two capacitors in series

Thread Starter

anhnha

Joined Apr 19, 2012
905
I simulated the circuit below including a voltage source 10V and two capacitors 1pF each in series.
I measured the voltage at the point between two capacitors and got the plot.
The voltage at that point increases linearly but at 100M seconds it is still very very small.
I guess that it takes infinite amount of time for capacitor C1 to get 5V.
Is that right?
How to explain it mathematically?
 

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Wendy

Joined Mar 24, 2008
23,408
Both caps charge at the same rate if they are the same value. Total capacitance is ½ the two caps (if they are the same value). In many ways it acts like a voltage divider, each cap will have half the total charge.
 

Jony130

Joined Feb 17, 2009
5,487
I simulated the circuit below including a voltage source 10V and two capacitors 1pF each in series.
I measured the voltage at the point between two capacitors and got the plot.
The voltage at that point increases linearly but at 100M seconds it is still very very small.
I guess that it takes infinite amount of time for capacitor C1 to get 5V.
Is that right?
How to explain it mathematically?
OMG are you serious?? If you don't know how to use a simulation program don't use it. Because here you have a typical example of GIGO.
To get "proper" result you need to turn on "skip initial operating point solution".
 

studiot

Joined Nov 9, 2007
4,998
I simulated the circuit below including a voltage source 10V and two capacitors 1pF each in series.
I measured the voltage at the point between two capacitors and got the plot.
The voltage at that point increases linearly but at 100M seconds it is still very very small.
I guess that it takes infinite amount of time for capacitor C1 to get 5V.
Is that right?
How to explain it mathematically?
Both caps charge at the same rate if they are the same value. Total capacitance is ½ the two caps (if they are the same value). In many ways it acts like a voltage divider, each cap will have half the total charge.
The diagram shows a direct supply.

I suggest you try the circuit with two capacitors and a 9V battery with real components and a real voltmeter.

You might find a suprise awaiting
 

#12

Joined Nov 30, 2010
18,224
If you have a resistor in there Eo = dE(e^-t/RC) where little e is a constant worth about 2.718

With no resistor, the voltage on the lower capacitor should instantly jump to 1/2 Vcc.
The formula for that is: Vcc/2 = 1/2 Vcc

If the 2 capacitors are not equal, it gets more complicated and the math works better (for me) if you think in terms of conductance.
 

Thread Starter

anhnha

Joined Apr 19, 2012
905
If you have a resistor in there Eo = dE(e^-t/RC) where little e is a constant worth about 2.718

With no resistor, the voltage on the lower capacitor should instantly jump to 1/2 Vcc.
The formula for that is: Vcc/2 = 1/2 Vcc

If the 2 capacitors are not equal, it gets more complicated and the math works better (for me) if you think in terms of conductance.
I understand this but just don't know why LTspice requires to skip initial condition to get the correct result.
If the "Skip initial operating point solution" is unchecked then the voltage across the bottom capacitor is almost zero.
 

#12

Joined Nov 30, 2010
18,224
I would think that the initial rise of applied voltage would be divided equally between the 2 equal capacitances, and, a simulator using an ideal voltmeter would measure half the applied voltage. So, I got (2) 1000 uf capacitors, connected them in series, hooked up my 10 meg input impedance volt meter and connected a new 9 volt alkaline battery to the most positive terminal of the capacitor string. The meter showed 4.84 volts and became lower as my meter discharged one capacitor.

That clears it up for me.
 

studiot

Joined Nov 9, 2007
4,998
So does the simulator show the effect of the 10MΩ?

What would happen if you used an electrostatic (non contact) meter?
 

BobTPH

Joined Jun 5, 2013
8,765
I think the problem with computing initial conditions is the way this is done. Capacitors are turned into open circuits. With both capacitors open, the node between the two has no connection to anything. In real capacitors there will be a parallel resistance or leakage which overcomes this problem. Try adding a large parallel resistance (10-100M) and see what happens.

Bob
 

alfacliff

Joined Dec 13, 2013
2,458
another possible problem with the origional circuit is that the caps are only 1pf each. making the rc time constant with non existant resistors (infinite resistance) way out of range for any simulators.
 

studiot

Joined Nov 9, 2007
4,998
This seems an odd question to be posted right after I described an experiment with real parts. Were you asking me? I don't now. I don't use simulators.
Nothing personal.

But I am achieving my objective of making people think about the circuit in both reality and the simulator.

If the centre point was perfectly isolated how could it charge to any voltage?
If the two caps were stacked up in the field of the battery you could argue that it is midway between, but if one cap was widely separated and turned at right angles to the field then...?

Equally if there is a resistance, either due to the voltmeter or capacitor imperfection then there is a charging path.
In this case you have clearly noted that the centre charges to midway then decays.
So it is a legitimate question to discuss whether the simulator reflects this real word edit : world action, and if not, how to make it so.
 
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alfacliff

Joined Dec 13, 2013
2,458
pc type smps use two caps in series to provide the initial supply voltge for the pwm chip. those are electrolytic and connected from the + rectifier out to ground. they supply voltage for a little while till the chip gets to running and powers off the supply. these are also the most likely to go bad from heat and age. just look for two small electrolytics near the 8 pin (usually) chip, check them and the values are usually in the pf range instead of several microfarads.
 

#12

Joined Nov 30, 2010
18,224
So it is a legitimate question to discuss whether the simulator reflects this real word action, and if not, how to make it so.
I'm with you on that. For my own edification, a battery and 2 capacitors shows me the practical solution. For other people, getting a simulator to give reasonable results is an important goal. I hope you can get this worked out. It will help a lot of people.
 

vk6zgo

Joined Jul 21, 2012
677
Nothing personal.

But I am achieving my objective of making people think about the circuit in both reality and the simulator.

If the centre point was perfectly isolated how could it charge to any voltage?
If the two caps were stacked up in the field of the battery you could argue that it is midway between, but if one cap was widely separated and turned at right angles to the field then...?

Equally if there is a resistance, either due to the voltmeter or capacitor imperfection then there is a charging path.
In this case you have clearly noted that the centre charges to midway then decays.
So it is a legitimate question to discuss whether the simulator reflects this real word action, and if not, how to make it so.
It makes my head hurt to think about Rhetorical questions relating to things that we know happen!:D

A related question is,how is it that we can charge two capacitors,put them in series & read the sum of the individual charges across the series circuit?

Why can we then discharge them?

Can LT Spice reproduce this?
 
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