# Simple switching circuit help - TIP41C

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Hey, I have a TIP41C here, and would like to make a super simple TTL switch with it.

Basically, when it gets >3V on it's base, it fully opens and lets the current flow.

I know how to work with the transistors, but not sure how to have it only react to >3V

If I could make it analog too, as in, it let more current thru as the voltage increased to 5V.

Cheers,
Dan You have some misunderstandings of how a transistor operates.
You need to understand the characteristics of this transistor. Firstly you need to know the current taken by the load.
This will determine the current you need to supply into the base for the transistor to turn on (fully turn on – saturate).

The load current is a LM317 regulator and a laser diode running at 100ma, so I wouldnt say over 150ma.

How are you going to provide the current for the base?

It will need about 150/20 = 7mA

You will need a voltage of about 5v. The base-emitter voltage will be about 2v, so 3v will be dropped across a resistor. The value of the resistor will need to be 3/0.007 = 430R

You will need a voltage of about 5v. The base-emitter voltage will be about 2v, so 3v will be dropped across a resistor. The value of the resistor will need to be 3/0.007 = 430R
The base-emitter voltage is 2V for 6A collector current. It's much less at lower currents. From the data sheet graph it will be about 800mV at 150mA of collector current. For lowest "on" voltage you should use a gain of 10, giving a required base current of 15mA. The base resistor would then be (5V-0.8)/15mA = 280Ω.

If you want to drive it from a TTL output, you will need to connect the base resistor to +5V and connect the TTL output to the junction of the resistor and the transistor base, since TTL can sink much more current then it can source.

This gives an inversion (a TTL high shuts the transistor off) so you will have to add a TTL inverter gate if you want the opposite polarity.

Make sure the TTL gate you are using can sink 15mA when the output is at logic 0.

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