# Hello can somebody tell me please collector capacitor

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#### michael1978

Mar 17, 2012
388
Hello can somebody help with t his is c1 and R2 RC time of i dont know;-) what do the capacitor??

#### davenn

Moderator
Sep 5, 2009
14,260
this looks like a homework/study style question

I have moved it to that section
you will get guidance rather than outright answers
We expect input from you regarding what you do know/understand about the circuit

#### michael1978

Mar 17, 2012
388
this looks like a homework/study style question

I have moved it to that section
you will get guidance rather than outright answers
We expect input from you regarding what you do know/understand about the circuit
no is not home study, is just my hobby,

#### Alec_t

Jul 7, 2015
3,590
what do the capacitor?
It stores charge.
What are you trying to do with that circuit?

#### davenn

Moderator
Sep 5, 2009
14,260
no is not home study, is just my hobby,

well it is ion the style of homework, so we will treat it as such

1) where did you get this circuit?
2) how much of the operation of the circuit do you understand?
3) what google searching have you done to try and identify what may be happening in the circuit ?

#### Cannonball

May 6, 2017
193
I wish I could help, but I have never seen a circuit like this. I have seen it in an npn circuit used for a signal ground.

I have never seen it on the load resistor in any circuit. Npn or pnp.

Good luck with this one.

#### Ratch

Mar 10, 2013
1,099
It stores charge.
What are you trying to do with that circuit?

No, capacitors store energy, not charge. A capacitor energized to 1000 volts contains the same net charge as one with no voltage across it. For every charge carrier that enters a capacitor, a matching charge carrier leaves it.

Ratch

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,719
Looks like you're trying to simulate a common base circuit, but then R1/C1 need to go to a positive supply, not ground.
But I have no clear idea what C1 is good for. It will decrease the collector impedance for higher frequencies, thus lowering gain.
Are you trying to change the frequency response? It would really help to know where you git that circuit from. Obviously you haven't made it up yourself, otherwise you'd know what C1 is good for.

#### michael1978

Mar 17, 2012
388
It stores charge.
What are you trying to do with that circuit?
you mean the capacitor dry the collector resistance?

#### michael1978

Mar 17, 2012
388
I wish I could help, but I have never seen a circuit like this. I have seen it in an npn circuit used for a signal ground.

I have never seen it on the load resistor in any circuit. Npn or pnp.

Good luck with this one.
you mean does not exsist such a connection?

#### michael1978

Mar 17, 2012
388
Looks like you're trying to simulate a common base circuit, but then R1/C1 need to go to a positive supply, not ground.
But I have no clear idea what C1 is good for. It will decrease the collector impedance for higher frequencies, thus lowering gain.
Are you trying to change the frequency response? It would really help to know where you git that circuit from. Obviously you haven't made it up yourself, otherwise you'd know what C1 is good for.
here i get
https://electronics.stackexchange.c...r-in-a-bjt-collector-with-a-parallel-resistor
and i will like to know how its work, or maybe dont exist such connection

#### Alec_t

Jul 7, 2015
3,590
No, capacitors store energy, not charge. A capacitor energized to 1000 volts contains the same net charge as one with no voltage across it.
Is this article wrong then, Ratch? Q=CV, so for a given C the charge and voltage are proportional. Energy E = CV^2/2.

Jul 7, 2015
3,590

May 6, 2017
193

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,719
but i see in one site this
There's one big difference: The original circuit uses a DC input source, not AC as you do.
In the original circuit the transistor, the voltage source and teh 2.5 k and 5 k resistors make a constant current source which in turn charges the 1 μF capacitor. The question is, how long it takes to charge the capacitor and what will be the final voltage across the capacitor is a 1 k resistor is parallel to the capacitor.

Honestly, I don't know why you are asking here as the are answers on the original website. If you do not understand then please ask specifically for the points that are unclear instead of letting us guess.

#### Ratch

Mar 10, 2013
1,099
Is this article wrong then, Ratch? Q=CV, so for a given C the charge and voltage are proportional. Energy E = CV^2/2.

Yes, it is wrong, as are so many articles on how capacitors work. Anytime you hear someone saying a capacitor or battery being "charged", that is wrong. The battery does not transport charge from one plate to the other. The capacitor separates charge by accumulating charge carriers on one plate and depleting an equal number of charge carriers on the opposite plate. No charge goes through the capacitor dielectric. The charge carriers that accumulate on one plate are not the same charge carriers that are removed from the other plate. Separating the charge causes an electric field to form and stores energy. The net charge change is always zero. Therefore, capacitors become charged with energy, not charge. It is correct to say that capacitors are energized, not charged.

Ratch

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,719
@Ratch : your signature expresses it perfectly

I totally agree that you are right in a physical sense as any charge on one electrode is compensated by an equal charge of opposite sign on the opposite electrode; and energy is stord in the electric field between these electrodes.

Nevertheless it is common usage to handle capacitors as "charge storage" elements. This is reflected in the definition of capacitance as C = q/V (Wikipedia). This is simply a convenient way for handling capacitance in the context of circuit analysis and design.

While you are physically correct, I doubt this helps the op understanding the problem. With the above definition it is comparatively easy to see that V = q/C and with q = integral (i(t) dt the problem can be solved. Regardless of the physical fact.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Ratch, I find the errors in your post so egregious that I am forced to ignore it all.

So many people write "say" when they more accurately mean "write", and "hear" when they mean "read".

There is no sound involved in most of this communication, so "hear" and "say" should be an anathema to anyone who prides themselves on literal accuracy. Words clearly have only a single meaning, independent of context, and I appoint myself sole arbiter of the correct meaning.

If people continue to use words I understand but feel are not perfectly accurate, I will argue their use, deflect the conversation, and bring attention to my pedantry.

:-D

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