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[on:]

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?

[Reply #1 on:]

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.

[Reply #2 on:]

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?

[Reply #3 on:]

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.

[Reply #4 on:]

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

[Reply #5 on:]

Well, an inductor has high impedance to high frequencies, but low impedance with low frequencies.

I understand you so far..

So without a resistor... well, never mind the graph.  Here's what the circuit would look like though:

I see that it burned up but I don't know why it burned up.

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?  

No, on the voltage graph I see a large, positive spike in voltage. So just add a 12v regulator right after the inductor to limit it to 12v right?
The other spike is negative voltage. Well, a diode will stop that right?

Also, yeah.. and inductor and cap combo make a good isolator.  

Oh, that's the name for what I'm trying to build..

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.  

I see. So you're making the battery the path of least resistance so that it doesn't suck the current and voltage from the camera. Won't the camera still be deprived of current though? As in, the current would be momentarily diverted to the the high amp load starving the low amp load when the high amp load is switched on?

[Reply #6 on:]

The picture is kind of a joke.  If you take an inductor and put it in parallel with a battery, you have immediately high resistance, and very soon after, practically zero resistance.  In essence, you short-circuit the supply.  The resistor is there to show a tangible load -- since, without it, you've only created a battery destroyer.  Or a space heater.

OK, so you're thinking putting a 12v regulator after the inductor would "regulate away" the positive spike.  Well, two problems with that.  First, the spike could very well be excessively high-voltage, subjecting the semiconductors inside the regulator to a good old fashioned pummeling.  It's not the method of choice, at least.  Second, depending on the transient response of the regulator, it could slip right through the feedback loop and on to the load.  Typically, fast diodes with high voltage tolerance (100-1000v) are used instead where transient suppression is necessary.  In reality, whether the load "sees" that positive spike or not is questionable.  But if the application deems that it does, chances are a linear regulator will be just as susceptible to damage.  Sorry, nothing's that easy. 

I see. So you're making the battery the path of least resistance so that it doesn't suck the current and voltage from the camera. Won't the camera still be deprived of current though?

Yes, which is why all parts of your circuit need their own local supply filters.  To ensure they are private reserves, it is sometimes necessary to isolate them from each other.  Since your low-power loads will have fairly consistent (and low) demands for current, one method is to keep its voltage supply at a higher impedance than the battery.  As also mentioned, another technique is to bulk up on capacitance such that the startup draw of the high-power load is entirely satisfied by local capacitance.  Of course, then the caps need to recharge, so some PSU isolation may be warranted anyway.  IMO, you're on the right track using inductors.  They provide high resistance during surges and low resistance during idle times -- perfect for allowing local supply caps to do their jobs.

[Reply #7 on:]

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?

An LC network to isolate the camera from short-lived changed in the supply voltage would probably do the job, but the inductor would be huge. I would use a power diode and series resistor of about 1 ohm instead of the inductor. The capacitor probably needs to be quite large, maybe 10000uF.

The idea is that when the 4A load is turned on and the power supply voltage dips briefly, the capacitor goes on supplying the camera. The diode prevents the capacitor discharging through the 4A load. The resistor limits the surge current through the diode when you first connect it up to the car battery and the capacitor charges. You will lose a little less than 1V in the diode and resistor.