Electronic – Voltage doubler for LED backlight

One reason you see less than 100mA when you connect the display up to a 5V supply through an 8 ohm resistor is because the forward voltage of the back light LED is higher than the 4.2V level. Another reason could be due to the meter that you put in series to measure the current. Meters generally measure current by looking at the voltage drop across a small value resistor and you would have effectively placed that resistance in series with the 8 ohms and thus reducing the current some amount.

For the buffer circuit try connecting it up like this:

This will allow the MCU to switch the NPN transistor on and off successfully. As you had it drawn before the LED back light load was being operated as an emitter follower. In such configuration the load was seeing a voltage that was at least a Vbe drop lower than the voltage the MCU pin was placing on the base circuit of the transistor.

With it redrawn as I've shown it the MCU can saturate the transistor and allow the collector circuit to sink the full current from the supply and through the LED and 8 ohm resistor.

If the MCU is able to place a 5V signal on its output pin then the base current of the transistor with 1K resistor will be:

Ibase = (5 - Vbe) / 1K = (5 - 0.7) / 1K = 4.3mA

The transistor current gain is Hfe. With a desired up to 100 mA of current in the collector the transistor will need an Hfe of at least this much:

Hfe = 100mA / 4.3mA ~= 23.

The transistor part numbers I have shown should have Hfe values that are higher than 23 and so the 4.3mA base current should allow the transistor to fully saturate.

One note is that this circuit is going to turn on the backlight when the MCU pin is placed into the high state. Thus logical '1' = High = On. Nice.

If you have a microcontroller (as I suspect) driving your display, you could probably use a spare pin with pwm to drive a simple capacitive voltage doubler. Thus you'd need only common parts (caps, diode, a bjt/mosfet) - no need for 555, opamp or whatever and no inductor so little added noise.

Edit : here is the classic voltage doubler circuit :

^{simulate this circuit – Schematic created using CircuitLab}

C1 should be much smaller than C2 (lower C1 gives lower ripple, but lower curtent capacity and slower voltage ramp-up), and critically diodes should have lowest possible Vf. You won't get exactly doubling from this circuit as it is, but for what I guess your application is that may not matter much.

Drive PWM pin by setting it to 127, and choose frequency matching C1 (or C1 matching the PWM frequency).