If a charged capacitor is connected to ground and one terminal connected to output. Will the capacitor discharge ?
if not why?
I came across an example of miller's integrator where a 1 Volt pulse was applied for 1 ms. The output of the integrator is a ramp function which rises to its peak volt for 1 ms and remain at its peak value there after, even when there is no pulse at the input of the integrator. For reference I found the example on page 85 of Microelectronics Circuits by Sedra smith[6th edition].

 

Hi,
A Capacitor have a lot of configuration to connect in a circuit.
Like using by bypass, filter .coupling.decoupling method.
Can you tell me in what method did you use or connect it in a circuit.

 

If a charged capacitor is connected to ground and one terminal connected to output. Will the capacitor discharge ?
if not why?
I came across an example of miller's integrator where a 1 Volt pulse was applied for 1 ms. The output of the integrator is a ramp function which rises to its peak volt for 1 ms and remain at its peak value there after, even when there is no pulse at the input of the integrator. For reference I found the example on page 85 of Microelectronics Circuits by Sedra smith[6th edition].

Hi,

An integrator is a special kind of circuit that is created to sum the voltages or currents on the input and provide an output that is the sum over time of the inputs over time. This means if you input a 1 volt pulse for 1 second you may see 1 volt at the output, but if you input 1 volt for 1/2 second and then later some time you input again 1 volt for 1/2 second (do the same thing twice but separated in time) you would see:
1*0.5+1*0.5=1 volt

which is the same as before because the circuit had 'summed' the two separate instances of an input. If you had twenty such inputs it would sum all of them, and because the integrator is made to sum the inputs the output result must stay at the sum and can not decrease unless we apply a negative input at some point such as -1 volt. If we input 1 volt for 1 second then later apply -1 volt for 1 second we will see the sum which is zero (0 volts) which is what the integrator should do.

But again the integrator is a special circuit designed just for that purpose, and if you look up the mathematical derivation of an integration you will find that it starts with a sum. So the integrator does the same as the math would suggest it should do.

Not all integrators are perfect however, some may loose some voltage over time because of their imperfections. That means with some circuits the output has to be used right away before the output can start to droop too much.
Not all integrators perform the integration perfectly either, some have a changing character that is based on what has already been summed and so they may not sum with the same weighing over the full range of operation. That is often found in circuits that dont need perfect integrators because the circuit is much simpler.

 

Hi,
A Capacitor have a lot of configuration to connect in a circuit.
Like using by bypass, filter .coupling.decoupling method.
Can you tell me in what method did you use or connect it in a circuit.

hi,
if you google the image of a miller's integrator you will know that it is connected in feedback to the negative input terminal of the op amp.

 

Hi,

An integrator is a special kind of circuit that is created to sum the voltages or currents on the input and provide an output that is the sum over time of the inputs over time. This means if you input a 1 volt pulse for 1 second you may see 1 volt at the output, but if you input 1 volt for 1/2 second and then later some time you input again 1 volt for 1/2 second (do the same thing twice but separated in time) you would see:
1*0.5+1*0.5=1 volt

which is the same as before because the circuit had 'summed' the two separate instances of an input. If you had twenty such inputs it would sum all of them, and because the integrator is made to sum the inputs the output result must stay at the sum and can not decrease unless we apply a negative input at some point such as -1 volt. If we input 1 volt for 1 second then later apply -1 volt for 1 second we will see the sum which is zero (0 volts) which is what the integrator should do.

But again the integrator is a special circuit designed just for that purpose, and if you look up the mathematical derivation of an integration you will find that it starts with a sum. So the integrator does the same as the math would suggest it should do.

Not all integrators are perfect however, some may loose some voltage over time because of their imperfections. That means with some circuits the output has to be used right away before the output can start to droop too much.
Not all integrators perform the integration perfectly either, some have a changing character that is based on what has already been summed and so they may not sum with the same weighing over the full range of operation. That is often found in circuits that dont need perfect integrators because the circuit is much simpler.

Thank you,
I now have better understanding of the integrator.

 

Again we can find an integrator image after several seconds of web searching and post it here:

The current thru the capacitor is completely controlled by the input voltage and R1:

Ic = Vi / R1

since the - input is a high impedance point and is not a source of current (neglecting bias currents).

Do note that for a positive input the output will decrease, in the same manor and for the same reason as the similar op amp inverting amp.

Should we apply a constant voltage at Vi there will be a constant current thru C, and a linear rise in voltage across the cap.

If we then apply zero volts to Vi we change the current to zero, and the voltage across the cap no longer changes.

 

I would be very interested in how one would get a dc current through a capacitor( if its not leaky)?

 

Hi,
your question is too puzzle. if a capacitor alone without connecting to any circuit,it will not be charge.
A capacitor that its not charge no current or voltage can get.

 

We are not talking about DC currents. Where did you get that quaint notion? The momentarily constant current is limited by the Vcc supply voltage. When the voltage across the capacitor approaches Vcc the current drops to 0 and no further increase appears across the capacitor.

