# Doubling 12V current

• 12-24-2001, 07:47 PM
alexchannell
Doubling 12V current
I bought 2 12v LDRs from Jameco, I was told it is not good to put silicon devices inparallel due to one reciving more of the load. Is this true? Is this why, in Sproggy's design the 5v regulators are conntected through inductors? Con someone explain the theory of this to be?
Thanks.
• 12-24-2001, 08:04 PM
Probedude_2000
Each one is slightly different so they won't share the output load evenly.
If you wish to parallel them, you should put a small value resistor in series with the output of each, and then tie all the resistors together. Something like 0.2ohm at 1W will work.
• 12-24-2001, 09:41 PM
alexchannell
How would that solve the problem? If one had an effective .1 ohm resistance and the other had a .12 ohm resistance the .1 ohm, would always flow respectivly more because of the .02 delta? how would resistors inline of each help?

LDR---/\/\/\-----|
LDR---/\/\/\-----|--------computer

is this correct?

[ 12-24-2001: Message edited by: alexchannell ]
• 12-25-2001, 01:10 AM
zootjeff
The problem isn't really going to be solved with resistors. If one device is set to provide 5.01 volts and the other device provides 5 volts you run into problems because the one that is supplying 5 volts turns off because it is already gotten to 5 and the other one brought it up to 5.01 So the feedback loop gets messed up. Inductors solve the problem and they have very little resistance.
• 12-25-2001, 02:41 AM
Probedude_2000
Jeff,
Inductors do NOTHING in a DC application. Alex said he bought LDR (low dropout regulators). He is not building a switcher.

Alex,
you're right, if both regulators are matches and if one resistor is 0.1 ohm and the other is 0.12, the 0.1 will flow more current, however both regulators are contributing to your output.

However if you tie two regulator outputs in parallel without the resistors, the higher output one will drive all the load (until it reaches current limit).

With a resistor in series with each regulator, the higher voltage regulator will supply all the current until the resistor drops the voltage down to the level of the lower regulator. The lower regulator then starts contributing to your load. The voltage drop across each resistor is IR where I is the current being drawn and R is your resistor. You can see that if your current draw is 1.0A, your voltage drop will be 0.2V with a 0.2 ohm resistor. You can resize your resistors based on what your regulators are set to to minimize dropping losses.

[ 12-25-2001: Message edited by: Probedude_2000 ]

[ 12-25-2001: Message edited by: Probedude_2000 ]
• 12-25-2001, 04:48 AM
zootjeff
Quote:

Originally posted by Probedude_2000:
[QB]Jeff,
Inductors do NOTHING in a DC application. [ 12-25-2001: Message edited by: Probedude_2000 ]
QB]
Geez dude, bite my head off! Sorry, I am wrong, I stand corrected. He asked, "Is this why, in Sproggy's design the 5v regulators are conntected through inductors?" I was attempting to answer that, I guess I failed.

And inductors have some finite resistance. If you are going to max out the load anyway, is it so important that you don't have your linear regulators current limiting? There is always this method: Page 12 of this http://www.national.com/ds/LM/LM340.pdf

Sorry,

-Jeff

[ 12-25-2001: Message edited by: Jeff Mucha ]
• 12-25-2001, 10:11 AM
alexchannell
Inductors do act in DC circuits, they impeade changes in current flow (which is generally an AC characteristic, but appears in DC as well). Thanks alot for the help the both of you!
• 12-28-2001, 05:02 PM
Probedude_2000
Jeff,
Sorry, didn't mean to come across that way - wasn't my intent.

Alex, in your case your DC is relatively constant, the affect of any inductance is going to be incredibly small unless your inductance is incredibly large.