Hi:
I set up a double RC circuit (see image) but for some reason, the voltage in the capacitors is not reaching 5V, it gets stuck on 4.22V. I already tested the capacitor in a single RC circuit and both capacitors are OK (they reached 5V). Is there any additional phenomena that is being overlooked?
Any help/suggestion is welcome!
Capacitors are not perfect, they leak. This can be modeled by a resistor in parallel with the capacitor. So the circuit becomes a voltage divider. The 1M resistors must be large enough that the leakage current is not insignificant.
Also, the device you are using to measure the voltage draws some current. How are you measuring the voltage?
If it is a standard multimeter with 10 Meg input resistance, then that forms a voltage divider with the two 1 Meg resistors. You would expect to measure 5.0*10/12 = 4.16 V from the 5.0 V battery, assuming accurate resistors (which they are most likely not).
jremington:
Also, the device you are using to measure the voltage draws some current. How are you measuring the voltage?If it is a standard multimeter with 10 Meg input resistance, then that forms a voltage divider with the two 1 Meg resistors. You would expect to measure 5.0*10/12 = 4.16 V from the 5.0 V battery, assuming accurate resistors (which they are most likely not).
Indeed, Im measuring the voltage with a multimeter (it uses a 9V battery)
Not only that, but when a cap is uncharged and the circuit is closed, both caps act as shorts. So once C1 starts to charge, slowly C2 begins to charge too. Now here is where is gets interesting. Once C2 is charging, C1 slowly discharges, adding more resistance to the circuit, and starts to split off the current to C2.
C2 will still charge, but now that both caps are charging at the same time, they are acting as current dividers. Eventually (less than a second) the caps will equalize, but they will never reach the source voltage.This is because as Jack C pointed out they leak and want to refill.
The more stages you add, the lower the voltage on the caps will be.
Ideally both caps will equal the source voltage, but nothing is ideal.
It sounds like the issue is the 10M input resistance of the DMM.
Simplest way to understand whats going on is to measure the leakage current of the capacitors.
I presume they are both electrolytics.
I set up a double RC circuit (see image)
Why?
What are you trying to achieve or prove?
Grumpy_Mike:
I set up a double RC circuit (see image)
Why?
What are you trying to achieve or prove?
Implementing control over the capacitor 2 charging time. Is there any difference if I substitute the electrolitic capacitor for ceramic ones?
There will be less leakage with a ceramic capicitor, but the actual value of capacitance is going to be a bit variable depending on the voltage across it.
The problem you are seeing is due to the comparitave low impedance of your measuring device. Try upping the capacitors by a factor of 10 or 100 and reducing the resistors by the same factor.
Still not sure what you hope to acheave with this circuit, what you will see is a double exponential curve.
Well, at 0.1Hz, the response is the same within a very small fraction of a dB. However, by 10Hz input, the output from one 1M/1uF combo is down by about =36dB, but down by =70dB using both.
So a much more rapid dropoff in response to higher frequencies.
But the OP was talking about and illustrating a simple DC circuit.
Clearly (at least to my eyes), the major influencing factor is the impedance of the test meter.
Agreed, the 10meg resistors and the leakage of the capacitors may/will distort results but the OP has now learned that measuring devices DO influence the measurement.
As a super simple experiment, he could rig up the circuit and, whilst measuring the voltage across R2C2 junction and ground, he could disconnect C1, observe change (if any) then disconnect C2 and observe change (if any) then from the results obtained, calculate actual capacitor leakage currents.
He will then , not only appreciate meter influence, but be aware of how much leakage capacitors introduce.
The process of learning is a combination of tuition reinforced by experimentation.
polymorph:
Well, at 0.1Hz, the response is the same within a very small fraction of a dB. However, by 10Hz input, the output from one 1M/1uF combo is down by about =36dB, but down by =70dB using both.So a much more rapid dropoff in response to higher frequencies.
Well, since I'm using a DC ppower source, I cannot vary the frequency of the system. Am i right?
What is it you are trying to do?
polymorph:
What is it you are trying to do?
Implementing control over the capacitor 2 charging time. I was just curious about your comment on the effects of the frecuency on the system.
So this is not a DC circuit then. You are interested in the short-term effect of applying a step voltage to this circuit.
polymorph:
So this is not a DC circuit then. You are interested in the short-term effect of applying a step voltage to this circuit.
You're right, the step is 5V, but regardless of that is a DC circuit, right?
Yes it is a dc circuit undergoing a transient change. Alternating circuits involve a reversal of polarity which yours clearly does not.
Are we studying electronics or semantics?