Question about a resistor before a capacitor vs after

[uk.casmith]

Hi,

I'm trying to understand why placing a resistor after the capacitor doesn't cause the power to drop on the other side of the capacitor.

Hard to explain this but if you see the photos it hopefully makes more sense.

In photo 1 you can see that I have the resistor before the capacitor and when I press the button to allow the capacitor to drain causing the other side of the circuit to drop to zero volts for a few miliseconds. However, when I move the resistor past the capacitor (photo 3) and press the button it's not causing the voltage drop?

Is it because the resistor is causing the capacitor to drain more slowly? Thus it's not affecting the otherside?

Thanks,

Cameron.

[radiogeek381]

Redraw your schematic(s) then remove the cap from the circuit.  That will show you what the "eventual" node voltages should be.

You could even simplify the analysis further, and remove the diode, but leave it in if that makes things clearer.

[uk.casmith]

I think I understand now. If the resistor is after the switch then the capacitor will always keep 5 volts because its the resistor that is dropping all the voltage.

If its before the switch then the capacitor keeps 5 volts when the switch closes to ground because the resistor is before the capacitor its dropping all the voltage because it has a path to ground now, and the capacitor has zero volts.

If that makes sense?

Cameron

[radiogeek381]

Hmmm... We're making progress.  But I'm not sure we're there yet.  Perhaps I misread the pictures of your setup.

I think the two circuits are like this:
```
                              A
+V -+-------R1------+-------LED-----------+--- GND
       |                      |                              |
       |                      |                              |
       |                      C                             |       
       |                      |                              |
       |                      |                              |
       +------R2-------+--------SW-----------+
                              B

                              X
+V -+------R1-------+------LED-------------+--- GND
       |                      |                              |
       |                      |                              |
       |                      |                              |
       |                      |                              |
       +------C--------+-----R2-----SW------+
                             Y
```

Did I get that right?   And you're measuring nodes B and Y with respect to ground?

If so, here's a quick way to analyze the circuit *after all transients have settled*.

0. Note that for node "B", when the switch closes B is connected to ground.
1. Remove the capacitor C, as at stasis there will be no dV/dT (no change in voltage) on any node (this is a purely DC/transient circuit -- no oscillations.)  Since A's voltage isn't changing, then there is no current through C.  If a branch isn't carrying any current, then we can safely ignore it in this analysis.
2. That leaves node "A" at the LED's forward voltage drop Vf. (assuming +V is higher than that, and that R1 is "reasonable." 
"reasonable" is a tough word here.  Diodes don't really "turn off" -- there will always be some current through the diode as long as the voltage across the diode is non-zero.  If R1 is really really big, the drop across it will be big, but not so big that node A makes it to ground.  If R1 is really really small, then the diode will *really* turn on.  At this point, almost all diodes are light emitting, though for a very brief time.  After that time, the diode will be an ex-diode (pushing up daisies) and A will be +V. A "reasonable" range for R1 would be whatever makes +V / R1 end up between a few 10s of microamps and 15 or 20 mA.


3. In the case of nodes X and Y -- they will both be at the same potential.
   a) At one extreme, they'll look like a voltage divider if V * R2 / (R1 + R2) is less than about Vf for the LED: as a no-calculation-approximation the LED is "off." (Note to pedanticists -- yes, Vf is a function of current, but we're not at that level of sophistication here.). X will be pretty near V * R2 / (R1 + R2).
   b) At the other extreme, if V * R2 / (R1 + R2) on its own is greater than Vf for the LED, then X will be pegged at Vf. 

If I misread the pictures, then the conclusions above are wrong.  However, the analysis approach is reasonable:

1. Assuming stasis, remove all capacitors.
2. Analyze the circuit assuming the LED is "off" or out of the circuit.
3. Analyze the circuit assuming the LED is "on" and the drop across it is Vf.

Did that help? Did I read the photos correctly?