# Newbie problem with transistor switching

First post, so I hope that I adhere well to all the rules.

I couldn't find any other posts that could help me, so here's my problem:

I am trying to build a simple transistor switch with a 3906 PNP transistor. The end-goal of this project is to design a circuit that illuminates one of 5 different LED's to indicate the charge of a Li-Ion battery cell.

I'm starting with the Red LED which should illuminate at ~3.3V but it illuminates at ~4.8V.

My 3906 datasheet says that the minimum base current is 50 nA, so I assembled the circuit below to generate 50 nA @ 1.6V on the base when the input reaches 3.3V. I assume the issue is that I'm biasing it badly. (I am trying to get my head around transistor theory and failing, me thinks).

Please forgive me for this terrible MS paint schematic done with a laptop mousepad.

• Why didn't you want to use a LM3914? – Ignacio Vazquez-Abrams Jun 14 '15 at 0:05
• Welcome to EE:SE! Your first question looks good. FYI, there's a schematic editor built into the question/answer editor - it's pretty usable with a laptop touchpad, so your days with Paint may soon be over :) – Greg d'Eon Jun 14 '15 at 0:20
• Your battery is shown with the wrong polarity (or the LED and transistor need to be flipped). The arrows in the LED and transistor should point towards the negative terminal of the battery. (And I assume the transistor is a 2N3906, not just a 3906.) – Peter Bennett Jun 14 '15 at 7:09
• Thanks for catching that! My schematic was definitely wrong. Also, I avoided using any IC's because this project (for me) is about learning transistors and how to use them as switches. – caleb colbert Jun 14 '15 at 15:00

This answer only uses BJTs, because you explicitly mentioned studying transistor theory. An easier solution would be to use a comparator and a voltage reference.

# Original schematic

I cannot reproduce the 4.8V switch on. Simulating on circuitlab shows a ~1.6V switch on (current through LED > 0.5mA).

Even if we fix the base biasing (to put $V_{BE}$ to ~0.6V), the switch on characteristic is quite soft.

simulate this circuit – Schematic created using CircuitLab

# Second stage

By adding another stage, the switch characteristic is improved:

simulate this circuit

• Thank you! I found a couple issues myself with my math (and my schematic had mis-oriented components). It does seem that I will need to use multiple transistors as in your response. The first stage will have to be the driver for my second stage transistor. Back to the drawing board! – caleb colbert Jun 14 '15 at 14:58