# LDO Vol. Reg. Calculations and findings.

Discussion in 'General Electronics Chat' started by vandaycalta, Jun 15, 2016.

1. ### vandaycalta Thread Starter Member

Mar 22, 2016
53
4
Thanks for taking the time to view my never ending questions!
I have been working as of late with LDO voltage regulators. I would like to display my findings and logic and look for your responses to see if I am correct in my practice.

Voltage in: 18volts
Current: .250 This has a .01 safety margin as I need at max .240.
Voltage regulator is 5v out and can handle 20v max. in

Example #1: 18v-5v x .250 =3.25 watts Way too much power for this LDO Vol reg.

Example #2 In this example I thought about using a resistor before the LDO vol reg to see if I could reduce my Power dissipation.

18 volts into a 28 ohm resistor of 2 watts(safety margin) (7v x .250A = 1.75 watts)yielding 7 volts and .250A after the resistor for a power dissipation of: 7v-5v x .250 =.5 watts.

Next I thought that I could dissipate more of the heat of the resistor by running 2 resistors in parallel and came up with this:
Equiv resistance of 28 ohms. 2 resistors(2 watts each includes safety margin) at 56 ohms each with 7 volts thru each gives me .875 watts thru each resistor for a total of 1.75w which checks out in my previous findings.

So i would end up with only .5w across the voltage regulator which will work for my LDO vol reg for theta JA which is 55 deg C/W which in this case is only 27.5 deg and the package I am using as per the data sheet and gives me plenty of head room with ambient temp as well as it is 150 deg C max.
Can you see any fault or error on my part in my calc. or findings and help to steer me in the right direction.

Val

2. ### dl324 AAC Fanatic!

Mar 30, 2015
7,802
1,861
Without knowing the specs for the LDO in question, can't say whether a safety margin is required. Are you certain that your load won't draw more than 0.24A? Will it handle the regulator protecting itself from over current (assuming it has built-in protection like most regulators have)?

I've seen zener diodes used more often than resistors. You could also add an external pass transistor and reduce the current in the LDO to a few tens of mA. Or you could add a switching pre-regulator to drop the LDO input voltage to a little more than it's drop-out voltage.

3. ### vandaycalta Thread Starter Member

Mar 22, 2016
53
4
Thanks for your response. The Vol. Reg. I am using is: 497-6429-1-ND from digikey.com I am sure I will not have a problem with .250A rating of this reg. My main question was heat dissipation and if I chose what I did was correct.

From what I have listed can you see that my calculations are correct?

Val

4. ### WBahn Moderator

Mar 31, 2012
23,389
7,099

If you are going to draw 250 mA of current from an 18 V supply down to a 5 V supply using linear circuitry, then you have got to dissipate (250 mA)(18 V - 5 V) = 3.25 W of overhead power. It doesn't matter how you do it.

So if you use a resistor ahead of the regulator to dump 1.75 W of power, then the regulator has to be dumping the remaining 1.5 W, not the 0.5 W you are coming up with.

So now that you know that your answer MUST be wrong, go back and take a look at how you are coming up with that 0.5 W for the regulator dissipation. What is the voltage across the regulator? You are claiming that it's 2 V. But if you have 7 V across the input resistor, 2 V across the regulator, and 5 V across the load, how does that jive with the total needing to be 18 V?

5. ### vandaycalta Thread Starter Member

Mar 22, 2016
53
4
Thank you WBahn for pointing out the fact that I had an error.

In looking this over I found the following:
18-5 x .25 = 3.25w
one resistor of 44 ohms as equivalent (I am running 2 resistors in parallel each is 88 ohms and dissipates 1.375 watts) for 2.75w.
V/ 44 ohms = .250A making V = 11
Now this puts 11 across the resistor leaving 7 volts across the regulator. I then used my original formula for power dissipation and came up with: 7v-5v x .250 =.5w The LDO for this reg. is .4 so I see that the 7 volts is above this at 5v + .4 or 5.4v min before dropout.
So now my voltage does =18v as 11+7=18 and my power equals 3.25w with 2.75w across the resistors and .5 across the regulator.

Am I closer? Thanks for your help!

Val

6. ### WBahn Moderator

Mar 31, 2012
23,389
7,099
Your approach is fine (as are your results), but you need to be much more careful with your equations and recognize that physical quantities have units (they are not just numbers). If I tell you that someone is 78 tall, are they taller or shorter than you? You don't know, because a number is not a height. The units are fundamentally part of the quantity and 78 inches is significantly different than 78 centimeters.

So the resistor doesn't put 11 across the resistor since 11 is just a number and there is a voltage across the resistor, namely 11 V.

Next, you need to adhere to the order of operations when you write expressions.

18-5 x .25 is not equal to 3.25, it is equal to 16.75. Since multiplication has higher priority than subtraction, that expression is equivalent to 18 - (5 x 0.25).

Another recommendation -- when you have a value that does not have a whole part (just a fractional part), it is best to put a leading zero before the decimal point. So use 0.5 instead of .5 because it is too easy to miss the decimal point when reading it.

But it looks like you are getting the concepts down pretty well.

7. ### vandaycalta Thread Starter Member

Mar 22, 2016
53
4
WBahn,
Thank you for your help in steering me in the right direction. Your guidance has helped me grow as I study this challenging field of study. I apologize for leaving out my units as yes I do see that it could be misconstrued. Again, thank you for all your help.

Val