was thinking about AC, middle voltage peak as 6V

Thanks

the midpoint between cap has to be 0v ? if not 0 then dc current is flowing through cap

If the midpoint is at 0 V, then the bottom capacitor has 0 V across it and the top cap has 12 V across it.

Remember, current flow in a capacitor is related to CHANGES in the voltage across it. As long as the voltage isn't changing, there's no current flowing.

Imagine I take one 1 uF capacitor and charge it to 20 V. I take the other 1 uF capacitor and charge it to -8 V. I now put these in series. The total voltage across both is 12 V. I now hook them up to a 12 V battery. Because the voltage across the battery is the same as the voltage across the two capacitors, no current will flow and they will just sit there looking stupid.

The point is that if you just give me a circuit that has a 12 V source connected to two 1 uF capacitors in series, there is no way for me to determine what the voltage is on the node between the caps unless I make assumptions which may or may not be valid. Simulators generally won't make those assumptions and so they aren't even able to start the simulation because, to do so, they need to know the initial operating point for all components and all nodes in the circuit. Having a DC path to ground for all nodes ensures that this is possible.

Having said that, some simulators have a mode of trying to determine a DC operating point even with floating nodes by starting all supplies in the circuit at zero and assuming that all nodes are at zero voltage and no current is flowing anywhere in the circuit. They then slowly ramp up the DC supplies to the specified outputs. This is generally a long process and it not guaranteed to work.

 

If the midpoint is at 0 V, then the bottom capacitor has 0 V across it and the top cap has 12 V across it.

Remember, current flow in a capacitor is related to CHANGES in the voltage across it. As long as the voltage isn't changing, there's no current flowing.

Imagine I take one 1 uF capacitor and charge it to 20 V. I take the other 1 uF capacitor and charge it to -8 V. I now put these in series. The total voltage across both is 12 V. I now hook them up to a 12 V battery. Because the voltage across the battery is the same as the voltage across the two capacitors, no current will flow and they will just sit there looking stupid.

The point is that if you just give me a circuit that has a 12 V source connected to two 1 uF capacitors in series, there is no way for me to determine what the voltage is on the node between the caps unless I make assumptions which may or may not be valid. Simulators generally won't make those assumptions and so they aren't even able to start the simulation because, to do so, they need to know the initial operating point for all components and all nodes in the circuit. Having a DC path to ground for all nodes ensures that this is possible.

Having said that, some simulators have a mode of trying to determine a DC operating point even with floating nodes by starting all supplies in the circuit at zero and assuming that all nodes are at zero voltage and no current is flowing anywhere in the circuit. They then slowly ramp up the DC supplies to the specified outputs. This is generally a long process and it not guaranteed to work.

Thanks..although only the first 3 paragraphs understood well

Thinking again, midpoint can’t be 0..if 0 then the dc that charged first cap to 12v must have passed through second cap too

 

If the midpoint is at 0 V, then the bottom capacitor has 0 V across it and the top cap has 12 V across it.

Remember, current flow in a capacitor is related to CHANGES in the voltage across it. As long as the voltage isn't changing, there's no current flowing.

Imagine I take one 1 uF capacitor and charge it to 20 V. I take the other 1 uF capacitor and charge it to -8 V. I now put these in series. The total voltage across both is 12 V. I now hook them up to a 12 V battery. Because the voltage across the battery is the same as the voltage across the two capacitors, no current will flow and they will just sit there looking stupid.

The point is that if you just give me a circuit that has a 12 V source connected to two 1 uF capacitors in series, there is no way for me to determine what the voltage is on the node between the caps unless I make assumptions which may or may not be valid. Simulators generally won't make those assumptions and so they aren't even able to start the simulation because, to do so, they need to know the initial operating point for all components and all nodes in the circuit. Having a DC path to ground for all nodes ensures that this is possible.

