Go Down

Topic: A few simple questions about inductors.. (Read 746 times) previous topic - next topic

David82

Using the image below as a reference,
1. What would the two graphs look like if switched from a, to b, then back to a (instead of c)?
2. Why is a resistor used in the circuit? Alternatively, what would the graph look like without it?
3. Could the first voltage spike be taken care of with a voltage regulator and the second with a diode, or would both be handled by the regulator?

dc42

Using a ideal inductor and ideal switch, when you switched from b to a you would get a spike of infinite voltage across the inductor. In practice, what happens is that you get arcing across the switch, which (together with the capacitance of the inductor windings) reduces the size of the spike.

Without the resistor in the circuit, the current in the inductor would rise linearly when switched to b, and remain constant at whatever value it had reached when switched back to c (this actually happens with superconducting coils).

If the diagram is a prelude to a discussion of switch mode voltage regulators, then the resistor represents the load.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

David82

I have an 85mA load (analog camera) that shares a power supply with a 4A load that is occasionally switched on. Even if the power supply is a car battery, the amp draw and voltage to the camera gets sucked away briefly when the 4A load is switched on causing the camera to flicker. I was thinking a 125mA, 10H inductor would maintain the current and voltage to the camera (or at least make the transitions smoother). Is this the right thinking or should I be looking at using a large cap instead?

SirNickity

#3
Jan 16, 2013, 01:42 am Last Edit: Jan 16, 2013, 01:46 am by SirNickity Reason: 1
2. Why is a resistor used in the circuit? Alternatively, what would the graph look like without it?

Well, an inductor has high impedance to high frequencies, but low impedance with low frequencies.  So without a resistor... well, never mind the graph.  Here's what the circuit would look like though:


3. Could the first voltage spike be taken care of with a voltage regulator and the second with a diode, or would both be handled by the regulator?

You mean, reducing the first spike by limiting the current?  That might work, but sizing it right would be difficult and may end up with nuisance tripping of any protection circuitry.  On the other hand, you could put a cap after the inductor to help smooth the output voltage as seen by any downstream load.  Fixing the negative spike is easy.  Put a forward-biased diode from the voltage source to the inductor to protect the source from back EMF.  Then, put another diode in inverse parallel across the inductor so it is not conducting during normal use, but will conduct the negative spike back through the coil until it dissipates.

Edit:  Sorry for the huge image.  Browser scaled it down to reasonable size when I originally found it.

Also, yeah.. and inductor and cap combo make a good isolator.  Otherwise, you can indeed have large loads drawing current from the sources of least impedance (nearby filter caps) instead of the battery.  Another option is to feed the low-current section through a series resistor, with its own local supply cap.  This will make the battery a lower-resistance (therefore, more attractive) supply in comparison.  The drawback is of course a slower rise time at the low-current supply pins.  Make sure any ICs are reset properly on power-up, or have brown-out protection and/or startup delays as appropriate.

Docedison

Two Quick things:
1. The R in the circuit represents the circuit resistance or the curve would look as you describe.
and
2. Use a large capacitor... > 5000 uF  and if the wire run was long I'd use a great deal more... 10X more.
Or more. The point is that the capacitor is invisible until the load hits and then it's invaluable.

Bob
--> WA7EMS <--
"The solution of every problem is another problem." -Johann Wolfgang von Goethe
I do answer technical questions PM'd to me with whatever is in my clipboard

Go Up