Calculating base resistor on transistor?

I just wanted to be sure of the right value.
I have 4 LEDs hooked up to a 3904 transistor. Each LED is lit at 3V, 20mA - I figure the LED resistance is 150Ω while lit - I guess. Series resistor 120Ω. Source is 15 volts. Load to the 3904 collector is therefore [120Ω resistor + (4*150Ω LEDs)] = 720Ω?

So, my base resistor is RB = 0.2 × RL × hFE , about 30k? (hFE is about 200)
Right?

Transistor Circuits

Thanks!

R2-D2 said:

I just wanted to be sure of the right value.
I have 4 LEDs hooked up to a 3904 transistor. Each LED is lit at 3V, 20mA - I figure the LED resistance is 150Ω while lit - I guess. Series resistor 120Ω. Source is 15 volts. Load to the 3904 collector is therefore [120Ω resistor + (4*150Ω LEDs)] = 720Ω?

So, my base resistor is RB = 0.2 × RL × hFE , about 30k? (hFE is about 200)
Right?

Click to expand...

No, you're not trying to use the transistor as a linear amplifier, only as a switch - you want the transistor to be switched hard on, so use a much lower value - 1K should be fine.

Thanks Nigel.
Yes, I am using it as a switching transistor from a 555 timer circuit.

Then why not just connect the LEDs directly to the output of the 555 which can supply up to 200mA?

Hero999 said:

Then why not just connect the LEDs directly to the output of the 555 which can supply up to 200mA?

Click to expand...

Oops, I didn't look at the circuit right. Rather, the LEDs are connected to an OR gate output (4071).

Then it needs a transistor.

Please post the schematic, as it's CMOS, you might not even need a base resistor, especially if the supply voltage is low and the output isn't connected to anything else.

schematic posted

Hero999 said:

Please post the schematic, as it's CMOS, you might not even need a base resistor, especially if the supply voltage is low and the output isn't connected to anything else.

Click to expand...

Part of the circuit...

Please post the whole circuit.

You might be able to save an IC by using diodes for the OR gate.

What's the supply voltage to the logic gate?

The schematic shows one of sixteen of those branches of LEDs. That's why I'm using four of 4071 OR gate CMOS chips (with sixteen transistors). The interconnections between outputs of the dual timer circuits into the OR gate inputs is a bit complex. I am applying 15 volts to all 4071 and the 556 timer.

I built one branch (as shown in schematic) on a breadboard with the timer circuit and tested it without a resistor to the 3904 transistor base. To my surprise, the transistor did not get hot without its base resistor. Then I added 1K.....then 10K to its base and LEDs were just as bright, no change in their intensity. Any higher than about 10K to its base then the LEDs begin to get dimmer. I think this is because there's more voltage drop across the transistor (C-E) as its base resistance increases which allows less voltage across their corresponding LEDs. I might just go with a 1K - seems to work fine.

Why people insistis in explaining the circuit instead of showing it? A recurrent thing in all forums.

You can remove the OR gate ICs and replace them with cheaper diode gates.

Yes, I'll do that. Definitely cheaper. Thanks for that idea and schematic.

Except I made a mistake with the value of the base, resistor, it should've been 10k.

Thanks Hero999!