Hi, I'm trying calculate value of voltage across two parallel capacitors after 1 second of charging and it's giving me trouble. I've tested that in simulation and my calculations (on paper) don't match with it. In simulation after 1 second the C1 (8uF) charges up to 395V, while C2 (2.2uF) up to 26V. Am I calculating wrong? Also the voltage supply is 560V.
Capacitor charging time
- Thread starter lukimaister
- Start date
Well 560V is just an example voltage supply, it could be any value, but my calulations would still be wrong.
Maybe the first calculation is correct, because 418V is relative close to the simulations 395V.
hi luki,
You know that voltage on C1 will rise exponentially over the 1 second period
So that exponential voltage is the voltage which is charging C2 thru the 4Meg resistor, not the input 560V.
Do you follow that.?
E
Note:
You know that voltage on C1 will rise exponentially over the 1 second period
So that exponential voltage is the voltage which is charging C2 thru the 4Meg resistor, not the input 560V.
Do you follow that.?
E
Note:
But the charge up voltage will be lower, due to the current that is passing thru thr 4Meg into C2
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Ok so C2 is charging from C1.
So if C1 is charging exponentially then would't it be impossible to calculate C2 charging time? Maybe if I use 0.7 time constant voltage from C1.
I think I've got it. I calculated 31.5V on C1, but both calculations are still off a bit for some volts, so it it still correct?
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If you are looking for an accurate answer, then you won't find it by using simple exponential charging curves. Each capacitor affects the charging of the other capacitor, so it will take at least a second order differential equation to accurately describe the circuit behavior. However, due to the large difference in magnitude of the time constants involved, one can expect that using the simple exponential method will provide a close approximation of the actual simulation result.
Hello,
I redrew your drawings to allow us to discuss this clearly.
First point you will notice is that the 1M resistor has been disconnected as i am sure you realize it does not affect the voltages across either cap.
Second, if you open R2 so that only R1 and C1 remain then calculate the voltage across C1 after 1 second you will get some voltage Vc, and it should be common sense that if you did include R2 and C2 it would present a 'load' on the voltage across C1 (Vc) and so the voltage across C1 would definitely be LESS than the voltage we would see without R2 nd C2. If the voltage across C1 comes out HIGHER than that calculated voltage, then something is not right.
Third, there is no reason to put the two resistors R1 and R2 in parallel in order to calculate the voltage across C1. The reason this is so is because the current that flows through R2 does NOT flow through R1 and so the current through R2 does not increase the current through C1 and thus the voltage across C1 does not increase as a result of having R2 in the circuit. If anything, R1 and R2 would act as a voltage divider that would make the final voltage across C1 LOWER than it would be if R2 were open circuited. This is a very important thing to think about. Resistors in parallel increase current, but since R2 is not in parallel with R1 it acts as a current sink which reduces the current through C1 and thus the charge voltage after some relatively short time (like 1 second).
Note if you just want to calculate the time constant then R1 can actually be thought of as being in parallel to R2 but the voltage will not come out right without also considering the voltage divider effect. In other words, do both the parallel combo PLUS the voltage divider effect and you should get a good approximation.
I must say though that how you attempt to solve this problem has a lot to do with what you have learned in the past about solving circuits with a number of resistors and capacitors because that would tell you how to proceed.
Because more than one R and C in a circuit make things more complicated we usually have to resort to some more advanced methods such as using Laplace Transforms or Ordinary Differential Equations. To get a shortcut to this we'd have to do some thinking about the particular topology of this circuit and see if we can come up with something. For example, since 100k is much less than 4Megs maybe we can calculated Vc1 and Vc2 as if there were only those single caps in the circuit and then because of the two facts:
1. C2 charges about 10 times slower than C1, and
2. C1 is also charging and a linear interpretation of that charging would be a linear ramp and so C2 would change very roughly about half as much as if C1 was fully charged the whole time,
we can approximate the voltage across C2 as roughly 1/2 of what it would be if C1 where charged the whole time, and then use that to 'adjust' the value of the voltage across C1.
But you see there is quite a bit of thought that has to go into this, and since we have general methods for this already and they will be applicable to really most every circuit we encounter, it is a good idea to learn those methods and use them and this also allows is to calculate the exact theoretical values for both voltages for any charge time period. Once we do that, if we really want a short cut to this particular circuit topology then we can just develop a formula that is exact and includes both resistors and both caps, although ti would always be limited to this one circuit topology without some extra modifications.
Think about all this and see if it makes sense to you and perhaps you can mention what kind of methods you were already taught to handle these kinds of circuits.
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