Assuming this circuit below:


simulate this circuit – Schematic created using CircuitLab

How do I get the time constant? I know that the time constant is just Rth * C, but I'm having a confusing time finding Rth. My first instinct is just add the two resistances but that seems wrong. Is there any way I can get Rth?

  • \$\begingroup\$ What two resistances? \$\endgroup\$
    – Andy aka
    Oct 22 at 10:41
  • 2
    \$\begingroup\$ @andyaka R1 and R2? \$\endgroup\$ Oct 22 at 10:43
  • \$\begingroup\$ That would be incorrect then. \$\endgroup\$
    – Andy aka
    Oct 22 at 12:24
  • 1
    \$\begingroup\$ You need to know whether the transistor is switching, or operating in the linear region (as a current source). If switching, you may find different time constants for switching on and off. \$\endgroup\$ Oct 22 at 12:32

It's not really appropriate to try and apply Thevenin's theorem to this circuit, because:

  • Assuming you intend for this transistor to be on or off, like a switch, those two conditions are so different that finding a Thevenin equivalent that fits both just doesn't make sense.

  • The transistor is not a linear element, and Thevenin's theorem requires that all the elements be linear.

That said, I can help you understand how this circuit would behave, and how you would determine the time constant in each of the two transistor states, on and off.

The first thing to realise is that R2 only affects the rate of charge of C1 indirectly. It controls the base current of Q1, but it is really Q1 that decides how much current flows from its collector to emitter. So R2 will certainly have some influence, but it's more complicated than simply multiplying by C1 to get some time constant.

Let's begin with a couple of really simple "equivalent" circuits, that represent the transistor as a switch, either fully conductive with 0Ω resistance from collector to emitter, or fully off, with infinite resistance between collector and emitter. It's an over-simplification, but it does illustrate the asymmetry of the circuit that makes Thevenin's theorem difficult (or impossible) to apply here:


simulate this circuit – Schematic created using CircuitLab

On the left you have SW1 open, meaning that the only two components here which are passing any current are R1 and C1. Assuming C1 starts completely discharged, it will charge up to 5V, and the time constant of that charging curve will be \$R1\cdot C1\$.

On the right, the switch is closed. This means that the capacitor will discharge immediately. That is, if we assume the switch has no resistance, the time constant will be \$0\Omega\cdot C1 = 0s\$.

Obviously this is impossible, and no switch, even a fully "on" transistor, has zero Ohms resistance. We will need a better model of this state of affairs than a simple switch, one that accounts for the transistor's behaviour when it's on.

So, back to your original circuit, with some extra detail:


simulate this circuit

The resistor R2, is providing current into Q1's base. The base will have a potential of 0.7V above the emitter, which is at 0V, so we can say that the base potential is 0.7V.

Therefore the voltage across resistor R2 is:

$$ V_{R2} = 5.0V - 0.7V = 4.3V $$

Base current \$I_B\$ will be the current through R2, which we can calculate with Ohm's law:

$$ I_B = \frac{4.3V}{10k\Omega} = 430\mu A $$

Now we need to know what the current gain of the transistor is, because that will determine (together with base current) what current will flow from collector to emitter.

A typical value of current gain \$\beta\$ for the 2N3904 is about \$\beta = 150\$, but this can vary a lot from device to device. I'll use it here because it's the approximate value that CircuitLab's model for this transistor seems to use.

The equation for collector current vs. base current can tell us what collector current to expect:

$$ \begin{aligned} I_C &= \beta \cdot I_B \\ \\ &= 150 \times 430\mu A \\ \\ &= 65mA \end{aligned} $$

By this model, collector current will be constant at 65mA. The model is naive, but it's good enough for a rough estimate of the time it will take the capacitor to discharge. Some of that current comes via R1, but it's a small percentage, and I have to neglect it because the maths would be a nightmare otherwise.

Because the capacitor is discharging with a constant current, the charge curve is a straight line, not an exponential curve you expect from a resistor-capacitor combination. We must use a different formula to find the time it will take for the capacitor's voltage to change from 5V to 0V given a constant current through it of 65mA.

Formally, the relationship is

$$ I = C \frac{dV}{dt} $$

but since the current is constant, we can skip the calculus, and it simplifies to:

$$ I = \frac{CV}{t} $$

$$ \begin{aligned} t &= \frac{CV}{I} \\ \\ &= \frac{100nF \times 5V}{65mA} \\ \\ &= 7.7\mu s \end{aligned} $$

I hope you didn't get too confused there! If you take away anything from all this, it should be:

  • R2 is not playing a direct role in some "time constant", because the transistor is in the way.

  • The transistor is also complicating matters for you, because it may not be considered as some constant resistance. In fact it varies its effective collector-to-emitter resistance to maintain a more-or-less constant current gain.

  • Since the base current is constant, we may (naively) assume that collector current is also constant, and that the capacitor's discharge curve is more like a straight line than an exponential. That means you can't use the usual exponential equations with some time constant.

  • When the transistor is off, its effective collector-to-emitter resistance is very very high compared to R1, and so can be ignored. So the capacitor charges via R1 in the usual exponential way, and you can use the time constant \$ R_1 \cdot C_1\$ as you would normally.


You have a transistor with about 430uA flowing into the base. The transistor will act more-or-less like a constant current sink until the voltage across the transistor drops below 200mV or so. The current is hFE dependent, but somewhere around 100mA with a large possible variation.

The resistor R1 fights that current sink, but only affects it by 5% or so until the very end, so we can pretty much ignore that except right at the end of the discharge it will determine the final voltage seen at the collector.

If the capacitor was originally charged to 5V, it will discharge to < 300mV in less than 10us typically (half that given the above rough estimate of Ic).

This is not a "time constant" per se, but if you like you could calculate the time to discharge to 63% = (1 – 1/e) of an original 5V voltage.


The resistance in the RC circuit is pull-up resistor R1.

So your RC circuit is R1 and C1.

If it's a switching circuit, where Q1 is saturated or off, then R2 has no effect on the RC timing constant. R2 passes current through Q1 base to Q1 emitter.

However, bear in mind that the RC circuit will not be discharged to 0 V by Q1 but to the Vce(sat) of Q1. This Vce(sat) is typically 0.2..0.3 V but check the datasheet and take the worst-case min or max voltage for your application. If the RC timing not being too short is most important for your function, calculate R and C based on the Vce(sat) max value, otherwise use the min value.

  • \$\begingroup\$ May i ask why R2 is not considered? \$\endgroup\$ Oct 22 at 10:44
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    \$\begingroup\$ @AedenSchmidt, that's saying you don't know how a transistor works, which is well outside the scope of an answer on this site. You can easily research detailed and plentiful documentation on an NPN BJT as a switch on the internet. \$\endgroup\$
    – TonyM
    Oct 22 at 10:50
  • \$\begingroup\$ @AedenSchmidt it could be considered but it is rather irrelevant as transistor discharging a capacitor is not an RC circuit so it does not have an RC time constant. Base current would be 0.5mA tops, which means that even with modest transistor current gain of 100 and assuming the transistor is a constant current sink it will discharge the cap in 10 microseconds. Typically the gain is 400 so 2.5 microseconds. \$\endgroup\$
    – Justme
    Oct 22 at 10:58

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