23:46 -0500, "tempus fugit"
toccata.no.spam@ciaccess.com> wrote:

Thanks again Default.

I checked the data sheet, and it says that the typical current limit on
the
TO220 (which is what I'm using) is 2.2A. Wouldn't that destroy the
regulator
(which is spec'd for 1.5A) in a short circuit situation? Or am I looking
at
the wrong spec (or interpreting it wrong)?

Thanks again.

The part is spec'd with a minimum of 1.5, typical of 2.2, and maximum
of 3.7.

The one you get will fall in that range. If you are a commercial
designer you use the minimum current, because they guarantee the part
to work at 1.5, the other currents are possible and most of the parts
they turn out are 2.2.

If you are in a position where you just bought 100,000 three terminal
regulators for a product - they don't want you sending them back if
they only output 1.5 - and they don't want to select parts for a
particular user . . .

That doesn't mean the regulator will be destroyed at 2.2 amps out.
With a good heat sink it should put out 2.2 amps or even 3.7 without
complaining. But if it only puts out 1.5 before it current limits -
it is still a good part.

The parts tend to get better with manufacturing experience, so it may
be that very few will output only 1.5 these days - but if you are
committing large amounts of money to a product, you would be wise to
observe the lower limit or call National and see what they are willing
to do for you.

From your hobiest/experimentor perspective, you don't have to worry
about the part, it is protected from over current and shorts.

You still need to worry about the power source. If you have a
transformer supplying power for the device it must be able to source
3.7 amps to the regulator without burning up. (that could actually be
higher - like 5 amps - depending on how the transformer is specified)

For arguments sake, you have a transformer that will put out 12 volts,
at 3 amps. That is 12 volts RMS AC voltage at 5 amperes. Use a full
wave bridge rectifier to filter it and you might have 17 volts of DC
(1.4142 * the RMS voltage, give or take a little for transformer
regulation and rectifier losses)

That doesn't mean you get to suck 17 volts at 3 amps from the filter
cap side of things. 3 amps/17 volts at the filter is close to 4.25
/12 at the transformer.

If your only concern is that the regulator and transformer survive a
temporary short, don't sweat it. If you are building something like a
battery charger where the over-current condition can last for hours it
is a different matter.

I learned that lesson a long time ago. Wanted some tunes at work and
built an amplifier, but didn't consider the difference in AC and DC
amps. My power transformer was sealed in a steel case and potted in
tar. I fell asleep one night listening to music in the darkened shop.
The transformer exploded and woke me up.

I also learned something from this discussion. I was looking at class
A audio amps and it occurred to me that one could be made by using a
single output transistor working against a constant current source
(for a push-pull output). That was a year ago - but I didn't try it.
Looking at the old regulator book I saw where they actually do it.
Neat idea.


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Click to expand...
 
Sorry, that's what I meant.
"default" <R75/5@defaulter.net> wrote in message
news:5ed4tvkhs84spbktvj6mfnto9c5m78dkc7@4ax.com...
I wasn't sure if there was a difference between AC and DC amps. So do you
multiply it by 1.414 just like RMS voltage? (3A*1.414=4.25A)?
Yes, except you have it backwards - as the voltage goes up with
rectification and filtering the current allowed on the output goes
down for the same power level at the transformer.

Multiply by the reciprocal of 1.414 - .707


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Click to expand...
 
On Sat, 6 Dec 2003 23:00:53 -0500, "tempus fugit"
<toccata.no.spam@ciaccess.com> wrote:

Thanks Rich
Well, there's still X volts at the output of the regulator,
but what you did was create a voltage divider.


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


it sounds like you've got
16 ohms in series with whatever the load is, and the load
itself changes.
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
Click to expand...

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
will vary with the current drawn by your intended load.



--
Peter Bennett, VE7CEI
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca
 
On Sat, 6 Dec 2003 23:00:53 -0500, "tempus fugit"
<toccata.no.spam@ciaccess.com> wrote:

Thanks Rich
Well, there's still X volts at the output of the regulator,
but what you did was create a voltage divider.


How? I've got the resistance in series with the output.
Click to expand...
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 ą5% tolerance in your meter, and ą5% tolerance in
your 16 and 100 ohm resistor, etc.. The actual reading you get should
fall within ą10% of those voltages at your 100 ohm resistor)



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tempus fugit wrote:

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.
Click to expand...
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
say that you've added an additional load. Draw a schematic, and
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,
your DUT and the parallel resistor is the load. From the
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
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
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


If I'm not mistaken, the LM317 adjustable three terminal regulator has
built in over current protection at something like 1.5 amps (will
protect itself and the transformer if the transformer can output 1.5
amps safely. Voltage drops as current tries to increase beyond its
design value. (load resistance decreases)

That's good news. I'll check the datasheet and see if it does.


You don't say that you need the current to be adjustable just want
short protection - the LM317 already does that.

No, I don't really need adjustable current, just wanted something to
prevent
me from having to replace the regulator IC every time I accidentally
touched
the hot out to ground.
Thanks again
Click to expand...
 
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.


"Peter Bennett" <peterbb@somewhere.invalid> wrote in message
news:aeh5tv0ed1iecn8rup2pavi75rormb6cad@news.supernews.com...
On Sat, 6 Dec 2003 23:00:53 -0500, "tempus fugit"
toccata.no.spam@ciaccess.com> wrote:

Thanks Rich
Well, there's still X volts at the output of the regulator,
but what you did was create a voltage divider.


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


it sounds like you've got
16 ohms in series with whatever the load is, and the load
itself changes.
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


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
will vary with the current drawn by your intended load.



--
Peter Bennett, VE7CEI
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca
Click to expand...
 
On Thu, 11 Dec 2003 23:40:25 -0500, "tempus fugit"
<toccata.no.spam@ciaccess.com> wrote:

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.
Click to expand...
I think you've got it.



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