Will this work using the Diodes to the Base of the Transistor? Do you see any potential issues?

We're using this as a dual level brightness control in my son's toy car.

Thanks!

rage

 

Yeh it probably will work.

I'm jist taking a wild guess at this right now,
but I think the 5K and the 22k may not make much difference, heres my theory,

I think the resistance in the emitter lead is multiplied beta times, when it comes to input impedances, so the emitter restistance would be so high with respect to your 2 base resistors, that the difference between the 2 base resistors may be too small to make much of a optical change in the lighting effect of the led.

What may work better would maybe be to put 2 different resistors switched into the emitter lead. but if you do that you mind as well get rid of the transistor altogether, and just use 2 seperate resistors, with a switch.

this is just a guess..

 

I meant to have that 5k be a variable resistor, so we could dial it in.

What about using a PIC later on to control the switch. I can't do PWM due to other programming that will be working on the chip, so could I simply switch PORTs using the above schematic?

I guess I really want to know if the diodes present any issues with the way they are positioned or is there really no issue there.

 

No issue


the diodes are positioned properly. the setup looks good.

 

1. you don't need the diodes at all
2. if you add another resistor from the NPN base to ground it will work much better

 

What THE_RB said.

You're using the NPN transistor as an emitter follower. The voltage on the emitter will be roughly 0.7v lower than the voltage on the base. It won't take much current at all on the base to have this happen.

[eta]
Here's my interpretation of what THE_RB was referring to:

Your 220 Ohm R1 limits maximum current to the LED. If the LED is rated for 2.1v @ 20mA, it'll get about 10mA when S1 is in the down position (+5V to Q1's base.)

If S1 is in the up position, R2 and R3 act as a voltage divider. Since the pot R2 and R3 have the same maximum value (10K Ohms), the range of voltage to the base is 5v to 2.5v.

The voltage at the emitter of Q1 will be about 0.7v less than the voltage at the base, so this give the emitter a range of about 4.3v down to 1.8v.

So, when S1 is UP, you have nearly full-range control of the brightness of the LED. I say "nearly", because as current through an LED (or just about any diode for that matter) changes, the Vf (forward voltage) changes.

 

I must say "big thanks", because you really took the time to get that spelled out properly.

So, there's no issue running the full voltage to the Base of the transistor? I thought a resistor of some size would have to be in there to protect the transistor from too much current.

Once again - well done!

 

I must say "big thanks", because you really took the time to get that spelled out properly.

Glad you understand it better.

So, there's no issue running the full voltage to the Base of the transistor? I thought a resistor of some size would have to be in there to protect the transistor from too much current.

When you're using a transistor as a voltage follower (as you are), the base current will be minuscule.

It's very different from when it's being used as a saturated switch, or common emitter configuration.