# Using Voltage Regulator for current limiting - John Fileds ?

Discussion in 'Electronic Basics' started by William A. Bong, Jan 10, 2006.

1. ### William A. BongGuest

I think that it was John Fields that posted a circuit using a voltage
regulator IC such as the 7812 as a current limiting device - John or anyone
else that could help out with this please repost it - I searched but it's
obviously no longer on my server.

I was thinking of using this as a current limiter to build a USB to Nokia
phone charger - I have
the hardware and connected it up straight, but it trips the power to my
notebook USB port requiring a reboot to reset the USB.

Bill.

3. ### Pooh BearGuest

There is a way to configure a standard regulator as a constant current source.
I suspect this is what you mean. Only ever did it once for EMI reasons.

For a positive output you need a negative regulator and you use it 'back to
front'.

A power resistor ( R ) is placed between Common/Sense and Out. I = R/Vref (
Vref = 5 for 7905 for example or 1.25 for LM337 ) . DC in + goes to Common /
Sense and the current out comes from 'In'.

Graham

4. ### Pooh BearGuest

And then I recall that there's a way to make a shunt regulator to IIRC ! Uses the
'opposite polarity' chip again in a wacky configuration.

Graham

5. ### Anthony FremontGuest

I think allot of voltage regulators can be used like this, but the LM317
seems one of the most popular for the task.

The datasheet is also a good reference source for cutesy circuits like
this.

The key is that the output pin and the ground pin will always have 1.25V
across them. By floating the ground and connecting it to the output pin
via a resistor, you can create a constant current source. The resistor
is sized using Ohms law (R=E/I). So for a 25mA current source, you
would need 50 Ohms of resistance between the pins. Be sure to calculate
the power being dissipated by the resistor and size it accordingly. In
this case, the power is 31.25mW so a 1/8W resistor should be plenty big
enough. If you want to supply 1A, then you would need a resistor of
1.25 Ohms that would be dissipating 1.25W, so use at least a 2W and
don't burn yourself. ;-)

6. ### Rich GriseGuest

A current limited regulator isn't what you're looking for here - all that
will do is make the power supply to your device droop, and it won't
operate properly. You need either a USB with a beefier supply, or an
auxiliary power supply, like a wall wart.

Good Luck!
Rich

7. ### kellGuest

Are you connecting your Nokia phone directly to the USB? It's actually
not such a bad idea, in principle -- the USB is a 5 volt power supply,
and you can charge a Nokia cell phone from a 5 volt power supply. But
if your Nokia is like mine, it doesn't draw a steady low current when
charging. Mine draws its power in pulses. And I'm sure these pulses
would be enough to trip a USB, which can only put out very small
currents.
Try putting a resistor in series with the USB and allowing a capacitor
to charge from the limited current coming through the resistor. Charge
resistor, and 1000 uF cap. With a correctly sized resistor, you may be
able to pevent the USB from tripping and yet still get the cap to
charge up between pulses. You should be able to make this work if the
total power drawn by your phone, over time, does not exceed the amount
of steady state power the USB can deliver. But there's no guarantee of
that.

8. ### Jasen BettsGuest

use the regulator to regulate the voltage in a resistor.

usually works more economically with a 5V or lower regulator.
as the 5V etc is wasted

Bye.
Jasen