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Here is a simple circuit for common emitter. Is the time constant for charging \$C_1\$ equal to \$C_1R_2\$? It doesn't seem logical that \$R_1\$ would influence the charging of \$C_1\$, since it's connected to ground.

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  • \$\begingroup\$ Except that the charging circuit goes through R2, C1, and R1. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 2 '14 at 18:13
  • \$\begingroup\$ Thanks very much, I'm newbie with this. So time constant for charging is equal to (R1+R2)*C1 .I guess time constant for charging circuit doesn't have any relations with voltage on ends of capacitance? \$\endgroup\$ – Fatty So Oct 2 '14 at 18:39
  • \$\begingroup\$ I haven't seen the math so I can't confirm that it's simply the sum of resistances, but KCL states that both resistors are involved regardless. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 2 '14 at 18:48
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Follow the current flow. When Q1 is off, C1 charges from the current flowing in the series circuit composed of R2, C1 and R1. Since R1 & R2 are in series you can simply add them together in this case. When Q1 is on you have a different case. Now C1 is discharging through the series circuit of Q1, C1 and R1. So in short - your charging time constant would be related to C1*(R1+R2) while your discharging time constant would be related to C1*R1 (assuming Q1 behaves as a perfect switch).

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  • \$\begingroup\$ Sorry - but i have to disagree. The relevant time constant is C*(R1||R2). \$\endgroup\$ – LvW Oct 2 '14 at 20:08
  • \$\begingroup\$ Sorry - it was too late to cancel the above comment. It`s just the opposite (see my sepatere answer). \$\endgroup\$ – LvW Oct 2 '14 at 20:16
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FattySo - to find the time constant of a circuit without the necessity to perform involved calculations it is, in most cases, helpful to ask: "What are the ways the corresponding capacitor can discharge"? That means, you have to consider the circuit as seen from the capacitor´s side. And in the case under discussion it is clear that the capacitor will drive a discharging current through R1 in series with R2 (assuming that the transistor´s output resistance is infinite). Hence, T=C*(R1+R2).

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Intuitively, the voltage across \$C_1\$ is affected by both \$R_1\$ and \$R_2\$ so the time constant is determined by both.

To find the small signal time constant \$\tau\$ you can use the open circuit time constant (OCTC) method. The equivalent small signal circuit to find \$\tau\$ for \$C_1\$ is:

schematic

simulate this circuit – Schematic created using CircuitLab

The \$g_{m}v_{\pi}\$ current source is 0 since \$v_{\pi} = 0\$ and is therefore an open circuit. \$r_{o}\$ and \$R_2\$ are in parallel, and this resistance is in series with \$R_1\$ through the small signal ground node so

$$R_{o1} = R_{1}+R_{2}||r_{o}$$

and the time constant is

$$\tau = \left(R_{1}+R_{2}||r_{o}\right)C_{1}$$

If \$r_{o} >> R_{2}\$ then \$r_{o}||R_{2} \approx R_{2}\$ and the time constant is

$$\tau \approx \left(R_{1}+R_{2}\right)C_{1}$$

Note that the other capacitors in this circuit (the transistor's \$C_{\pi}\$ and \$C_{\mu}\$) are not shown since they are open circuits by the OCTC method. However, the longest time constant for a common emitter circuit (which limits its bandwidth) is usually the one from \$C_{\mu}\$ due to the Miller effect.

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  • \$\begingroup\$ I suppose this is also the right answer, if you take Miller in account. It was mistake, I mean the downvote. But I think the Miller effect is in my conclusion ignored. Sorry for not mentioning that. \$\endgroup\$ – Fatty So Oct 2 '14 at 19:28
  • \$\begingroup\$ @FattySo My answer does not depend on the Miller effect. The Miller effect is ignored (it affects \$C_{\mu}\$, but I did not compute its time constant as you did not ask for it). I was simply adding some extra information by pointing out that the bandwidth of the circuit will probably be limited by \$C_{\mu}\$ rather than \$C_1\$. \$\endgroup\$ – Null Oct 2 '14 at 19:33
  • \$\begingroup\$ I haven't looked at it closely, but from here it seems that Ro1=R1+R2( R1 and R2 are not connected in same point, they are separated by Ro1, or open circuit). But thanks for highlighting octc metod, I'll try to use it on some other examples. And sorry, for bad grammar. \$\endgroup\$ – Fatty So Oct 2 '14 at 20:17
  • \$\begingroup\$ @FattySo Ah, yes, mentally I was thinking of \$R_1\$ as the transistor's \$r_o\$ (which is assumed infinite here), which would be in parallel with \$R_2\$. \$\endgroup\$ – Null Oct 2 '14 at 20:32
  • \$\begingroup\$ @FattySo Added \$r_{o}\$. \$\endgroup\$ – Null Oct 2 '14 at 21:03

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