The 200Ohm resistor is throwing me off. Is it supposed to be 200Ie, or 200Ib when I do KVL?

I have tried simulating the circuit, but that does not alleviate my confusion.

Ultimately, I am trying to solve for Vce and Vo, but solving for Ib is my first sticking point.

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  • \$\begingroup\$ To address your 200 Ohm emitter resistor question, 'Is it supposed to be 200Ie, or 200Ib when I do KVL?', the voltage is found with 200 * Ie. Ie is the sum of Ic and Ib, so Ib has already been accounted for. \$\endgroup\$
    – Dave H.
    Nov 5, 2021 at 2:55
  • \$\begingroup\$ So when I do KVL I get: -5V + 10,000 x Ib + 200(1 + hFE) x Ib = 0 ; is that correct? \$\endgroup\$ Nov 5, 2021 at 3:08
  • \$\begingroup\$ Looks right to me. Though check out Jonk's answer as it takes into account saturated mode vs. active mode. \$\endgroup\$
    – Dave H.
    Nov 5, 2021 at 3:26
  • \$\begingroup\$ active mode: ib=142.384 uA, Vce=4.866V, vo=2.876V \$\endgroup\$
    – Syed
    Nov 5, 2021 at 7:41

3 Answers 3


For KVL on the input loop, you need the sum of the voltages to be 0. You can also look at it as the voltage across the 10k base resistor + Vbe + the voltage across the 200 Ohm emitter resistor should equal the 5V of the battery.

Here's what you know:

  • The voltage across the 10k base resistor is 10k * Ib.
  • The voltage across the 200 Ohm emitter resistor is 200 * Ie
  • Vbe was stated as 0.7V

So you have: 10k * Ib + 200 * Ie + 0.7 = 5

The current Ie is the sum of Ic and Ib, so you can do this substitution:

10k * Ib + 200 (Ic + Ib) + 0.7 = 5

Ic is the product of Ib and Beta, which is given as 100.

Another substitution gives you:

10k * Ib + 200 (Beta * Ib + Ib) + 0.7 = 5

You've been given a Beta of 100, so substituting that gives:

10k * Ib + 200 (100 * Ib + Ib) + 0.7 = 5


10k * Ib + 200 (101 * Ib) + 0.7 = 5

You now have an equation with only one variable, Ib. Combine like terms and solve for Ib. Then use that and Beta to find Ic and eventually Ie.

  • \$\begingroup\$ Why does beta is needed Sir as such cannot we solve the circuit by just two loops laws and one current law ? Variables are the currents only isnt ? I always think this \$\endgroup\$
    – Orion_Pax
    Apr 14, 2022 at 0:02
  • \$\begingroup\$ Oh i guess i got it , its because variables are four in total three currents and one voltages , we have just three equations from Kirchoff? \$\endgroup\$
    – Orion_Pax
    Apr 14, 2022 at 0:11

When you are facing a BJT circuit, it's usually:

  1. Active mode: Apply the given \$\beta\$ value as an input for analysis.
  2. Saturated mode: \$\beta\$ is an output of the analysis, not an input to it.

So, it's common to first assume (1) above and see if that works out okay. If it does, you are done. If not, then it is usually because the collector has been forced to be between the base and emitter voltages and you are in case (2) and you perform a different analysis.

Active mode assumption

Your input loop is fine. Start in the lower-left corner. Call that \$0\:\text{V}\$. So, running around the loop you add \$+5\:\text{V}\$ and then subtract \$I_{_\text{B}}\cdot 10\:\text{k}\Omega\$ from that and then subtract the base-emitter junction voltage (which varies a little depending on the collector current but for which most people just assign a constant value) and then subtract \$I_{_\text{E}}\cdot 200\:\Omega\$ to get back to the same place you started.

You know, in active mode anyway, that there is a relationship: \$I_{_\text{E}}=\left(\beta+1\right)\cdot I_{_\text{B}}\$.

You can plug that in and then solve for \$I_{_\text{B}}\$.

With that, you can compute \$I_{_\text{C}}=\beta\cdot I_{_\text{B}}\$ and then compute the voltage drop \$I_{_\text{C}}\cdot 500\:\Omega\$. Using that, and your knowledge of the \$+12\:\text{V}\$ supply rail, you can figure out the collector voltage. If you find that the collector voltage is equal or higher than the base voltage (which you also need to compute), then you are in active mode and all of your calculations are done.

Saturated mode

Otherwise, you've determined that the BJT is saturated. Here, you make an assumption about the collector-emitter voltage (often only a couple of hundred millivolts) and from that compute the current in the \$500\:\Omega\$ collector resistor as \$I_{_\text{C}}\$. You have a new relationship: \$I_{_\text{E}}=I_{_\text{B}} +I_{_\text{C}}\$.

And you use that now to solve your loop equation for \$I_{_\text{B}}\$, instead.

  • \$\begingroup\$ Sir usually why grounding is shown in transistor circuits , what is useful in doing grounding ? Its not necessary right ? \$\endgroup\$
    – Orion_Pax
    Apr 14, 2022 at 0:04

The simple solution uses hFE with Rin = hFE * Re thus Vb= 0.6 + Ib * hFE * Re is a KVL divider of Rs, Vbe & Rb load= hFE*RE ignoring Vbe. yet using Vbe=0.6V for 0.5mA to 1mA =Ib, you are pretty accurate to compute Vbe and thus Vb then Ib.

  • \$\begingroup\$ I don't understand what you mean even a little bit. Can you elaborate? Also, the problem states to use Vbe = 0.7V \$\endgroup\$ Nov 5, 2021 at 2:36
  • \$\begingroup\$ I don't care if you use 0.7 or 0.6 the result is not much different, yet 0.7 is wrong since Ib cannot exceed 0.5mA \$\endgroup\$ Nov 5, 2021 at 2:39
  • \$\begingroup\$ Tony, you gotta give me a little more here. Can we start with KVL of the input loop? I got -5V + 10,000 x Ib + 200(1 + hFE) x Ib = 0 ; is that correct? \$\endgroup\$ Nov 5, 2021 at 3:07
  • \$\begingroup\$ This is trivial.KVL 5 -Ib * 20k - 0.7V - 100 * Re(=200) =0 WHat is Ib? \$\endgroup\$ Nov 5, 2021 at 3:28
  • \$\begingroup\$ You messed up the equation. It is supposed to be (1 + 100) * Re \$\endgroup\$ Nov 5, 2021 at 3:54

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