# Adding voltage to a circuit?

Discussion in 'General Electronics Discussion' started by XRZ, Oct 12, 2014.

1. ### XRZ

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Jul 30, 2014
Well, if multiple power sources can be in one circuit. Can there be a voltage source that just "adds" voltage to the circuit?
A voltage dump if that makes sense. Adding voltage with the least current possible to a circuit. Is there such a thing?

2. ### Gryd3

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Jun 25, 2014
What application?
Adding voltage without adding current can't be done... as current would be the result of the new voltage and the current circuit.

3. ### davennModerator

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Sep 5, 2009

None of your post really makes any sense ??
can you please be a lot clearer in what you are thinking about
draw a circuit ... show us what you mean by ---> if multiple power sources can be in one circuit
and by ---> Adding voltage with the least current possible to a circuit.

Dave

4. ### XRZ

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Jul 30, 2014
Well, I don't have anything specific just another question that helps me learn a bit about certain circuits in electronics.
Say we have a power source that could apply 20VDC at 1A to a circuit that has load(L), now could we add another power-supply or source that just adds more voltage to the circuit at low currents. So we have 20W in our circuit, can 15W(15V at 1A) be additionally added? Where only the voltage is increased additional the to the first supply.

5. ### Colin Mitchell

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Aug 31, 2014
The voltage you are adding must not have the negative lead connected to the negative of the other voltage. It must be a separately derived voltage (supply).
Now you can add the new voltage (supply) by putting the negative of the new voltage (supply) on the positive rail of the old voltage and the new voltage will be added to produce a new higher voltage.

6. ### davennModerator

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Sep 5, 2009
I suspect Colin misunderstands what you are thinking

no doesn't work like that

if you have a circuit that is drawing 1 A at 20VDC ( that means the load resistance is 20 Ohms) and you add say another 10 VDC supply so now you have 30 V across that load, the current through the load is going to increase to 1.5 A and the load ( depending on what it is) is probably going to burn up

its basic Ohms Law

Dave

7. ### BobK

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Jan 5, 2010
Also note that if you add voltage sources in series, the current limit will be that of the lowest source.

Bob

8. ### XRZ

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Jul 30, 2014
Well, what about a constant voltage source that is independent of the load's resistance?
I understand its basic ohm's law, but some circuit designs seem to confuse me as to how more voltage can be outputted into the circuit with lower current. Using a converter for example.
But then again... a converter is not relevant to ohms law?

Last edited: Oct 13, 2014
9. ### Gryd3

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Jun 25, 2014
V=IR .
Typically, one of these values is pre-determined... like the resistance of a circuit and cannot be changed.
This leaves I and V that can be manipulated... Not at the same time though or the above formula would be false.
If you increase voltage, the current will increase as well. The only way around that is to somehow decrease the resistance of a circuit.
Same thing with manipulating current... if you add 200mA, the voltage will go up. Once more the only way around this is to change the circuit's resistance.

Here's the catch with a converter... Where do you connect it to the circuit?
If you give a converter 12V, it spits out 6V.... now the circuit will pull the current from the 6V side of the converter, which will have an inverse affect on the 12V side of the converter. (ie, pulling 1A from the 6V side, will only pull 0.5A into the 12V side)
This requires two formulas (or one if you merge them together)
V=IR on the circuit side. (Voltage = Current x Resistance)
P=IV to determine the current draw at the 12V side of the converter. (Power = Voltage x Current)

10. ### XRZ

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Jul 30, 2014
Well, that's the thing that's what I'm focusing on. Using converters to output high voltages at lower current to a circuit.

Example: 1A at 12V input, 0.4A at 30V output to the circuit to add a higher voltage to a circuit(lets use the values of my previous example): We have 1A at 20VDC and would like to "add" higher voltage from another power source. Say a 12W power source, using a converter output 0.4A at 30V to the circuit or load. Which should have a total power of 12W + 20W, or am I'm doing something wrong?

Last edited: Oct 13, 2014
11. ### Gryd3

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Jun 25, 2014
You are assuming that you can control the voltage and the current into the circuit at the same time.
If you build an example circuit that has a 1KΩ resistance, you should be able to calculate the current through the circuit when using a 12V source... now if you decide to add an additional 6V to the circuit, you can then calculate the new current through the circuit. (18V through the circuit)
The 'converter' you mention will only change the input voltage and current to the converter itself... the output of the converter (when set to a specific voltage) will always put out the amount of current the circuit is demanding.

Some devices that are only capable of putting out a certain amount of current... like 200mA for example... if the resistance of the circuit is too low, the current demanded by it will be too high, at this point one of two things happen. The voltage source dips, which will in turn reduce the current draw by the circuit, or the supply will cut out resulting in NO current through the circuit.

12. ### XRZ

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Jul 30, 2014
Right, and therefore that's wrong since ohms law must apply. But what if there are two power sources? One applying high current while the other just applies high voltage?

And the circuit's demand current can be calculated from the resistance of the circuit. But what if the power supply or input power to the converter is 12W, yet the output voltage is set to 12V, the resistance is 1Ω current can't be 12A... so what is the outcome here?

I think this kinda answers my question above but not sure. Can't it just have high current? With low current?

Last edited: Oct 14, 2014
13. ### davennModerator

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Sep 5, 2009
As already explained several times .. it doesn't work that way
Ohms law determines the current flow in a circuit for a given voltage and resistance

14. ### XRZ

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Jul 30, 2014
Bear with me here, I'm starting to learn ohms law a bit better.
What if the resistance is 1Ω and the voltage is 12V yet the power supply is 12W? The current can't be 12A... Does that mean current is equal to square root(12/1) ? And voltage too to equal 12W?

15. ### davennModerator

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Sep 5, 2009
No, the power dissipated by the resistance is 12W

Yes it is ... why would you think otherwise ?
that's what Ohms law states ( assuming the power supply can supply 12 Amps ... if it cannot, then the power supply voltage will sag ( drop) )

I = V/R = 12V / 1Ω = 12A

Dave

Last edited: Oct 14, 2014
16. ### XRZ

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Jul 30, 2014
Got it, thanks.

17. ### davennModerator

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Sep 5, 2009
good

so just to make sure you have

if you have 2 x 1Ω resistors in series across 12V, what is the current through the resistors ?

Dave

18. ### XRZ

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Jul 30, 2014
2 resistors in series that mean's 12/(1Ω + 1Ω) = 6A.

19. ### davennModerator

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Sep 5, 2009
cool well done

so you see for a given voltage, if you double the resistance, you will 1/2 the current
if you 1/2 the resistance ... go from 1Ω down to 0.5Ω you will double the current

now if you keep the same resistance but double the voltage, then you will double the current

Dave