LM350 Voltage Drop

Discussion in 'The Projects Forum' started by PCBoy, Jan 30, 2012.

1. PCBoy Thread Starter New Member

Dec 14, 2011
26
0
I got an LM350 voltage regulator setup with these specs:

Input - 34V
Output - 15V

I'm using a 560 ohm resistor that connects the ADJ and the Output pin, and I have a 10K potentiometer that connects the ADJ pin to the ground.

The voltage is good, I get my desired voltage by turning the pot. But the thing is, when I connect a 12V 25W load, I get a voltage drop of approx. 2.5V. It's not a heatsink issue because I have a huge one and the regulator doesn't shut down and sustains 12.5V.

Could the choice of resistors be the cause? The datasheet says to use 120 ohms, and a 5K pot.

2. Audioguru New Member

Dec 20, 2007
9,411
896
The datasheet says to use 120 ohms between the ADJ pin and the OUT pin because if this resistance is higher then some LM350 ICs will have an increased output voltage without a load. The datasheet says the "minimum load current" is 10mA. 120 ohms with 1.25V across it has a currrent of 1.25V/120 ohms= 10.4mA which is a little more than the minimum load current that is needed so it works fine. That is not your problem.

Your 12V 25W load needs a current of 25W/12V= 2.1A. Its resistance is 12V/2.1A= 5.7 ohms. With a 15V supply it draws 15V/5.7 ohms= 2.6A.

The datasheet for the LM350 says that it reduces its maximum output current when it has more than 10V from its input to its output. You have 34V - 15V= 19V from input to output. A graph in the datasheet shows that a "typical" LM350 reduces its max current to about 2.1A which yours is doing.

3. bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
Audioguru is right on the money. All linear regulators have safe operating area protection which forces the current limit value lower as the voltage drop across the device gets larger.

4. PCBoy Thread Starter New Member

Dec 14, 2011
26
0
I see, thanks for the reply.

How can I reduce the voltage before the LM350 without using another LM350? (LM350's here are quite expensive for my college budget)

Could I use a 24V Zener Diode before the LM350?

5. ifixit Distinguished Member

Nov 20, 2008
639
110
A power zener would very expensive.

You could add a 3.9Ω 30W resistor in series with the input to the LM350.

Example...
At a current of 2.6A the voltage drop would be 10.14V. So 19V - 10V = 9V across the LM350 and it won't limit power dissapation because the power (26W) is now being dissapated in the resistor instead.

Do you have enough budget for a power resistor? Three 12Ω 10W in parallel would do also.

Regards,
Ifixit

6. PCBoy Thread Starter New Member

Dec 14, 2011
26
0
Thanks! I'll be trying that now.

7. PCBoy Thread Starter New Member

Dec 14, 2011
26
0
Hmm, problem. I asked the local shop, they said they only have 5W 12 ohm resistors. If I'm correct, I'd need 6 of those in parallel.

Is there any other way?

8. PCBoy Thread Starter New Member

Dec 14, 2011
26
0

I'm building a +- 15 Supply and a variable supply.

The variable and the +15 are using LM350's and are suffering this problem.

For the negative part, I'm using a 7915 regulator with a pass transistor.

All supplies will have the same test load (12V 25W).

I was wondering, if I build a separate pcb board with 6 24 ohm resistors in parallel.. can I hook that board to my supplies such that each supply shares the "voltage dropper". Or would that affect the current drawn already?

(disregard the 12 ohms, I forgot to change it but the question stands)

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Last edited: Jan 30, 2012
9. Audioguru New Member

Dec 20, 2007
9,411
896
The +15V power supply needs a +18V input, not +34V.
The -15V power supply needs its own -18V input.
You forgot to tell us the max output voltage of the variable power supply but if it is +15V then it can use the same +18V input for the +15V power supply. its max output current will be reduced when the voltage is set to 5V and less.

It is silly using huge and very hot resistors to reduce the input voltage.
You can't get 10W resistors anyway.

10. PCBoy Thread Starter New Member

Dec 14, 2011
26
0
What do you suggest I should do if I won't do the resistor thing?

Btw the max for the variable is 24V. So min input for that is 27 right?

11. Audioguru New Member

Dec 20, 2007
9,411
896
Instead of using silly resistors, why not use the proper voltages for the inputs?

The datasheet for the LM350 shows that its maximum output current drops when it has an input to output voltage of 10V or more. So with a 27V input its maximum output current drops when its output voltage is set to 17V or less.

12. PCBoy Thread Starter New Member

Dec 14, 2011
26
0
I have a 24V transformer, that passes through a diode bridge to make 24Vdc then it goes through a filtering cap that makes the 24V go upto 34V which I have now have to reduce to work with the LM350s.

13. Audioguru New Member

Dec 20, 2007
9,411
896
The 24VAC from the transformer is too high. It should be 14VAC then it has a peak voltage of 19.8V and the bridge rectifier reduces it to 17.8VdC.

To get 27VDC you need a 20.5VAC transformer that has a peak voltage of 29V and the bridge rectifier reduces it to 27VDC.

To get -17.8VDC you need another transformer that is 14VAC.

14. PCBoy Thread Starter New Member

Dec 14, 2011
26
0
Like I mentioned before, I'm on a college budget and purchasing a new transformer is really out of my choices.

15. sureshparanjape Member

Feb 10, 2012
64
2
I am a beginner too and in the process of using LM350 for power supply.
Can it be all right if one reduces input voltage, before giving it as input to LM350 so that the difference in Vin and Vout is not large? If it can be, what precautions should be taken?
If the power supply is to be used for hobby, one wouldn't know load that would be there . Are there parameters for a load of a circuit? How to determine them before so that such power supply can handle them?
I like the quote that is there- if you ask a question, you may sound stupid; if you don't, you remain stupid! I have chosen the first option.
sureshparanjape

Last edited: Mar 17, 2013