I'm sure an analog of this question has been asked before, but I've spent the last day researching this and I've decided it's time to reach out for some help.

I'm working my way through Electronics for Dummies, 3rd edition. I'm practically finished with my Masters in Computer Science, and this has been the book that I've gravitated towards to break into the world of electronics. There is a specific circuit in Chapter 10 involving a 9 volt battery, two LED's, an NPN transistor, and a voltage divider consisting of a 1 megaohm potentiometer, a 10k ohm resistor, and a 470 ohm resistor (I think the potentiometer counts as part of the voltage divider - please correct me if need be, I really want to learn.)

Schematic for circuit

Breadboard with components

Even with the pot dialed all the way up (to 1 megaohm), the white light still shines super brightly. I'm not sure if I've setup the circuit wrong - I followed the instructions and the illustrations in the book to the best of my abilities. It's my understanding that the white light can shine, but that it should not be bright, and that it should not brighten until I dial the pot all the way down (to 0 ohms).

enter image description here

When the pot is dialed all the way down, the red LED does come on and shines with respect to the setting on the pot. But I thought there shouldn't be any current flowing between the white LED and the base leg of the transistor (except for leakage?) until the voltage rises by about 0.6-0.7 volts? The white light shining brightly with the pot at 1 megaohm has me concerned that I'm missing a major concept here.

I've checked my resistors with my multimeter, all correct there. I checked continuity between the collector and emitter of the transistor, and there is none - so that should mean the transistor is fine, right? Even the potentiometer is testing fine, from 0 to 1 megaohm.

Any comments or suggestions would be very welcome. Apologies if this is extremely stupid (I have no idea) and thanks to everyone who reads this.

Edit: Answered by StainlessSteelRat below, I had the collector and base legs reversed. Thanks again! =D

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    \$\begingroup\$ What is the exact part number of the transistor you used? \$\endgroup\$
    – The Photon
    Mar 17, 2020 at 18:20
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    \$\begingroup\$ Is your 10k really 10k? And which LED us read and which is white in your schematic? \$\endgroup\$
    – winny
    Mar 17, 2020 at 18:21
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    \$\begingroup\$ When you pull out the 10k, does the white LED go out? \$\endgroup\$
    – rdtsc
    Mar 17, 2020 at 18:34
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    \$\begingroup\$ Are you sure you do NOT have the base and the collector reversed? \$\endgroup\$ Mar 17, 2020 at 18:47
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    \$\begingroup\$ @StainlessSteelRat Ding ding ding - I had the collector and the base legs reversed. Thank you so much for the sanity check - if you repost as an answer I’ll pick it as the right one. \$\endgroup\$
    – cjdw94
    Mar 17, 2020 at 18:58

1 Answer 1


I think the potentiometer counts as part of the voltage divider

You can use a potentiometer as a voltage divider but here it is not used that way. Here the voltage across the potentiometer is somewhat constant (9 V (battery) - 0.7 V (Vbe of the NPN) - 2 V (LED) = 7.3 V.

Yes I am ignoring the 10 kohm resistor, I pretend that the potentiometer can vary between 10 k ohm and 1.01 Mohm.

As that 7.3 V across the potentiometer is somewhat constant, the current through the potentiometer will vary with the resistance. At 10 kohm we get 7.3 V / 10 k ohm = 730 uA, at 1.01 Mohm we get 7.3 uA.

So the potentiometer in this circuit behaves like a "variable current source". That current flows into the base of the NPN. The NPN transistor will then amplify this current by a factor 100 to 500 depending on the NPN. That will then vary the intensity of the LED on the right.

  • \$\begingroup\$ Thank you for the explanation about the voltage divider part. I really appreciate it and it does paint a clearer picture on that for me. \$\endgroup\$
    – cjdw94
    Mar 17, 2020 at 18:42

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