1. We will be in Read Only mode (no new threads, replies, registration) for several hours as we migrate the forums to upgraded software.

Voltage Regulator, Help!!

Discussion in 'The Projects Forum' started by NM2008, Mar 8, 2008.

1. NM2008 Thread Starter Senior Member

Feb 9, 2008
135
0
Hi,
I am trying to set-up this voltage regulator circuit The problem I am having is, the circuit does not seem to give enough current for example at 4v, an LED across the load will pull voltage down to 1.9.

I want to draw 7v at 1.5A from 12v.

With the 1k pot I was only able to achieve 4v, But according to this circuit I should be able to get 7v, I do not understand why this is. I then changed the 1k pot to 5k, I got 6.5v but again, an led across the load dropped it to 1.9v.

The only change I made to this circuit is, (as I did not have a 23ohm resistor to hand)
I used a 15ohm instead of 23ohm.

Thanks in advance any help given,
Regards NM

2. beenthere Retired Moderator

Apr 20, 2004
15,808
295
Why are you using a load resistor? The regulator will work just fine without it, and you won't be wasting all that heat.

317's are a bit happier with a 5K adjustable resistor to set the output voltage.

The voltage went down with the LED across the load resistor because LED's have no resistance when they start to conduct. You could just pitch the load resistor and source the LED through a properly-sized current limiting resistor. Take the forward drop of the LED from the source voltage, and base the resistor on that (I = E/R).

3. SgtWookie Expert

Jul 17, 2007
22,201
1,809
There are two major probabilities:
1) The 12VDC source you are using is not capable of supplying sufficient current for the regulator to maintain it's output voltage.

2) You do not have a heat sink on the LM317, and it is going into thermal overload protection mode and shutting the output current down.

"Rload" is simply a representation of a load. It should not be a part of the actual supply circuit itself.

You should not have increased R2 to 5k Ohms. As I originally drew it, R2 being a 1k pot gives a range of about 1.4V to about 10.2V, which is right at the limits of what an LM317 can supply when the input is 12V DC.

If the circuit will be operated with no load, R1 must remain at 120 Ohms in order to ensure a minimum of 10mA current flow through the regulator, which is a stated design requirement in the LM317 datasheet in order to guarantee a stable output. Of course, this limits the useful output of the supply to 1.49 amps. If the supply will never be operated under no-load conditions, the values of R1 and R2 can be porportionately increased, otherwise they should remain as they are.

There is also the possibility that the LM317 has been overheated and is no longer good, or that one of your caps has gone bad and is shorted.

Also, be aware that an LED must have the current flowing through it controlled, either by a constant current circuit, or by a suitably-sized current limiting resistor. This supply was designed to regulate voltage output, not current.

Since you say that your LED measured 1.9V across it, that is it's Vf (forward voltage drop). I will assume for a moment that it is a standard red LED, with a maximum current rating of 20mA.

Re-install the 1k Ohm pot and set the regulator to output 7V even.

Let's calculate the current limiting resistor you will need for your single LED.
First you need to determine the actual forward voltage of the LED, which you've already done. Normally, one would use (as a rule of thumb) 100 Ohms per volt across the original test voltage. This will limit current through an LED to a maximum of 10mA, even if it has a very low Vf.

Then, you subtract the LED's Vf from the supply voltage:
7V - 1.9V = 5.1V remaining.
Using Ohm's Law, calculate what resistance is needed to limit current to 20mA at 5.1V drop:
R = E/I (Resistance = Voltage / Current)
R = 5.1V / 20mA
R = 5.1 / 0.02
R = 255 Ohms
The nearest standard E12 value of resistance is 270 Ohms - see chart:
http://www.logwell.com/tech/components/resistor_values.html
If you have access to E96 values, you could use a 255 Ohm resistor.
You don't want to use less resistance than the formula indicates; your LEDs will be brighter, but they won't last long at all.
To calculate the actual current from the selected resistor:
I = E/R (Current = Voltage / Resistance)
I = 5.1/270
I=0.0188888... or roughly 18.9mA.

You can run several LED's together in a string, and use one resistor to limit the current through them - however you will need to subtract the Vf of each LED from the supply voltage before you can calculate the resistor you will need to use.

You should not attempt to power multiple LEDs or multiple strings of LEDs in parallel using a single resistor, as each LED will have a slightly different Vf, and will receive a different amount of current.

4. NM2008 Thread Starter Senior Member

Feb 9, 2008
135
0
Thanks for taking time to answer my post,
In the end I discovered a mix up with the regulator input and output, once changed the circuit worked perfect!
Also thanks to SgtWookie for info on the regulator circuit.
Regards NM