 

I would be very interested in how one would get a dc current through a capacitor( if its not leaky)?

There is nothing in physics preventing a DC current from passing thru a capacitor. You create a DC current source, connect it across a capacitor and a DC current will flow.

There are many applications where a capacitor is used to block DC and only pass an AC signal, but these are voltages not current. Initially at turn on the capacitor does indeed pass a DC current until it charges to the same input DC voltage, then only passes the AC portn of the signal.

 

Sorry, Don't know where your capacitors are from but I have NEVER seen one which will pass a DC current. Every capacitor that I have ever seen/used that passes a DC current is FAULTY and should be replaced. How can you say you can block DC current then say current passes through a cap???.
I will agree that a DC voltage applied to a capacitor will result in a Charging Current till plate volts equals applied volts. At that time current will be effectively zero for a GOOD cap. Current still does NOT pass "through" a good cap. Never no how!!
But maybe these things work differently in the Northern Hemisphere.

 

Sorry, Don't know where your capacitors are from but I have NEVER seen one which will pass a DC current. Every capacitor that I have ever seen/used that passes a DC current is FAULTY and should be replaced. How can you say you can block DC current then say current passes through a cap???.
I will agree that a DC voltage applied to a capacitor will result in a Charging Current till plate volts equals applied volts. At that time current will be effectively zero for a GOOD cap. Current still does NOT pass "through" a good cap. Never no how!!
But maybe these things work differently in the Northern Hemisphere.

I guess we can't rely on you for any deep understanding of physical phenomena.

 

Any capacitor that cannot pass a DC current is faulty.

Final answer.

 

Well, its seems that physical phenomena works differently for you guys. Can you actually demonstrate a DC current passing through a capacitor?
Love to see that one done!

 

That charging current is DC Current, is it not?
If you pass a constant current through a capacitor, DC current explicitly, the voltage on the capacitor rises or falls depending upon the polarity of the current. Of course, upto the point of the compliance voltage, of the DC Current source!
You are talking about the blocking DC capacitor application. Even there, till the capacitor is charged, DC current flows through!

 

Well, its seems that physical phenomena works differently for you guys. Can you actually demonstrate a DC current passing through a capacitor?

Well this guy seems to get a DC current to flow:

I will agree that a DC voltage applied to a capacitor will result in a Charging Current till plate volts equals applied volts.

Just because something may happen for a only limited time does not mean it never happens not once ever.

I would advise you to seek a deeper understanding of circuits such as that in post #6 as they are using a capacitor in a manor you seem not to understand.

 

Do you really know how capacitors are made??? They consist of basically two conductive plates SEPARATED by a DIELECTRIC. A DIELECTRIC is an INSULATOR. INSULATORS do not allow current to pass through.
The original capacitor(condensor) was a Leyden Jar. It was a GLASS JAR with a metal foil on the inside and another on the outside which could the be used to store a static charge. Now how can glass conduct a DC current?? (Refer Wikipedia) Capacitor construction has progressed but still use a DIELECTRIC to separate the positive and negative connections. THEY DO NOT PASS A DC CURRENT.
I have worked in Electronics for over fifty years and have yet to see a good capacitor pass a DC current.
The circuit you mention in post 6 does NOT pass DC current through the capacitor but Charges it up to the volt difference between Vo and Vi.

 

Do you really know how capacitors are made??? They consist of basically two conductive plates SEPARATED by a DIELECTRIC. A DIELECTRIC is an INSULATOR. INSULATORS do not allow current to pass through.
The original capacitor(condensor) was a Leyden Jar. It was a GLASS JAR with a metal foil on the inside and another on the outside which could the be used to store a static charge. Now how can glass conduct a DC current?? (Refer Wikipedia) Capacitor construction has progressed but still use a DIELECTRIC to separate the positive and negative connections. THEY DO NOT PASS A DC CURRENT.
I have worked in Electronics for over fifty years and have yet to see a good capacitor pass a DC current.
The circuit you mention in post 6 does NOT pass DC current through the capacitor but Charges it up to the volt difference between Vo and Vi.

You seem to only have a problem with DC current. What about AC current? Wouldn't the exact same argument, namely that capacitors are two conductive places separated by an insulator which does not allow current to pass through, mean that capacitors don't allow current to pass through, period?

If you are going to claim that capacitors allow AC current to pass through them, then how are you going to reconcile that with the notion that insulators do not allow current to pass through?

 

That charging current is DC Current, is it not?

No, it is not, which is what this whole semantic pissing contest is about. Allow me to explain and mention other semantic piss-points. If a DC source (***not*** DC current source, the C already stands for Current, so "DC current" means "direct current current") of some voltage is connected to a capacitor with a different voltage across its plates, there will be *transient* electron flow (***not*** current flow; current is the flow of electrons, so "current flow" actually means "electron flow flow") until the capacitor charges up to potential of the DC source voltage, at which time the DC component of the energy will be zero.

ak

 

All current is DC current. It only moves in one direction at one time.