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transistor npn circuit, current wont flow if R5 is removed, why?

Why won't current flow in this circuit if R5 is removed?

When I remove R5, current does not flow.

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    \$\begingroup\$ R5 isn't needed. You likely removed the wrong resistor by accident - maybe R2. \$\endgroup\$ Apr 3, 2023 at 17:43
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    \$\begingroup\$ Unless you “remove” it by replacing it with a short, the LED will light either way when the switch is closed. \$\endgroup\$ Apr 3, 2023 at 17:46
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    \$\begingroup\$ Perhaps show us a photo of your actual wiring, it's entirely possible something is not wired up as you think it is. \$\endgroup\$ Apr 3, 2023 at 17:48
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    \$\begingroup\$ It's not clear to me that R2 and Q1 are needed either, the cathode of the LED could just be connected to ground and you'd get the same effect. But there could be some purpose that's not immediately obvious I guess. \$\endgroup\$
    – John D
    Apr 3, 2023 at 17:54
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    \$\begingroup\$ Yikes, if you replaced R5 with a wire then you've shorted out your 9V power supply. Current will flow (a lot !), just not through the transistor/diode. \$\endgroup\$
    – td127
    Apr 3, 2023 at 19:50

3 Answers 3

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Here is a simulation of this circuit with and without the extra R5 resistor:

schematic

simulate this circuit – Schematic created using CircuitLab

As you can see, the current through the transistor is the same in both cases. The R5 resistor serves only to waste an extra 81 mW for no reason.

Your circuit has three resistors connected to the same (9 V) node. Are you sure you pulled the right one?

edit: I see from your comments that instead of removing the resistor from the circuit as you have written, you removed the resistor and then added a wire. In the future, please make sure to accurately report all the steps you've taken when encountering a problem.

Now on to the answer. To model the actual circuit you've built, we'll need to consider parasitics. Your 9V battery consists of six 1.5-volt alkaline cells in series, with significant internal resistance of anywhere from 1 to 100 Ω. Additionally, your wire has some small parasitic resistance as well -- approximtaely 25 to 50 mΩ per linear foot depending on its cross section. In other words, this is a more-accurate model of the circuit you've built:

schematic

simulate this circuit

As you can see, you've short-circuited the battery. The internal resistance forms the top part of a voltage divider with the wires forming the bottom part. Because it's a 9 V battery, and based on the fact that it didn't vapourize your wires, its internal resistance is probably on the high end. It may have started to get warm, and probably drained pretty quickly.

The LED, on the other hand, sees at best a few microvolts -- certainly not enough to turn it on.

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    \$\begingroup\$ In fact the OP said that when he said "removing R5" he meant "replacing it with a wire".... \$\endgroup\$
    – Blup1980
    Apr 4, 2023 at 8:00
  • \$\begingroup\$ Thanks for the heads up, @Blup1980. It makes sense that there was something else going on. \$\endgroup\$
    – Matt S
    Apr 4, 2023 at 14:28
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In a correctly wired circuit, removing R5 will not prevent the LED from lighting when the switch is closed. Check your wiring, and the health of Q1.

R5 assures a complete and rapid turn-off of Q1 when the switch is opened. It has no other purpose. For that job, it is a relatively low value and consumes unnecessary current. I would increase it to 10K.

It can be located either where it is or from the Q1 base (the right side of R2) to GND. If you move it, it will consume less battery current and have a slightly better turn-off benefit.

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    \$\begingroup\$ But as the circuit is drawn, there's no need for a rapid turn-off of Q1. When SW1 opens, there's no energy storage element to keep the LED on nor is there anywhere for base current to come from. Unless there's something about the circuit not in evidence from the question. \$\endgroup\$
    – John D
    Apr 3, 2023 at 23:29
  • \$\begingroup\$ You might want to check his wiring, too. S1 turns everything off, there's no need for a pull-down. There's no need for the transistor either. But this answer has to address the question as written, which it doesn't. \$\endgroup\$
    – TonyM
    Apr 3, 2023 at 23:54
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It really helps to play with such circuits using a current-limited power supply with built-in voltage and current meters.

Absent that, connect a multimeter to the battery using crocodile or similar clips, so that you can have constant feedback as to the health of the battery.

The battery - in this case - has a fairly low short-circuit current, so shorting it like you did won't cause much damage, other than wasting the battery. In that case, monitoring the voltage is sufficient: any overloads will cause the voltage to drop significantly.

Had you tried this with, say, a small lead-acid 12V battery, or with unprotected lithium cells, you'd have had a fire on the desk in rather short order, with pretty but smoking light-emitting wires... And the voltage on the battery wouldn't have changed all that much while all this excitement was going on.

Using an overload-protected analog ammeter would be best for such experiments. Such ammeters have oversized back-to-back diodes across the meter movement, so that overloading them won't destroy them.

On the other case, if you'd be using a multimeter's DC current function for such monitoring, a sufficiently low-impedance battery will blow the meter's 10A protection fuse. That fuse is quite expensive on good multimeters, since it protects you from having a small explosion once the current is exceeded. For use with fairly low voltages, even the cheap 10A glass fuses in low-cost multimeters are enough. But perhaps for the price of some fuses you could get or make a cheap current-limited supply for experiments.

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  • \$\begingroup\$ Many cheap multimeters have an unfused 10A or 20A input. Sometimes this can be safer than a fuse opening - such as monitoring a CT with 5A secondary. \$\endgroup\$
    – PStechPaul
    Apr 5, 2023 at 1:16

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