# Current consumption w/LDO Volt. Reg.

Discussion in 'General Electronics Chat' started by wind77, Jul 26, 2016.

1. ### wind77 Thread Starter New Member

Jul 21, 2016
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0
Hello all!
Question for you.
If I have say, a LDO 5v volt. reg., that is powering my circuit and I want to find the total current used by the circuit, is it a correct for a measurement to be taken to check/ verify current on the ground pin of the LDO VOLT REG? Or is there a better way?

Lisa

2. ### #12 Expert

Nov 30, 2010
16,664
7,311
Yes, the "waste" current is in the ground leg, but it's generally less than 100 microamps. The datasheet for your exact chip will tell how much current to expect.

3. ### wind77 Thread Starter New Member

Jul 21, 2016
15
0
Little confused though on what you said. What do you mean by "waste current"? If I have a microcontroller and some other type of IC both powered by the volt. reg. and the micro uses 10mA and the other IC uses 12mA and I measure each of those at their ground pin and see said current at there ground pin as just mentioned, will I see this total (22mA) at the ground pin of my regulator? And If this is true and my Voltage reg. is rated for 1A max then I would surely be w/in the limits of this Volt. Reg.

Lisa

4. ### #12 Expert

Nov 30, 2010
16,664
7,311
The voltage regulator does not waste the entire amount of current the loads use. If you want to measure the total current of all the loads and the regulator chip, measure current into the input to the regulator chip. If you want to measure how much the regulator uses for its own purposes, that current is in the ground leg of the regulator chip. A regulator chip (in this decade of time) usually wastes less than 100 ua in the ground leg of the regulator. Your total current should be less than 22.1 ma.

Jul 21, 2016
15
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TY # 12

6. ### crutschow Expert

Mar 14, 2008
13,475
3,361
Just to be clear, the ground current from your load goes directly back to the power supply ground.
None of it goes through the regulator ground pin.
The regulator ground pin is mainly to provide a reference point for the regulator output voltage.

7. ### wind77 Thread Starter New Member

Jul 21, 2016
15
0
Crutschow,
Yes, that is correct and what I was thinking.

Lisa

8. ### Papabravo Expert

Feb 24, 2006
10,338
1,850
A very small point about the meaning of LDO. If the difference between the input voltage and the output voltage is large relative to the LDO threshold, there is nothing stopping the regulator from dissipating a large amount of heat and getting very warm. All it is telling you is that the difference can be made small, with respect to the required difference in the first generation of linear regulators. It does not mean that the difference will necessarily be small.

9. ### AnalogKid Distinguished Member

Aug 1, 2013
4,685
1,297
Not just a better way, the only way - measure the current going *into* the overall regulator circuit. This will be the sum of the current used by the load and the current used by the regulator circuit to power its own operation.

Note that, depending on the regulator IC and its external components, measuring the current into the IC might not be good enough. For example, some ICs have an undervoltage lockout function that shuts down the output if the input voltage is too low. The trip point is set by an external voltage divider that uses very little current, but that current is not going into the regulator chip and will not be captured by a chip current measurement.

ak

Jul 21, 2016
15
0