Question About CCCV And Ohms Law

[EasyGoing1]

I've been trying to wrap my head around the concept of a power source that provides constant current / constant voltage as it relates to ohms law ... allow me to explain ...

Say we have a very simple circuit ...



Ohms law says that the current in that resistor should be 6 amps ...

But what if we have a clamped source current at 5 amps? IOW, the power supply will not supply more than 5 amps of current ... according to Ohms law, the voltage drop across R would have to be 15 volts ... but if the power supply is supplying a CONSTANT voltage of 18 and a CONSTANT current of 5 amps ... how is this possible? How does it work? Obviously the power supply isn't going to magically supply less voltage simply because it only provides 5 amps to a 3 ohm load ... or does it? If so, WHY?

The only thing that makes it potentially possible in my head ... to have 18 volts and 5 amps feeding a 3 ohm resistor (which in lab practice should be absolutely do-oable) ... without violating Ohms law, is the fact that the power dissipation over a load with 18 volts and 5 amps is 90 Watts ... and assuming this load can handle 90 watts, then things should be fine ... but what of V=IR ???

I cant reconcile this in my head and I lack the equipment to set up a test so can anyone chime in on this?

Thank you,

Mike Sims

[Andrew F. Ali]

Re: Question About CCCV And Ohms Law

You are correct, the voltage across the 3Ohm load will 15V and Current will be at 3Amps

[budm]

Re: Question About CCCV And Ohms Law

Constant current source power supply: the output Voltage will vary to maintain the constant current through the load when the load resistance is changed, the output Voltage is NOT constant.
The Constant Voltage power supply will maintain constant output Voltage, when the load resistance is changed the current will change but the Voltage will stay the same.
Do not get confuse between constant Voltage power supply and the constant current power supply.

[budm]

Re: Question About CCCV And Ohms Law

BTW "CONSTANT voltage of 18 and a CONSTANT current of 5 amps" What is the load resistance? It should be 3.6 Ohms. How did you measure this 5A of current with 3 Ohm load with 18V supply to get your 5A reading?

[stj]

Re: Question About CCCV And Ohms Law

the problem is the use of words.
they arent "constant" current / voltage - they are programmable "maximum" current / voltage.
they cant go over, but they can drop bellow.

[Andrew F. Ali]

Re: Question About CCCV And Ohms Law

BTW "CONSTANT voltage of 18 and a CONSTANT current of 5 amps" What is the load resistance? It should be 3.6 Ohms. How did you measure this 5A of current with 3 Ohm load with 18V supply to get your 5A reading?

Therefore, if the supply is Constant Voltage at 18V then the Current I amps will be 6amps through a 3Ohm load.
If the supply is a Constant Current at 5A then the Voltage V volts will be 15V across the 3Ohm load

[keeney123]

Re: Question About CCCV And Ohms Law

I've been trying to wrap my head around the concept of a power source that provides constant current / constant voltage as it relates to ohms law ... allow me to explain ...

Say we have a very simple circuit ...



Ohms law says that the current in that resistor should be 6 amps ...

But what if we have a clamped source current at 5 amps? IOW, the power supply will not supply more than 5 amps of current ... according to Ohms law, the voltage drop across R would have to be 15 volts ... but if the power supply is supplying a CONSTANT voltage of 18 and a CONSTANT current of 5 amps ... how is this possible? How does it work? Obviously the power supply isn't going to magically supply less voltage simply because it only provides 5 amps to a 3 ohm load ... or does it? If so, WHY?

The only thing that makes it potentially possible in my head ... to have 18 volts and 5 amps feeding a 3 ohm resistor (which in lab practice should be absolutely do-oable) ... without violating Ohms law, is the fact that the power dissipation over a load with 18 volts and 5 amps is 90 Watts ... and assuming this load can handle 90 watts, then things should be fine ... but what of V=IR ???

I cant reconcile this in my head and I lack the equipment to set up a test so can anyone chime in on this?

Thank you,

Mike Sims

Think of the battery as a variable resistor that supplies voltage. For constant current one can vary the resistor to maintain the current. So if I were to vary the load to reduce the current the only thing I can do to maintain the current is vary the resistance of the voltage source to maintain the current. With constant Voltage one now needs to have enough current to maintain the voltage source. This is not a infinite thing. There is a limit to how much power any supply can provide before the voltage is effect and there is a limit to how much current a power supply can provide before it affects the supply. The two of these have a crossover point. When you try to supply to much current for the voltage then the Voltage will be affected. When you try to supply to much voltage for the supply then the current will be affected. You can not have both constant voltage and current at the same time from a supply at any resistance. This constant voltage and constant current at the same time is a theoretical point. Kind of like a point of singularity.

[EasyGoing1]

Re: Question About CCCV And Ohms Law

BTW "CONSTANT voltage of 18 and a CONSTANT current of 5 amps" What is the load resistance? It should be 3.6 Ohms. How did you measure this 5A of current with 3 Ohm load with 18V supply to get your 5A reading?

I didn't measure anything ... I was merely generating discussion and trying to understand this whole CCCV thing ...

[EasyGoing1]

Re: Question About CCCV And Ohms Law

Thank you, everyone who took the time to reply.

So what I'm just now understanding, is that a CCCV power source is:

Either supplying a constant current at a given level OR a constant voltage at a given level, but never both, and always in compliance with Ohms law, since it's impossible to violate it in the first place. Voltage is adjusted in CC mode and current is adjusted in CV mode. Seems simple enough.

[televizora]

Re: Question About CCCV And Ohms Law

I like the simple explanations:
--If the current is a water flowing via a pipe, then the voltage is the water pressure, the resistance is the obstacles and the diameter of the pipe and the water debit is the current itself.
So - less pressure on the same pipe means less water debit.
Same applies for electrical circuits. The difference in the potentials between the points is the thing, that is pushing the particles thru the conductors.(the wires)
The nature tend to balance itself and this leads to compensation processes like the wind or...electricity.
About your example. If the resistance is a constant(pure resistor this means), then less voltage means less current. The voltage drop on the resistor is U=I.R, but I=U/R So less voltage means less current if the keep the resistance the same.
If the voltage of the source is 18V, and the resistance is 3Ohm, then the current will be I=U/R = 18/3=6A The voltage drop across the resistor will be U=I.R = 6.3=18V. If the source voltage is lower, less current flows. The voltage drop is less.

Either supplying a constant current at a given level OR a constant voltage at a given level, but never both, and always in compliance with Ohms law, since it's impossible to violate it in the first place. Voltage is adjusted in CC mode and current is adjusted in CV mode. Seems simple enough.

Switched-mode PSU-s compensates for the load. The utilize PWM. The energy they give is proportional to the duty cycle of the PWM, which is automatically controlled, so drops in the voltage are compensated. But this is only till you reach the full capacity of the supply. Then - overload protection. This is only true for the switched-mode supplies. Classical ones with diode bridge and transformer don't compensate the variations of the voltage due to variations of the resistance of the load.