Basic voltage question in circuit.

[.RC.]

I am just learning still and after watching a youtube video on astable multivibrator circuits I was playing with this simple circuit on a breadboard and was probing voltages with an oscilloscope.



I was wondering where the extra voltage is coming from

If you look at the circuit I have drawn,  I put the probe ground at ground as shown.

Then with the switch closed I measure the voltages at point A, B and C.

A is 0V as expected.   B is 10V and C is 10V after the capacitor charges.

I then open the switch which removes the negative side.   Point A, immediately jumps to +5V, B remains at 10V and C jumps to +15V.

Then the voltages at Point A slowly rise above 5V to 10V and Point C drops from 15V back to 10V, so there is 10V everywhere as expected.

[wasedadoc]

Capacitors resist change in voltage across them.  They try to keep the voltage across themselves from changing.  The larger the capacitance value the more they slow down the change in voltage.

With the switch closed the capacitor charges up to 10 Volts.  Its bottom end is at 0 Volts.  You open the switch and its bottom end is only connected to the 1k resistor.  That is now at a higher voltage than 0.  So the top side of the capacitor jumps up.

[.RC.]

Thanks, I think I now understand why the Point C Jumps to 15V.   The capacitor had +10 on one side and 0 on the other.   Then all of a sudden Point A turns to +5V, which means the negative side of the capacitor is at +5V, The other side of the capacitor was at +10V and needs to increase to +15V to maintain 10V across the positive and negative pins of the capacitor.

But I still do not know where the +5V at point A comes from as it seems to appear absolutely instantly and point B remains rock solid at 10V.  Or is it a leftover of the current that was flowing, then when the switch is made open the voltage rises instantly.

Much like water hammer.   You have water flowing in a pipe, then if you turn off a tap instantly the pressure rises as the water has momentum.

[wasedadoc]

Thanks, I think I now understand why the Point C Jumps to 15V.   The capacitor had +10 on one side and 0 on the other.   Then all of a sudden Point A turns to +5V, which means the negative side of the capacitor is at +5V, The other side of the capacitor was at +10V and needs to increase to +15V to maintain 10V across the positive and negative pins of the capacitor.

But I still do not know where the +5V at point A comes from as it seems to appear absolutely instantly and point B remains rock solid at 10V.  Or is it a leftover of the current that was flowing, then when the switch is made open the voltage rises instantly.

Much like water hammer.   You have water flowing in a pipe, then if you turn off a tap instantly the pressure rises as the water has momentum.

When point C jumps up to a higher voltage than 10 volts there is voltage across the top 1k resistor.  That makes current flow in that resistor.  The current also flows through the capacitor and the bottom 1k resistor.  As the two resistors are equal value and have the same current flowing through them, the voltage across each must be the same.  That condition is met with 5 Volts across each.  The bottom end of the bottom resistor jumps to 5 Volts, the top of the capacitor jumps to 15 Volts and the top resistor has 5 Volts (15-10) across it.

Try changing the ratio of resistor values and observe what happens.

[.RC.]

OK, I think I am getting it.

It is the 10V input supply that causes the jump of the bottom to 5V because of the resistor value.  Change the resistor value (I have not done this yet) it will be a different result. Change to a 500R and then would it be 3.3333333333V or would it be 2.5V  or the recipricol (6.66666 or 7.5 as it is a lower resistance). hmmm some sort of voltage divider.

[bdunham7]

It might be conceptually easier to imagine the capacitor is a 10V battery.  Then figure out what the current flows are in the resistors with the switch closed and then open, then calculate the voltage drop across those resistors in both cases.

[.RC.]

I ended up downloading LTSpice and running it in that. That way I can see what is happening.

I also changed the top resistor to 750R and called it R1

Bottom resistor to 1K and called it R2.

The capacitor was changed to a floating (not connected to ground) 10V voltage source and called V3.

The switch is closed.   Current flows from the initial 10V voltage source, through R2, out the switch to ground.

Zero current flows through R2.

Open the switch.   Current now flows though the V3 voltage source, via R1 and R2.

