LM317 Power supply

Hi,
I had a thread on the LM317 Regulator sometime back.. It was closed with only theoretical details..
when Now I started building it for higher voltage, I have big problems..



First of all I have a transformer which gives 32v 2.5A when rectified (Leave the current, I have it spare)
I need an output of maximum 18v 1A..
The available pots for my voltage range for R1 are 500Ω, 1kΩ and 4.7kΩ.. Heard 3.3kΩ is also there but was not in stock..
For R2 I tried with 100Ω 1/4watt and R2 1k but burned my resistor and voltage control was only between a few volts range..

Can somebody help me on R1 and R2 values with R1 only from among the listed above for 1.25-18v output..
Also I need 2 pots for R1 one being 500Ω in series with the other for fine control but not absolute necessary..
Will the circuit handle the input of 32vdc?? I have another tapping on the transformer giving 17vdc but that wont give space for LM317 regulator till 18v..
Could somebody help me out??

What is the safe range for R2 values?? Wont a 100 ohm work??

I think you may have the LM317 connected incorrectly, or perhaps D3 is backwards.

The LM317 attempts to maintain a nominal 1.25v (called Vref in the datasheets) between the OUT and ADJ terminals by sourcing current from the OUT terminal. Vref may range from a low of 1.2v to a high of 1.3v and still be within specifications.

Let's take the worst-case scenario; Vref being 1.3v. Since I=E/R, 1.3v/100 Ohms = 13mA current, and 16.9mW. You should not have "smoked" your 100 Ohm resistor unless something else was wrong.

Now, let's look at what happens with a "typical" Vref of 1.25v. 1.25/100 Ohms = 12.5mA. If the resistance from the ADJ pin to GND is zero, the voltage at the OUT terminal will be 1.25v.

So, what if you want 2v output? Increase the resistance from the ADJ terminal to ground.

Since the LM317 will keep Vref at 1.25v by supplying 12.5mA through R2, you can calculate the resistance value needed for that additional 0.75v:
R = E/I = 0.75 / 12.5mA = 60 Ohms.
Above that, every 80 Ohms will give 1v higher output. To get 18v output, you will need 1340 Ohms resistance from the ADJ pin to GND when R2=100 Ohms and Vref=1.25v.

Note that power dissipation in the LM317 will be VERY high when sourcing current from high input voltages. You will need a big heatsink with fan cooling, and even that probably won't be enough.

I think you may have the LM317 connected incorrectly, or perhaps D3 is backwards.

Click to expand...

Is the circuit diagram correct?
LM317 has an Adj, Vout, Vin config right??
I checked my board twice.. will check it again if anything missed my eye..

Above that, every 80 Ohms will give 1v higher output. To get 18v output, you will need 1340 Ohms resistance from the ADJ pin to GND when R2=100 Ohms and Vref=1.25v.

Click to expand...

I thought that the 100ohm resistor doesnt make it meet the minimum Vout of 10mA.. When I supply an input of 17v instead of the 32v it works fine and no smoking.. Somebody here told me that LM317 needs a minimum 120ohm resistor to work good.. Is it true??

Anyway first thing Im going to trace back my board and second thing I'm going to use my center tap for the supply of 17v.. Getting a cool maximum of 12v regulated is better than an 18v hot with melting components and smoking resistors..

jj_alukkas said:

Is the circuit diagram correct?

Click to expand...

It looks OK. Note that R3, the 2.5k current limiting resistor for LED D4, needs to be rated for 1W.

LM317 has an Adj, Vout, Vin config right??

Click to expand...

Yes, when the pins are pointed down and the printing on the face is towards you.
Note that the tab is also connected to Vout.

I thought that the 100ohm resistor doesnt make it meet the minimum Vout of 10mA..

Click to expand...

The 10mA load is required for the LM317 to achieve guaranteed regulation over the temperature range. Since Vref can vary from 1.2v to 1.3v between individual regulators, nominal being 1.25v, the maximum resistance between the Vout and ADJ terminals to achieve that 10mA minimum load for the entire Vref range (with no other load connected) is 1.2v/10mA=120 Ohms. If there is any other load connected to the output of the regulator, that can also serve to meet the 10mA minimum load requirement.

When I supply an input of 17v instead of the 32v it works fine and no smoking..

Click to expand...

The smoking might be from the LM317.

Somebody here told me that LM317 needs a minimum 120ohm resistor to work good.. Is it true??

Click to expand...

It's a maximum of 120 Ohms from Vout to ADJ to achieve guaranteed regulation over temperature. However, current through the load will accomplish nearly the same thing.

The regulator will attempt to keep the voltage between Vout and ADJ at a nominal 1.25v by supplying current from Vout.
That same current then needs a path to ground.

Keep in mind that if you use a low-value resistor from Vout to ADJ, that the power requirements in the resistance from ADJ to ground will increase as Vout increases. For example, if you have a 100 Ohm resistor from Vout to ADJ, and Vref is 1.25v, there will be 12.5mA current to be sunk from the ADJ terminal. If you are attempting to output 18v, then from the ADJ pin to ground will measure 16.75v. The power dissipation in the pot will then be 16.75v x 12.5mA = 209.4mW. This may be too much for small pots.

Anyway first thing Im going to trace back my board and second thing I'm going to use my center tap for the supply of 17v.. Getting a cool maximum of 12v regulated is better than an 18v hot with melting components and smoking resistors..

Click to expand...

Linear power supplies work best if the regulator doesn't have to drop much voltage across itself. Power dissipation gets out of hand when trying to output low voltage at high current from a significantly higher voltage.

For example, let's say you have a 5v @ 1A load that you are powering from 17v.
The voltage drop across the LM317 will then be 17v-5v=12v at 1A; that's 12 Watts. You'd need a large heat sink.
If you tried powering that same 5v@1A from 32 volts, the drop across the LM317 would be 32v-5v=27v @ 1A = 27 Watts. You would have a really hard time getting rid of all that heat. In either case, the load would only be dissipating 5 Watts.

The smoking might be from the LM317.

Click to expand...

The chip still works fine and I clearly observed the resistor lighting up like an LED.. 317 is still regulating..

Linear power supplies work best if the regulator doesn't have to drop much voltage across itself. Power dissipation gets out of hand when trying to output low voltage at high current from a significantly higher voltage.

Click to expand...

So for now Im quitting the 32v channel.. 14-15v regulated would be enough for now.. SO ill use 17v tap..

I disconnected the diodes and tied the pot's free pin to ground.. have to power it up now and will reply..

After all I think its the very high heat dissipation which burned my resistor.. The cirrcuit is fine.. no probs.. double checked.. And as i said, it works cool on 17v from the same xformer.. no probs at all..

Which resistor burned up? It seems to be R2, but it's not clear to me.

R2 which runs b/n vout and adj burned.. The pot is ok.. I think its the heavy voltage problem.. lower make it fine.. But 317 is built for 40v isnt it??

According to the datasheet, the absolute maximum differential between the Vin and Vout pins is 40v.