I am working on trying to relearn some basic circuit analysis and I am stuck on the attached problem. I have a circuit using a transistor and photoresistor to turn on an LED when dark. I have separated the circuit into three loops with the goal of calculating the current in each loop and the voltage drop across each resistor. However, when I solve my system of equations I end up with negative current values which doesn’t seem right to me.

I’ve double checked my equations and I believe I have the correct sign for my variables and equations, but I could be wrong. Is there something I missed that is resulting in negative current values or is there something simple I missed when setting up each of my loops?

Please let me know if there is important information I need to add. Thanks!

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  • 1
    \$\begingroup\$ I don't want to read that red colored stuff. Can you ... you know ... like use the latex-like syntax we have on the site (kind of MathJax or something)? \$\int x\:\text{d}x=\frac12x^2 +C\$? Readable stuff? But in general, if there's not much light then \$R_2\$ supplies base current to the BJT and it pulls its collector down turning on the LED. Otherwise, if there's lots of light then the BJT doesn't get nearly enough recombination current and it kind of turns the LED off. It's a horrible circuit -- no hysteresis for example. But it's common enough. \$\endgroup\$
    – jonk
    Commented Oct 6, 2022 at 10:55
  • \$\begingroup\$ Don't use KVL through a node as BJT "collector" (your KVL2). Just add KCL equations if needed. \$\endgroup\$
    – Antonio51
    Commented Oct 6, 2022 at 11:31
  • \$\begingroup\$ @Jonk ... Yes. Horrible circuit ... Values of currents very imprecise ... or "false". \$\endgroup\$
    – Antonio51
    Commented Oct 6, 2022 at 12:19
  • 1
    \$\begingroup\$ Learning4Fun, There are two fundamental areas of operation for a BJT: active mode and saturated mode. When the BJT is operating in active mode then \$\beta\$ tends to be high (and can be used as an input parameter for analysis) and the collector acts like a current source. But when the BJT is operating in saturated mode then \$\beta\$ is low (and is an output from analysis, not an input to it) and the collector acts like a voltage source. An approach is to assume active, perform analysis, and see if results make sense. If not, then switch to saturated analysis, instead. \$\endgroup\$
    – jonk
    Commented Oct 7, 2022 at 7:20

2 Answers 2


Thou doth calculate too much.
You can solve it in a faster way.


Let's assume, for a start, that the BJT is operating in the active region: with a 0.7V VBE, we immediately see that the current in R2 will be 0.7/1.5meg = a fraction of microamp. We can neglect it and assume that all the current flowing into R1 will end up in the base of the transistor. Now, what is this current? From

IB = I1 = (VCC - VBE)/R1 = (6 - 0.7) / 10 k = 0.53 mA

Now, with a beta of 100, this base current will result in a collector current of 53 mA.
Let's see if it's tenable: We know the diode will drop about 2.1V, and RC will drop 270*53 mA = 14.3V. We don't even have to bother to imagine what VCE could be because we have already exceeded the supply voltage. What does this mean? That our initial hypothesis of being in the active region is wrong: the transistor is saturated and the collector current is limited by the collector components.

Now, assume the BJT is saturated and its VCE is close to zero. You can assume it's 0.2V or even 0.1V if it's driven hard. Let's be conservative and assume 0.2V. We then have

Vcc = RC IC + Vd + VCEsat
6 = 270 * IC + 2.1V + 0.2 V

from which we deduce

IC = (6 - 2.3)/270 = 13.7 mA

With the approximations used, the transistor shows a beta(forced) of about

beta(forced) = 13.7/0.53 = 26

Which we could interpret as being driver hard. So, by using VCE = 0.1V we can recompute IC = 14.07 mA.
Which is reasonably close to what the simulation shows (LTSpice shows 13.65 mA.)

Your KVLs are completely wrong. You used differences of currents instead of single currents, and it seems to me you completely forgot about the LED drop.

I call I1 the current in R1, I2 the current in R2, IB the current in the base, and IC, RC the current and the resistance in the collector branch. It helps with sanity checks along the way.


However, when I solve my system of equations I end up with negative current values which doesn’t seem right to me.

You are right to think that. It is probably the fact that BJT is "saturated" and "beta" is thus reduced.

Just a remark: I never use a KVL equation that contain a voltage as Vce or Vc.
Just add the needed KCL equations.

Here is what I did. Maple sheet. Vbe linearized (Vbo and rbe ... beta ... to be adjusted).
The variables values listed at the end are: i1, i2, ib, ir1, Vc.
Checking is to be done with a simulator to confirm.

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rbe=1/(40*ib) ...

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Here is a DC Dynamics simulation made with microcap v12.
One can note a dispersion of values, which is obvious.

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For reference, I add a "modified" circuit with some "parameter" behavior. Dynamics DC Analysis.
One can see the behavior of R2 (R1 is an LDR).

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Here is a Maple sheet with the use of Ebers-Moll equations. Only results, (without the LED).

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  • \$\begingroup\$ `It is probably the fact that BJT is "saturated" and "beta" is thus reduced.' is the key when analyzing using the value for \$R_1=1.5\:\text{M}\Omega\$, for why negative values would be received using an active mode value for \$\beta\$! So +1 right out the gate. \$\endgroup\$
    – jonk
    Commented Oct 7, 2022 at 7:15
  • \$\begingroup\$ Not only, @Jonk. By "linearizing", many other parameters are removed, leaving only the "strict" minimum. it is for this reason that certain "values" seem reasonable and others rather "fanciful" like the current in the 1.5 MOhm (calculated with Maple 72 uA versus 514 nA with microcap). Will try with non-linear equation for Vbe. \$\endgroup\$
    – Antonio51
    Commented Oct 7, 2022 at 7:43
  • \$\begingroup\$ @Jonk I have solved with Ebers-Moll equation. See new results in answer (without the LED). EE&O. \$\endgroup\$
    – Antonio51
    Commented Oct 7, 2022 at 9:18
  • \$\begingroup\$ @Antonio51 thank you for your detailed response. For your calculations done with maple how was a 𝛽 value of 45 chosen (or calculated). For Rbe = 1/(40*Ib), where does this equation come from? Is this an assumption or approximation? My EE background is quite basic (1 semester 4 years ago in college) so I have a limited understanding of concepts. Both comments on the quality of the circuit are noted. The circuit came from a DIY electronics book and I attempting analysis based on curiosity. Again thank you for the help! \$\endgroup\$ Commented Oct 8, 2022 at 18:26
  • \$\begingroup\$ No worries. It is a pleasure. Remember that I have used a "linear" transistor (for Vbe). Beta = 45 (guess) was chosen so that Vce = ~ 0.5 V (and not negative ... ) is in the saturation zone. If I remember well, Rbe = 1/(40*ib) is the value of dynamic resistance calculated with d(Vbe) / d(ib) at the quiet point ib (d is the derivative operator). Will add in answer. \$\endgroup\$
    – Antonio51
    Commented Oct 8, 2022 at 19:55

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