Help with biasing resistor

[w1zard]

Hi
Could someone give me a hand with calculating a biasing resistor please?

I want to use a transistor as a switch to drive an LED, as in the attached circuit diagram. I have used various ways of calculating this, some which work, but I'd like to know the correct method to approach this calculation.

Thanks

 

[ericgibbs]

hi,
The way that you have shown is an inefficient way to drive most loads, including an LED.
Try if possible to place the load/LED in the collector circuit of the transistor.
You MUST have some form of current limiting of the current thru the LED, else it could burn out.

I have attached a circuit showing one option when using a transistor to drive an LED.
Again its inefficient, as the LED requires only the series resistor, no transistor or bias resistor is required.

 

[w1zard]

Thank you for the reply. I should have mentioned this is a simplification of the circuit - the base of the transistor is actually being driven by a 5V logic high or low, rather than being strapped to the positive permanently, hence the transistor in the circuit.

Can you clarify a couple of points for me?

Why is it preferred to put the LED/resistor on the collector?
What calculations need to be done done to calculate the resistor at the base? It what circumstances can it be excluded?
Why is the Hfe assumed to be 10 or 20 - the datasheet gives a higher value than this?

Thanks

 

[ericgibbs]

Thank you for the reply. I should have mentioned this is a simplification of the circuit - the base of the transistor is actually being driven by a 5V logic high or low, rather than being strapped to the positive permanently, hence the transistor in the circuit.

Can you clarify a couple of points for me?

Why is it preferred to put the LED/resistor on the collector?
What calculations need to be done done to calculate the resistor at the base? It what circumstances can it be excluded?
Why is the Hfe assumed to be 10 or 20 - the datasheet gives a higher value than this?

Thanks

Hi,
Point #1. When the LED is in the emitter circuit its almost impossible to drive the transistor into saturation [ which ideally you want to do in a switch].
Note your point about the logic drive to the base, always have a series resistor when in common emitter mode.

Point #2, The higher Hfe's shown in the d/s is for the transistor operating in a linear configuration. ie : as a linear amplifier,,, when used as a switch and its in Vce saturation the gain is quite low.

OK.

EDIT:
Added your circuit, to show the losses.
Almost 0.8V across the transistor, so reduced LED current.

 

[w1zard]

Thanks for all the information! Out of interest, what software are you using to draw the circuit diagrams?

 

[ericgibbs]

Thanks for all the information! Out of interest, what software are you using to draw the circuit diagrams?

I use LTspice, its a free download.
Its main purpose is as a simulator of electronics circuits.

There is a Yahoo LTspice user group online which has lots of examples and libraries.

 

[w1zard]

Thanks - I'll check it out

 

[audioguru]

When the transistor is an emitter-follower with the same supply voltage as the source signal then the base resistor is not needed.
The input is a logic high voltage (not +5V from a power supply) that might be as low as 2.4V if it is from an old TTL IC. Then the LED will not light.