I'm DC analysing a cascode configuration of BJT and my teacher do the KVL of the circuit like the Black arrow in the picture. But I don't understand why the KVL straight from the node to the ground, I've thought that KVL must apply to the loop only or is there anything I didn't understand here?


Solve of the DC analysis

picture 1 is problem and 2 is how my teacher solve the DC analysis of this cascode series of BJT


Solve of the DC analysis

picture 1 is problem and 2 is how my teacher solve the DC analysis of this cascode series of BJT

  • \$\begingroup\$ The loop is closed between Vcc and 0V \$\endgroup\$ Aug 8, 2017 at 18:54

1 Answer 1


Without seeing the equation your teacher wrote, it is difficult to guess why you are perplexed.

You are right that KVL needs a loop, but the loop can be implicit. Or, as a consequence of KVL, you may apply the principle that says that the sum of voltages along two different paths between two points in a circuit must be the same. This latter is just another way to state KVL.

In your case, I guess your teacher could have written something along these lines:

voltage at the base (referenced to ground) = sum of voltage drops along the path defined by that arrow

The two points between you are considering the two paths I mentioned above are the BJT base and ground.

EDIT (after the OP added more context to his question)

So it seems I guessed right. The last equation starting with \$V_{B1}\$ is exactly what I suspected your teacher had written:

$$ V_{B1} = V_{BEQ1} + (R_{C1} + R_{C2}) \cdot I_{EQ1} $$

This is KVL plus Ohm's law applied to the loop that runs around these nodes: Q1 base, Q1 emitter, connection point between Rc1 and Rc2, ground.

In other words, applying KVL as I explained above (equality of drops along two different paths), that equation says that the voltage between the base of Q1 and ground (\$V_{B1}\$) is equal to the sum of voltage drops along the path that goes from the base to ground running through: (1) base emitter junction and (2) the two collector resistors. The drop across BE junction is \$V_{BEQ1}\$ whereas the drop across those two resistors in series (using Ohm's law) is \$(R_{C1} + R_{C2}) \cdot I_{EQ1}\$

The first two equations, where \$V_{B1}\$ and \$V_{B2}\$ are calculated are NOT based on KVL, but they are an approximation. They are both applications of the voltage divider formula, which here would not be applicable, because R1, R2 and R3 are not strictly in series.

Why is that formula used then? Your teacher is making the common assumption that the current drawn by the bases of the BJTs are negligible compared to the "divider" formed by those 3 resistors. This assumption in jargon is described saying that we assume the divider to be stiff. In other words, we assume that we designed the values of R1, R2 and R3 so that the current through them is much higher than that flowing in the bases.

Under the stiff divider assumption you can (with good approximation) treat R1, R2 and R3 as a true voltage divider and apply the classic formula to obtain the voltages at the junctions between those resistors. Just as your teacher did.

Rationale of the analysis

First your teacher obtained the (approximated) value of \$V_{B1}\$ and \$V_{B2}\$ using the divider formula under the stiff divider assumption.

Then he obtained with KVL and Ohm's law an exact equation from which he could isolate \$I_{EQ1}\$, having only known quantities in the 2nd member (\$V_{B1}\$ is known as per the previous approximation; \$V_{BEQ1} \approx 0.6V\$ if the BJT is in its active region, as it is assumed to be, because otherwise the circuit wouldn't work as an amplifier).

  • \$\begingroup\$ here is my problem and how my teacher solve it (sorry for the bad hand writing. I don't know why I can't post picture so could you please take a look at the comment below? \$\endgroup\$
    – Thai
    Aug 8, 2017 at 15:34
  • \$\begingroup\$ @ThànhTháiNguyễn: "Below" doesn't work on this site. Questions move up and down with votes or user sort preference. Add additional information into your question instead of into "answers". I think someone is helping you out as I type. \$\endgroup\$
    – Transistor
    Aug 8, 2017 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.