# Noob Question About Current From Voltage Regulators

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#### Rory Starkweather

Nov 13, 2014
77
If I have a 2 Amp regulated supply, will the attached circuit have to drop the whole 2 amps, like voltage, or will it just take what it needs?

#### Bluejets

Oct 5, 2014
6,342
Attachment not there.....

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
Current can't be "dropped" the way voltage can. Current is a flow. All the current that flows into a circuit also flows back out.

With a linear regulator, the current coming in is equal to the load current plus a small amount of current that the regulator uses internally for its own operation, which comes out the ground pin of the regulator and is called the ground current. For a 78xx regulator the ground current is about 5~10 mA.

The load (the circuit that is being powered from the regulator) will draw whatever current it needs, and the current going into the regulator will be the sum of that current and the regulator's ground current.

#### Rory Starkweather

Nov 13, 2014
77
Attachment not there.....

No duh? The attachement is the load. I don't have enough to send to everyone.

#### Rory Starkweather

Nov 13, 2014
77
Current can't be "dropped" the way voltage can. Current is a flow. All the current that flows into a circuit also flows back out.

With a linear regulator, the current coming in is equal to the load current plus a small amount of current that the regulator uses internally for its own operation, which comes out the ground pin of the regulator and is called the ground current. For a 78xx regulator the ground current is about 5~10 mA.

The load (the circuit that is being powered from the regulator) will draw whatever current it needs, and the current going into the regulator will be the sum of that current and the regulator's ground current.

What if I don't use a regulator? Suppose it is just an external power source and a resistor?

I have a nice variable voltage supply, but it is rated a 2 Amps. Will the circuit have to drop the whole 2 amps?

#### Arouse1973

Dec 18, 2013
5,178
No duh? The attachement is the load. I don't have enough to send to everyone.

Nov 13, 2014
77
TYVM

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
What if I don't use a regulator? Suppose it is just an external power source and a resistor
You mean you have a power source, a series resistor, and the load? The load will draw current, which will cause a voltage drop across the resistor according to Ohm's Law. The current will be the same at all points in the circuit, but because the resistor will drop some voltage, there will be less voltage across the load, so the load may draw less current than it would otherwise. (It depends on what the load is; you should have told us that.)

Ohm's Law applied to the series resistor says:

V = I × R where
V is the voltage dropped across the series resistor, in volts;
I is the current through the resistor (the same current flows at all points in the circuit), in amps;
R is the resistance of the series resistor, in ohms.
I have a nice variable voltage supply, but it is rated a 2 Amps. Will the circuit have to drop the whole 2 amps?
You can't "drop" current, as I said before. The same current will flow through all points in the circuit. If your power supply is rated at 2A that just means that you shouldn't draw more than 2A from it otherwise it may overheat, shut down, or be damaged.

#### Rory Starkweather

Nov 13, 2014
77
You mean you have a power source, a series resistor, and the load? The load will draw current, which will cause a voltage drop across the resistor according to Ohm's Law. The current will be the same at all points in the circuit, but because the resistor will drop some voltage, there will be less voltage across the load, so the load may draw less current than it would otherwise. (It depends on what the load is; you should have told us that.)

Ohm's Law applied to the series resistor says:

V = I × R where
V is the voltage dropped across the series resistor, in volts;
I is the current through the resistor (the same current flows at all points in the circuit), in amps;
R is the resistance of the series resistor, in ohms.

You can't "drop" current, as I said before. The same current will flow through all points in the circuit. If your power supply is rated at 2A that just means that you shouldn't draw more than 2A from it otherwise it may overheat, shut down, or be damaged.

This is what I needed to know, as usual. TYVM.

I was trying to keep it simple. The single resistor was the load. Does that matter?

#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,831
Only thing that matters is you have an adjustable voltage power supply capable of providing (up to) 2 A to whatever load resistor you connect to it. The current the power supply provides to the load is I = E/R where E is the voltage in volts, R is the resistance in ohms, and I is the current in amperes, with a maximum value of 2 A determined by the maximum current the power supply can provide. So, yes, the resistor matters. If you don't connect the load resistor, the power supply provides zero current.

#### Rory Starkweather

Nov 13, 2014
77
Only thing that matters is you have an adjustable voltage power supply capable of providing (up to) 2 A to whatever load resistor you connect to it. The current the power supply provides to the load is I = E/R where E is the voltage in volts, R is the resistance in ohms, and I is the current in amperes, with a maximum value of 2 A determined by the maximum current the power supply can provide. So, yes, the resistor matters. If you don't connect the load resistor, the power supply provides zero current.

TYVM

The problem is that I do calibration references. 2 Amps is way to much.

I figure that, if I can get the current lower I will be able to work with 0 to 1 VDC, or 100 pA to 1000 mA a little better.

Good input, though.

#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,831
It might help if you could show us how you "do calibration references." Just because your power supply can provide 2A doesn't mean it has to! OTOH, if you need a "calibrated" current source in the range of 100 pA to 1000 mA with a compliance voltage between zero and 1 VDC, a variable voltage power supply may not be the best choice. Please describe in some detail what you want to do; maybe someone here will have a suggestion on how to do it. Try to avoid the pitfall "if all you have is a hammer, everything looks like a nail." Sometimes you need another tool.

