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I'm attempting to build the following circuit to better understand how to use a NPN BJT as a switch.

What I've Tried

I'm calculating the current across the LED (in order to light it up) as:

5.2V* - 1.7V (LED drop) = 3.5V
3.5V / 17mA = 200Ohms

*NOTE - The power source is 5.2V because I'm using 4 AA rechargeables in series (at 1.3V each).

The Problem

The problem I see when I push the button to turn the circuit on is that the transistor becomes very hot. I noticed this the first time because I could smell something. Then I touched it. Ouch! :) I'm assuming I'm providing too much current on the be (base to emitter) circuit.

Things I've Tried / Additional Problem

However, when I attempt to add resistance into the be circuit then the LED doesn't light up, even when my resistor gets down to a value as low as 47Ohms.

Questions

1. Is it possible (due to ratios of current needed) to even power both sides of the circuit from the same power source? Or is it ridiculously difficult or something and not done?


2. Can you help me understand the additional calculation(s) I should be using to power the circuit so my LED will light when I push the button?

^{simulate this circuit – Schematic created using CircuitLab}

When you close the switch, you are applying 5.2 volts across the base/emitter junction, which normally doesn't like more than 0.7 volts - this will destroy the transistor.

One way to use a switch and transistor to contol an LED is:

^{simulate this circuit – Schematic created using CircuitLab}

R2 will limit the base current when thet switch is closed. R3 pulls the base low when the switch is open, to ensure the transistor is not conducting.

Pressing the switch will provide base current through R2, allowing the transistor to conduct, drawing current through the LED and R1. R1 limits the LED current to a safe value.

Your circuit is not designed correctly for what you're trying to do. You currently have the BJT connected in parallel with the LED ciruit, instead of being in series with it. Because of this, in you circuit, when the BJT is turned on, all of the current is flowing directly from +5.2V to GND through the BJT (basically like a short circuit), which is why the BJT is getting so hot. Since there is 0 resistance between +5.2V and GND through the BJT, none of the current is going through the LED (thus staying off). Additionally, the switch SW1 is not connected to be able to turn on the BJT, rather it is connected to provide power to everything all at once.

I think this is closer to what you're looking for:

^{simulate this circuit – Schematic created using CircuitLab}

Note how this allows the switch SW1 to control the BJT via its base pin, which in turn allows the BJT (which is now in series with the LED circuit) to pass current through it from 5.2V, through the D1, through R1, through Q1 and finally to GND.

I tried to make this as similar-looking as possible to your original circuit to make it clearer what was incorrect. However, note that you'll still want to include a pull-down resistor and series-resistor on the Q1 base pin (similar to Peter Bennett's post).