I'm having a trouble selecting the proper transistor to switch this LED circuit


simulate this circuit – Schematic created using CircuitLab

The LEDs are 3.6v at 20 mA.

I tried figuring the calculation for the transistor and really the maths escaped me so I just tried out a BC547 I had around. What really confused me was when the switch was open, the LED's glowed dimly. (When closed they glowed bright as expected). I was surprised as from what I'd read transistors are either full on or off. The dim glow was effected by me touching the resistor or other spots on the bread board.

If it helps, I'm actually testing this with the aim to replace the switch was a capacitive touch sensor such as this one: http://www.mouser.com/ds/2/42/iqs904_datasheet-8785.pdf

Finally I'm aiming to build this in SMD even though I'm prototyping in through-hole parts at the moment.

Any advice would be greatly appreciated.

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  • \$\begingroup\$ Replace the transistor with MOS device. \$\endgroup\$ – AKR Sep 8 '13 at 6:49
  • \$\begingroup\$ Reverted to OP's original version so we can see what is being referenced without actually having to follow the link. \$\endgroup\$ – Chris Stratton Sep 8 '13 at 7:03

You have a few problems. First, you say the LEDs are 20mA standard? Yet you list 15Ω resistors. That's ((5 - 3.6v) / 15Ω) = 93mA EACH. They won't last long.

Second, transistors are not just on/off. That's only if you use them as a switch, in saturation mode. They can and are designed to have a wide range of use.

Third, when SW1 is opened, you leave the base pin and resistor floating. It's neither 0v, or 5v, ie "Off" or "On". It can float between the two, and cause the transistor to switch on. Basically like a little antenna.

Fourth, you have the floating pin and resistor on a breadboard, which has high capacitance. It acts like a capacitor, storing then releasing energy when it can, which is why it varies when you bring your finger near the resistor. Again like a little antenna.

Try adding a 10k resistor after the 100Ω resistor, to ground. This creates a weak pull-down, which will turn off the transistor unless the SW1 is closed, which will override it.

As for the specific IC you linked to, they essentially do the same. Instead of a NPN transistor, the "typical use/reference design" they provide uses a N-Channel Mosfet, but includes a 180k pull down (lower left). A NPN like you have will work just as well, BUT they don't list a maximum current that can be sourced from the load pin. I'd assume, based on the Backlight pin (tested at 3mA at 5v with a max 0.5v voltage sag, see datasheet page 17), it's only a few mA, probably 5mA at best. You might need a high gain transistor, or a darlington pair.

enter image description here

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  • \$\begingroup\$ I guess the answer for this is to use a MOSFET (and get the basics of the circuit correct) I've posted a new question regarding the specifics of the MOSFET as this would be a different question: electronics.stackexchange.com/questions/81682/… \$\endgroup\$ – user28659 Sep 8 '13 at 22:10
  • \$\begingroup\$ @user28659 no, not exactly. The reference design is just that, a reference. You adjust to see fit. A transistor, of the appropriate size and base resistor, will work just fine. A fet will work too. \$\endgroup\$ – Passerby Sep 8 '13 at 23:37

Transistors are not "either on or off" - they can also be used in a linear mode, where the collector current is (roughly) some multiple of the base current. This mode is used in audio amplifiers, amongst many other things.

In this case, to ensure that your transistor truly switches off, you need a resistor from base to ground (emitter).

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  • \$\begingroup\$ Using a BJT as a switch is actually a degenerate mode of operation; it just happens to be an extremely handy mode. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 7 '13 at 23:54

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