The voltage increases in reference to ground across V3 by 4.28V so the negative side is +4.28 and the positive side is +14.28.

Why 4.28V.    Because of the voltage divider formula 750/(750+1000)*10 =  4.28V

But it does not resemble a voltage divider circuit.

Also two resistors in series add up the resistance, so a 10V source, 1750 ohms resistance means 0.0057A flowing, and the law that says current flowing in, equals current flowing out.    So is there is 10V source at one side of R2, and 0.0057A flowing, Ohms law says 4.28V.

But I am still trying to wrap my head around it.

[AnalogTodd]

I ended up downloading LTSpice and running it in that. That way I can see what is happening.

I also changed the top resistor to 750R and called it R1

Bottom resistor to 1K and called it R2.

The capacitor was changed to a floating (not connected to ground) 10V voltage source and called V3.

The switch is closed.   Current flows from the initial 10V voltage source, through R2, out the switch to ground.

Zero current flows through R2.

Open the switch.   Current now flows though the V3 voltage source, via R1 and R2.

The voltage increases in reference to ground across V3 by 4.28V so the negative side is +4.28 and the positive side is +14.28.

Why 4.28V.    Because of the voltage divider formula 750/(750+1000)*10 =  4.28V

But it does not resemble a voltage divider circuit.

Actually, you need to think about things as suggested--think of the floating 10V source as a battery. You can stick a number of batteries in series to add their voltages, each one has its own negative terminal but that terminal is not necessarily system ground. It is ground from what the battery can tell, but it isn't the actual system ground. From that point, think of this as two separate systems--one when the switch is closed, the other when the switch is open. With the switch closed, you can figure out that R1 has 10V at each end (relative to system ground) and as such zero current flow. R2 will have 10V across it and run 1mA. Now open the switch and review the schematic again.

To start, remove V1 and look at what you have with V3 and the two resistors; it's a voltage divider. You've got those equations down, and figured out that with V3 across the two resistors where the midpoint is because of Ohm's law and Kirchoff's laws. Now, add V1 back in (you can do this because of superposition). Try and find a current path from the top of V1 looping back through to its bottom (assume your switch is truly open, not 1Meg as shown). There is no path for current to take.

Also two resistors in series add up the resistance, so a 10V source, 1750 ohms resistance means 0.0057A flowing, and the law that says current flowing in, equals current flowing out.    So is there is 10V source at one side of R2, and 0.0057A flowing, Ohms law says 4.28V.

But I am still trying to wrap my head around it.

Voltage source V1 forces the midpoint of the two resistors to be at 10V. No current flows from V1 into R1 or R2, (since current in equals current out), so those currents remain constant at what you calculated for the voltage divider circuit, and the voltage drops on them remain constant. Because you have 4.28V across R1, that forces the top of R1 to be at 4.28V above the positive side of V1 (which is 10V), giving you the 14.28V. Because V3 forces the bottom of R2 to be 10V below the top of R1, it sits at 4.28V.

[.RC.]

Thanks.

If I visualise it using the water analogy it become more clear.

V3 is a pump that takes whatever is at it's bottom and adds 10V to it to maintain the 10V difference between it's top and bottom.   So if there was +2V at it's bottom, in reference to ground there is now +12V at the top.   But there is still 10V across the battery.

The switch goes open.  Previously there was 0V below R2 and now V1 wants to rise the voltage below R2 and at the bottom of V3 to 10V if possible.

This causes voltage to rise at the bottom of V3, This in turn causes the voltage at the top of V3 to rise as it wants to maintain 10V across itself and the example of batteries in series, the voltage adds up.   So now there is a higher voltage then 10V at the top of R1, and a lower voltage at the bottom of R1.   So now current can finally flow across R1 and then the voltage stabilises because we can think of the circuit as simply two resistors in series and calculate the current flowing.  We know the V3 is always 10V across itself alone.  Then calculate the voltage drop across R1, and we know it must be that amount higher then V1.   Then back at V3, that voltage figure we just calculated, must be 10V lower at the bottom of V3.

Well that took awhile but the video I watched , never explained any of that.