View attachment Difference.pdf

I have attached two circuits..

One is an led driver using npn transistor and other one is led driver using a resistor..

I wanted to know what is the difference..

If i will not use the transistor and use just the resistor what is the harm?

I just want if my AND gate output is 1 the led should glow else not glow.

The transistor is behaving like a switch or amplifier?

I am confused coz transistors always confuse me..

 

please. export your .docx to a pdf. and upload the pdf.

 

i have changed it to .pdf

 

youre right in your first post, say youre using your chip, but you want to operate say, a 10mm LED.. and that takes 30mA of current, you will not be able to drive the led directly from the port of the chip, with the resistor (circuit B) you would have to use circuit A to make sure it got enough current.
however, say your using a smaller led, e.g.: surface mount or 3mm, and you run it with 5mA current and this will be bright enough (colour dependent) the chip will be able to run the led directly, using the resistor and the led.. the critical information is on your chips datasheet.. use this example
looking at "Recommended operating conditions" page 3.. we can see that the High level output current is 1mA. where as the low level output current is 20mA. this means that that chip would be better current sinking, than current sourcing (i.e.: when the gate is low/off. the led is on.)
does this all make sense?

 

but using circuit b my led is on... led configuration is 3.6v, 140ma and resistor is of 10 ohms.

 

140mA!?!?!?!?!??!
WHAT

 

Texas Instruments recommends that the output high current of an old 7408 is only 0.8mA. Then its output high voltage is a minimum of only +2.4V and is typically +3.4V.
Its output low is 16mA at a maximum of 0.4V.

 

like i said that was just an example.. im more concerned with this 140mA hes apparently driving!

 

A bunch of LEDs with a maximum current of 140mA will glow dimmly with a current of only a couple of mA. Your vision's response to brightness is logarithmic (forgetting about the additional attenuation of the iris).

 

yeh i guess i was thinking of a single led what with the original circuits posted.. but the voltage would be quite high to support a system like that..
food for thought i guess

 

The transistor is behaving like a switch or amplifier?
I am confused coz transistors always confuse me..

Both. It is acting as a current amplifier, since the current driving the base-to-emitter is about 1/10th of the current passing collector to emitter. In your circuit, it is controlled as a switch; the digital input to the base is either on or off. The resistor to the base is chosen to saturate the transistor (turn it fully on at 1/10th the load current) while still limit the B-E current.

You may want to consider using a MOSFET, a special transistor that is more ideal for use as a switch. They require almost no current to the base (called a "gate" on a MOSFET) and have very low internal resistance when switched on. So heat sinking and such concerns are reduced. You'll need a logic level MOSFET to operate at 5V.

 

so suppose i am using 10 ohms resistor for load then the resistor connected to base should be how much??

 

chrisw1990

i dont know how it is litting up but by using circuit b it is litting up and giving 140mA to the led.
the voltage at pin 3 of 7408, which is output pin, is approx 5V.

 

can you take a picture?

 

I dont know how it is lighting up but by using circuit b it is lighting up and giving 140mA to the led.
The voltage at pin 3 of 7408, which is output pin, is approx 5V.

IMPOSSIBLE!
I guess you did not look at a datasheet for the 7408. Its supply is supposed to be +5V. Its output is only +2.8V when it is overloaded with an output high current of only 10mA. Its recommended output high current is only 0.8mA then its typical output high voltage is only +3.4V.

I figure that the transistor and LED have a current of 40mA, not 140mA.
For the trransistor to have an output current of 140mA then its base current must be about 14mA which cannot be supplied by a 7408.

 

audioguru

so then i shall use an led with forward current less than 40mA??

 

The amount of current in an LED determines how bright it will be. Your LED has a high 140mA current rating but at only 40mA it might be the same brightness as an LED with a rating of 40mA.

 

chrisw1990

i dont know how it is litting up but by using circuit b it is litting up and giving 140mA to the led.
the voltage at pin 3 of 7408, which is output pin, is approx 5V.

1. Can you get 7408 these days? Are you using 7408 or some other logic family 7408, like, 74hc08, 74ALS08, etc., and if so, which make? That can make a difference - for example, 74hc part can give about 4mA+ with output guaranteed above 4.5V for output High. An ST part can give you a couple of hundred mV more! In any case, it is only driving the base, so anything above Vbe would be sufficient, though at lower voltages the base resistor becomes tricky.
2. What transistor are you using, and, the base resistor? With a high beta or darlington transistor, you can get just about, 140mA of led current, with a 3.6V LED and a 10 Ohm resistor. The transistor Vce drop would be around 0.3V when driven into saturation, and, can give you 110mA! The 140mA 3.6V led is 1/2W, and,you need to heatsink it sufficiently.