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#### Bluejets

Oct 5, 2014
6,342
No duh? The attachement is the load. I don't have enough to send to everyone.
Careful there...we offer our help from a vast knowledge base and if you choose to use words like "attached circuit" it is meant normally to mean a circuit diagram.

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
Good advice there from Hop as usual. But I should explain why he used "I = E / R" instead of "I = V / R" for Ohm's Law.

E was the original letter used for voltage. It stands for EMF, "electromotive force". I learnt Ohm's Law as "I = E / R" too. But at some time around the 1970s or 1980s I think, the convention was changed so voltage is represented by V. That's the only difference.

#### Rory Starkweather

Nov 13, 2014
77
Careful there...we offer our help from a vast knowledge base and if you choose to use words like "attached circuit" it is meant normally to mean a circuit diagram.

No. It is not. I do not have the capability to send images over the Internet, and I don't want it. My power is deliverd ever day by goats. In buckets. Remember that I Live in Misery, the armpit of the universe.

When I say 'attachment' it can only mean something that is real, not something that can be put on paper.

In this case, the 'Attached Circuit', or the 'Load' is a 1.5 kOhn resistor with a power rating or 10 Watts.

Since I only work with with currents and voltages less than 1 now, power dissipation is not a problen

#### Rory Starkweather

Nov 13, 2014
77
Good advice there from Hop as usual. But I should explain why he used "I = E / R" instead of "I = V / R" for Ohm's Law.

E was the original letter used for voltage. It stands for EMF, "electromotive force". I learnt Ohm's Law as "I = E / R" too. But at some time around the 1970s or 1980s I think, the convention was changed so voltage is represented by V. That's the only difference.

I've been doing this for 20 years now. You may have been doing it longer, but I do understand Ohm's Law. And even Ebers-Moll.

You don't seem to understand. I am neither 12 years old, nor stupid. I am just stuck with what the Marine Corps taught me. Like current, for us, does not flow in the same direction as it does for you.

Probably

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#### davenn

Moderator
Sep 5, 2009
14,179
Rory,

What Kris and Hop are trying to do is get you to state things more accurately, so that we are all on the same page. Its saves so much confusion and having to ask additional questions to figure out what is being spoken about
Don't take it as a personal attack, as you did above. Rather take it as good advice so that you have a chance to learn new things. This will help you keep up with today's state of play and make everything much easier for everyone who are asking and answering questions

Dave

#### Rory Starkweather

Nov 13, 2014
77
It might help if you could show us how you "do calibration references." Just because your power supply can provide 2A doesn't mean it has to! OTOH, if you need a "calibrated" current source in the range of 100 pA to 1000 mA with a compliance voltage between zero and 1 VDC, a variable voltage power supply may not be the best choice. Please describe in some detail what you want to do; maybe someone here will have a suggestion on how to do it. Try to avoid the pitfall "if all you have is a hammer, everything looks like a nail." Sometimes you need another tool.

Hi.

We're working on, oddly enough, vacuum tube testers. We are using a lot of D'Arsonvals (many haven't been made for 50 years.) We don't want to blow them up.

Current leakage in tubes, bad capacitors, effects of other components on the tuned, or even untuned parts of the tubes.

It actually has a relationship to modern circuitry. For instance, I spent hours today trying to figure out how to reduce a 2 Amp input to 1 Amp. There is no real answer out there. The best suggestion I could get was to use an LM317. Probably one of the least efficient chips in the world.

I think a couple of JFETs could do the job better and more cheaply, but I don't have the right data. And it looks like I won't get it without doing the research myself, and, for that, I need . . . well. You know.

I do have a limited budget, though. and thanks to a lot of you, I have wasted most of it on things that have no value for this project.

Well, as someone once said, "Caveat Emptor."

#### Rory Starkweather

Nov 13, 2014
77
Rory,

What Kris and Hop are trying to do is get you to state things more accurately, so that we are all on the same page. Its saves so much confusion and having to ask additional questions to figure out what is being spoken about
Don't take it as a personal attack, as you did above. Rather take it as good advice so that you have a chance to learn new things. This will help you keep up with today's state of play and make everything much easier for everyone who are asking and answering questions

Dave

TYVM

I'm not trying to be obnoxious, but I hate hearing, "Is it plugged in?"

I know all of you know a lot more about electronics than I do, but no one really answers any questions.

It seems to be all like, "Use my pet frog instead of the Thyratron tube. It will work better."

I'm not up for that.

Bottom line is that I have a lot of questions here, on a lot of different subjects, but this seems to go back to Democratic (not the political party) electronics.

Tell me it's true,and I will go back to experimentation, and never bother you again.

#### davenn

Moderator
Sep 5, 2009
14,179
You are not being obnoxious
and because you cant/wont post circuits/images we have no way to confirm what you are doing or what you may be doing wrong
Because of that all the questions have to start with the basics to try and figure out what is going on.

Others will post circuits and pics of the things they are building and or working on, that makes it VERY easy for others on here to diagnose possible problems and even to point out major mistakes in their wiring

You have previously admitted to poor eyesight, so it is reasonable for us to expect that you have made obvious wiring errors.

We all love to help people with their electronics .... none of us get paid for it ... we do it for the love of the hobby. And with that we need the person asking for help to help themselves and us as much as possible by supplying as much information as possible. We are not with you, we cannot pick up or look directly at the circuit you are working on, therefore you need to explain everything as clearly as possible

What are you using to access the internet ?
a PC compatible computer ? what operating system ?

Dave

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