# Non-changing current with C-EMF

Discussion in 'General Electronics Discussion' started by XRZ, Jul 30, 2014.

1. ### Gryd3

4,098
875
Jun 25, 2014
You should grab
There is always a catch
In an ideal world, a multi-meter would have an infinite resistance when measuring voltage, 0 resistance when measuring current. Oscilloscope leads would have 0 capacitance...
With many components, there are characteristics such as slew-rate http://en.wikipedia.org/wiki/Slew_rate that dictate how fast something will react.
That of course could be used to your advantage to to prevent c-emf from propagating through certain components.

5,165
1,087
Dec 18, 2013
Every component is ideal. Remember in the real world every component has LCR.

Gryd3 likes this.
3. ### XRZ

78
0
Jul 30, 2014
@Gryd3, @Arouse1973

As long as current is constant 90% of the time(which I think it would be with a CCS) I think that's "idea" for me, nothing can be 100%.
I know there are losses, and efficiency issues... But as long as current is constant most of the time that's great for this experiment. I know that the constant current source would react to maintain current, but how fast? I guess it's fast. Equal or a bit less than counter-emf's induction.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
In theory there is no difference between theory and practice. In practice there is.

Arouse1973 likes this.
5. ### XRZ

78
0
Jul 30, 2014
"Practice = losses"
Well most components nowadays are high in efficiency so I assume I can have the current constant like 95% and more possibly...?

6. ### XRZ

78
0
Jul 30, 2014
So...how fast would the constant current source react to induced-counter emf?
Also in both my example the range of voltage would be the applied EMF + counter-emf for the constant current correct? So for example if the applied voltage prior to counter emf was 2V and CEMF is 1V the voltage range would be 1V - 3V?

@BobK @(*steve*) @Gryd3 @Arouse1973

7. ### Gryd3

4,098
875
Jun 25, 2014
I can't give you an answer, as it is heavily dependant on the components in the circuit, and in the constant current supply..

8. ### XRZ

78
0
Jul 30, 2014
@Gryd3 Yea that's right... say you had such a system and you picked the components(by choosing the components you feel are right), could you guess/ or approximate it?

I mean, I know we can't put a time frame here(could we guess it?), is it possible to have the constant current source change it's voltage fast enough to maintain current? Having it react fast enough to respond to the change in voltage due to the induced-counter emf?

9. ### XRZ

78
0
Jul 30, 2014
Anyone know the price range of such power supplies, or how much the components could cost at the end(estimations anyone)? If I'm interested in building/buying them?
@Gryd3 @Arouse1973 @(*steve*)

10. ### Gryd3

4,098
875
Jun 25, 2014
@XRZ, it's a lot of theoretical work that would most likely be useless... I think it's time for a real-world test.

As far as buying them is concerned. They could be anywhere from \$3-4 to \$100s...

The biggest difference for you when you look at them is the current they are set to, and the upper limit of the input/output voltage.

11. ### BobK

7,682
1,688
Jan 5, 2010
"Fast enough" is a useless description. Any constant current source will track changes in the load and compensate within some amount of time. This might be 1uS or it might be 1S. Either of these are "fast enough" for someone's application. What is your requirement?

Bob

12. ### XRZ

78
0
Jul 30, 2014
One of the reasons I asked about "adding" voltage to a circuit(in another thread), is because I was wondering of an idea to just have a high voltage source(with exceptionally low current) aside from the constant current source that's purpose is just to cancel out with induced counter emf.

So that now there are multiple power source, which I assume can add up? P1(constant current) + P2(voltage source) = total Power in the circuit?

13. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
Nope. What you say makes no sense.

14. ### XRZ

78
0
Jul 30, 2014
Can't there be a secondary power source that just adds high voltage?
A secondary power supply that has a high voltage and low current to deal with counter-emf.

15. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
Nope. What you say makes no sense.

16. ### XRZ

78
0
Jul 30, 2014
What's wrong? What don't I understand that seems to make no sense?
I'm pretty sure you understood what I meant, but to a circuit it's nonsense?

Last edited: Oct 14, 2014
17. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
There is no way to "add" a high voltage, low current source and a low voltage, high current source to get both high voltage and high current at the same time.

XRZ likes this.
18. ### XRZ

78
0
Jul 30, 2014
How is this possible?
A diagram:

The first conductor and second, are in a changing magnetic field. From the diagram above it seems that they increase current in opposition while having the same induced counter EMF.

Yet in this diagram without the PS, the cancel out:

Of course both of them in a changing magnetic field.

19. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010

The main issue that I see is that you absolutely want the current to vary in an electric motor.

If you always had the stall current flowing, even when the motor was running with no load, you would be very sad.

20. ### BobK

7,682
1,688
Jan 5, 2010
Yes, that has come up before. People think back EMF is a bad thing. It is not, electric motors would not be practical without it.

Bob

Arouse1973 likes this.