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The circuit diagram I've attached has the direction of current flow represented by red arrows. It's the solution given for the 2006 Pennsylvania State Circuit lab competition for Science Olympiad from here: https://scioly.org/wiki/index.php/Test_Exchange_Archive#Circuit_Lab

I can't wrap my head around why Q1 has current going through it, but Q2 doesn't. It seems like current going down through R3 should engage the base of Q2 and current should flow down through the LEDs and through Q2...

Please help, so confused!

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  • \$\begingroup\$ Would it be due to the resistance of C-E for Q1 is much lower than B-E for Q2... \$\endgroup\$
    – Solar Mike
    Mar 24, 2019 at 20:40
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    \$\begingroup\$ All the battery symbols are upside down. The large plate is positive. The circuit won't work. If the batteries were the right way up we'd need to know the resistor values to determine if the transistors are on or not. \$\endgroup\$
    – Transistor
    Mar 24, 2019 at 20:41
  • \$\begingroup\$ This is supposed to be a circuit for a solar path light and the two batteries on the right are being charged. I forgot to mention that the one on the far left is a solar panel. \$\endgroup\$ Mar 24, 2019 at 20:46
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    \$\begingroup\$ Q1 is probably turned on hard enough that it is in saturation. This would pull the base of Q2 low enough that it wouldn't be able to turn on. (And if Q2 did turn on hard, you'd almost certainly get excess current in the LEDs, because there's no current-limit resistors). \$\endgroup\$
    – TimWescott
    Mar 24, 2019 at 20:51
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    \$\begingroup\$ @SpehroPefhany me, too. I'm sure it made sense in the 1750's or whenever the convention was established. That's a schematic drawing of a Volta Pile there, kid! \$\endgroup\$
    – TimWescott
    Mar 24, 2019 at 21:11

2 Answers 2

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Assuming the + and - of the batteries are correct and not the polarity shown by the battery symbols (the long line is the +, short line is -) and also assuming that the battery voltages are such that D1 isn't destroyed then:

It depends on the values of R1, R2 and R3 what mode of operation Q1 will be in.

If Q1 operates in active mode then it will have a large enough Vce to enable some current (coming from R3) to flow into the base of Q2. As the LEDs are indicted as not lighting up, we can assume that this is not the case.

If Q1 operates in saturation mode then Vce will be much smaller, typically less then 0.4 V which will prevent any current to flow into the base of Q2. For a current to flow into the base of Q2 there needs to be more than about 0.6 V across the base-emitter of Q2.

This isn't a well designed circuit. If Q2 would conduct then it also determined the current through the LEDs which (depending on Q1 and resistors) could be high enough for the LEDs to conduct too much current. There should have been resistors in series with the LEDs.

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  • \$\begingroup\$ The batteries on the right are pairs of NiCd (\$1.2\:\text{V}\$ each.) (Per the contest question.) So, charged, they are at about \$2.4\:\text{V}\$. Given the \$V_\text{CE}\$ of \$Q_2\$ of a few tenths of a volt, it's likely it "worked okay" at night without too much fear of burning them out. The main problem I'd worry about is one of the two LEDs hogging most of the current with the other one being noticeably dimmer. \$\endgroup\$
    – jonk
    Mar 24, 2019 at 21:25
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Ignoring the fact that the battery symbols appear to be drawn incorrectly, let's assume that Q1 is sufficiently biased so that it is fully saturated. If so, Vce of Q1 will be close to zero and so the input voltage to Q2 will be lower than the minimum voltage (typically about 0.6 to 0.7V) for Q2 to begin conducting. In other words, Q2 will be in cutoff and no current will flow through the two LEDs (shown as diodes).

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