# Voltage Source & capacitor, Voltage source & Diode

Discussion in 'General Electronics Chat' started by rage_speed, Jun 29, 2009.

1. ### rage_speed Thread Starter New Member

Jun 29, 2009
4
0
Hi All,

I want to know what will be the Output voltage when a Capacitor is connected to a voltage source in series. The circuit is not closed

Figures

|------||----o
Vin ...............vout1
|___________o

Similarly instead of a capacitor if a diode is connected what will be the output

|------ -> ----o
Vin ................ vout2
|____________o

|------ <- ----o
Vin ................vout3 ->,<- are diodes
|____________o

Thanks and Regards,

Anand Krishna G

2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515

There is no 'output' voltage when nothing is connected. Even the connection of measuring devices completes a circuit.

However taking your bottom terminal as ground or reference in each case a voltage does appear at the top terminal.

You did not specify if the voltage source is AC or DC?
If it is DC you did not specifiy the + and -

3. ### steveb Senior Member

Jul 3, 2008
2,433
469
I see nothing wrong with your terminology, and your question is clear to me. Further, the I feel the question can be answered even if you don't specify the type or polarity of the voltage source. Also, the voltage is there even if the circuit is not completed. In a sense the circuit is completed by the air, but even in vacuum it is meaningful to say there is voltage without completing the circuit. (hopefully, no one will try to start the "tree falling in the forest" debate.)

The case of the capacitor requires that you know the initial conditions of the voltage on the capacitor. If it is not charged, then the voltage is zero and the output voltage equals the source voltage. Otherwise, the charged voltage adds/subtracts from the source voltage.

The case of the diodes is a little tricky because there is a built in back voltage when there is no current flowing. The value depends on the type of diode. The back-voltage will add/subtract from the voltage source.

All three of these cases can be a little tricky to measure in reality. Any meter has resistance, even if it is very large. All three circuits have a high, but finite resistance in the circuit element (capacitor, unbiased diode). This creates the effect of what people call a "floating voltage" which is just a voltage that is very sensitive to the environment (temp, humidity of air, resistance of meter, stray fields, fingers touching the leads etc.).

Last edited: Jun 29, 2009
4. ### farhill New Member

Jun 14, 2009
2
0
suggest to do simulation by yourself. It will be great helpful for you to understand what happens to a circuit...

5. ### rage_speed Thread Starter New Member

Jun 29, 2009
4
0
Hi All,

@ steveb

I have few more doubts.

1) Does the capacitor charge because of the source ? If it does then how ?
2) How will the capacitor circuit react to a step input ?
3) How will the diode circuit react to a step input? If the source voltage < |back-voltage| what will be the output.

I hope I am clear with the questions.

@ studiot

Its a DC source.

|------||----o
+
Vin ...............vout1
-
|___________o

|------ -> ----o
+
Vin ................ vout2
-
|____________o

|------ <- ----o
+
Vin ................vout3 ->,<- are diodes
-
|____________o

6. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
I am guessing that you have got to the stage where you are studying 'walking through' analysis of a simple network and your question arises becauses of this.
I am further guessing that what you have published are little subcircuits of this and upstream is a switch. The questions ask

'What happens when the switch is closed'

So if the upper left terminal is suddenly connected to a +ve voltage, V

Vo3 will remain at zero as the diode will block

Vo2 will take up V as the diode appears as a short circuit or will pass a pulse or step

These two are easy to understand because action in the diodes does not depend on the rest of the circuit. The diodes perform the same action regardless of the what the rest of the circuit is or even if there is one (open circuit).

The capacitor is more difficult. You cannot have capacitor action without a complete 'circuit'. i.e. Both ends of the capacitor connected to something. The action in the capacitor is partly determined by the rest of the circuit, unlike the diode.
The left hand side of the capacitor will rise to +V.
In a perfect world with perfect insulators nothing more would happen and Vo1 would remain uncharged. But we cannot measure this.
In the real world, lacking perfect insulators, the capacitor will slowly polarise with Vo1 tied to zero volts through whatever resistance to earth there is.
A pulse would see the capacitor as a short circuit and appear at the top end of this resistance. We say that the voltage across a capacitor cannot instantly change so if the left hand side is suddenly raised so is the right hand side.

Hope this helps.

7. ### steveb Senior Member

Jul 3, 2008
2,433
469
1. Theoretically, the capacitor does not charge assuming ideal components and perfect insulation on the open terminals. In practice, it might be possible to get some charge on an open capacitor, I'm not sure. I'm tempted to say no because whatever air resistance is available to charge the capacitor (with very large RC time constant) should be more than counteracted by leakage in the capacitor dielectric. At that point the leakage resistances are going to factor in to the actual output voltage.

2. The capacitor does not react to the step input. The output voltage simply follows the input voltage with a shift equal to the capacitor charged voltage. In the capacitor is not charged, then the output equals the input, at least in theory.

3. I'm going to guess that the diode back voltage is still in play even with a fast step transition. Basically, my original statement that the type of voltage source should not matter still seems right to me, at least if we are talking about ideal circuits as you have drawn them. In a real case, the principles should still be right, but there are complications with the "floating voltage" as I described. Leakage air resistance, imperfect dielectrics, stray fields etc can be significant in these cases.

Anyway, it's important to distinguish between theory and practice with these circuits.

8. ### steveb Senior Member

Jul 3, 2008
2,433
469
These statements do not seem right to me. I won't go into the reasons because we have two experienced people here with differing opinions and the two of us debating about it won't resolve it or help the OP. This is a good area for the OP to explore further with feedback from other people, or even some experiments and simulations.

9. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Don't forget, Steve, that I think these are sub-circuits. We've had quite a few posters doing their homework where there is small network containing nothing more complicated than resistors, capacitors, inductors diodes and a switch to a source. the question is deduce what happens when the switch is closed or opened. Can't remember which threads though.

So in each case Vout would be passed to anothe sub circuit of the network.
This makes sense to me.

In the context of such questions on academic courses, diodes are short circuit in the direction of current and open circuit in the reverse.

We agree on the capacitor.

10. ### steveb Senior Member

Jul 3, 2008
2,433
469
I see. Of course, this type of information is important to know in answering the question.

I'm trying not to go too deeply here because there are a few variations one can consider. You just pointed out one i didn't think of. But also, there are a wide range of other assumptions one could make with regards to whether or not ideal components and environments are assumed and what models are assumed in any of the variations.

Various people are likely to envision a different interpretation of the question and end up with different answers in these tricky cases. Even when completely ideal conditions are specified, it's often difficult to not let our real experience mislead us about the ideal case.