Having said that, some simulators have a mode of trying to determine a DC operating point even with floating nodes by starting all supplies in the circuit at zero and assuming that all nodes are at zero voltage and no current is flowing anywhere in the circuit. They then slowly ramp up the DC supplies to the specified outputs. This is generally a long process and it not guaranteed to work.

Just tried it : two 2.2Uf and 12V

midpoint about 0.25V..Strange but okay ....Thanks....two days ago, would have thought it is 6V

 

hi Z,
Theoretically the junction/ midpoint would be Vsupply/2.
In a practical application is possible that the leakage current through the two capacitors will not be the same value.

What would you consider as a possible solution if that was the case.?

E

what's answer?

 

Just tried it : two 2.2Uf and 12V

midpoint about 0.25V..Strange but okay ....Thanks....two days ago, would have thought it is 6V

Electrolytic caps have a large variation in capacitance and significant leakage, so that will affect the voltage.
What did you measure the voltage with and how long did you leave it on that junction?
The meter resistance will drain away the charge on the capacitor.

 

Electrolytic caps have a large variation in capacitance and significant leakage, so that will affect the voltage.
What did you measure the voltage with and how long did you leave it on that junction?
The meter resistance will drain away the charge on the capacitor.

just changed to 0.1Uf of no polarity cap

from about 3V at middle of caps, 3mins later down to 6mv..guess going to 0

first cap charging? or 2nd discharging through multimeter?
Anyways, trivial matter

but if 1 cap is charging then same current to charge it should charge 2nd cap

 

hi Z,
Power up the supply voltage.
Measure across the top cap for a few minutes and then quickly measure across the lower cap for a few minutes.
What do you see.?
E

 

hi Z,
Power up the supply voltage.
Measure across the top cap for a few minutes and then quickly measure across the lower cap for a few minutes.
What do you see.?
E

Electrolytic?

 

hi,
As a self teaching experiment, try both cap types.
Later on when you are building power supplies, you may have to use two caps, of the same specification' in series, in order to meet the power supply design spec.

E

Updated: cap types

 

hi Z,
Power up the supply voltage.
Measure across the top cap for a few minutes and then quickly measure across the lower cap for a few minutes.
What do you see.?
E

Hi E
using 0.1U no polarity
lol....as the first cap discharges from about 9V(power supply is 12V : maybe discharges from 12v but shows 9v) : discharges from 9 to 0 then switch meter and see that the other cap was charging so it is now it's turn to discharge and the first cap that discharged charges

yeah : dunno why asked what cap when can use both



caps are having fun

 

hi,
If the caps where 'identical' in their specification, you will see the same measurement result when measuring the top and bottom caps.

As a practical example, say you had to use two 100uF 25V working caps in series across a 40V supply, to give 50uF at 50V.
You can see if the caps are not identical in their specification, one of the caps could have greater that 25V across it.!!!

As you said earlier, balancing resistors across each cap would be required,

E

 

hi,
If the caps where 'identical' in their specification, you will see the same measurement result when measuring the top and bottom caps.

As a practical example, say you had to use two 100uF 25V working caps in series across a 40V supply, to give 50uF at 50V.
You can see if the caps are not identical in their specification, one of the caps could have greater that 25V across it.!!!

As you said earlier, balancing resistors across each cap would be required,

E

Guess what??

using the 12V and after a while..Measure the voltage at mid point, what is answer before starts discharging??

I think half the voltage : 6V....my half was about 7V maybe due to imbalance of capacitance

 

Huh????

Put two 1 uF capacitors in series across a 12 V battery. Are you saying that "theory" predicts that the voltage across each capacitor is 12 V?

I am sorry but "theory" predicts the value at midpoint..practice backs up theory

The midpoint has to be half the supply voltage

Started with caps discharged (470UF so slow discharge)..with 470U, midpoint about 5.6V

 

I think half the voltage : 6V....my half was about 7V maybe due to imbalance of capacitance

hi Z,
I agree.
If the supply was DC, the difference would be due to leakage current imbalance.
If AC supply, a difference in the actual capacitance will also contribute to give a different junction voltage.

E