# Basic resistor-transistor help

Hi everyone!

First of all: Since im getting into pure electronics lately instead of embedded systems, and I do not want to spam this forum with just electronics questions:
Are there nice and clean fora for electronics only?

Then, the actual question:
I want to build this contactless voltage sensor, as an excercise. Now, I want to power it by a 3.3V button cell instead of a bulky 9V battery.

My question now is: Can I divide the values of all the resistors by ±3, since the voltage is about 3x smaller? Im not sure if this is right since Im not that good with transistors

[Edit: I just noticed the ‘General Electronics’ forum, so if one of the mod’s could post it there, that would be awesome!]

That line of thinking is roughly correct. However, you are neglecting the transistor base-emitter junctions, which each require 0.6 to 0.7 V to be functional. Three in series adds to about 2 V, so the circuit may not work at all on a 3V battery. Also, the red LED requires about 2 V, so the series resistor would need to be much lower for the same current.

When you want to use a low voltage supply more often, you should look into using FETs instead of bipolar transistors.

Thanks for the replies everyone!

I tried it on a breadboard with a 3v button cell, and it works!
Today, I tried to solder it on a very small piece of perfboard, but now it doesn't work anymore the led just lights up all the time, with a little difference when I touch it with my hands...

And about the transistors being in series: aren't they parallel? The drop would then be about 0.35V (1/drop = 1/0.7 +1/0.7 +1/2.7).

And since I have BJT's lying around, Im using those, however, Ill look into FETs if I use low voltage supplies more often. Thanks for the tips!

They are not in parallel.

Also, adding components in parallel does not change the voltage drop. Look up "Kirchoff's Laws".

As far as Ive learned years ago, voltages in series can be added, and when they are in parallel, their reciprocal can be added to get the reciprocal of the voltage drop.

And how come it works then?

No, this isn't quite true.

The series-sum parallel-reciprocal relation you are thinking of works for resistances and inductances.

For voltages you need to solve the entire circuit. Being able to add the base voltage of transistors or diodes depends on an approximation that the drop is approximately 0.7V.

There are two paths through the bipolar junction transistor, the path adding up in series is Vbe, the base to the emitter.

Your soldered board is likely failing (always on) because of leakage currents due to surface contamination. Did you use neighbouring holes for the highest resistance transistor?

Surface resistances are very small, so normally don't cause a problem, but for this circuit they do.

There are a few ways to reduce this leakage
-clean it
-increase the separation between connections
-make holes in the board

Alternatively you have damaged the transistor due to soldering temperatures and it's leaking internally

You need to post a picture of what you made along with a schematic so we can spot where you went wrong.

I'm guessing the schematic is the same as in the article? (Except for lower resistances)

Michiel_:
Are there nice and clean fora for electronics only?

A good "general electronics" forum site can also be found here:

Be aware, that while it is a good forum, it can be a bit daunting to "newbies" - so lurk a bit, and notice how things operate before you dive in with questions and/or help.