Hi!
I'm quite ignorant with transistors so I'd need help to make this transistor to work (as a switch) because myself I'm hitting my head into the wall since friday.
I have a 12V power supply and a square wave 0-5V as a input on the base.
The bjt is of course in common emitter configuration. How can i calculate
the resistor between the input and the base and the resistor between supply and collector?
My input signal is anyway particular: it's not always guaranteeing 5V. It's the voltage given from the pins of a PC parallel port and this voltage changes depending on the load it's driving, so i guess the right question for the base should be: how much current i should shoot into the base? (then i will find the right resistor for it)
I've tried with both 4.7K but the transistor doesn't turn on.
In the end.. another circuit i should drive with this bjt is a 12V relay (with ofc a diode not to bias the collector) in this case how much current should go into the base? (or what resistor i should use with this 0-5V input?)

Ty for your attention

 

In general, beta for most transistors is about 100 or so, which means that you want to inject about 1/100th of the current into the base that you want at the output. If you're using the transistor purely as a switch, that means you want to saturate the output, or try to cause more current to flow from collector to emitter than your current limiting resistor will allow.

For example, if you want 'on' to mean that 100 mA flows through your output resistor (call it R2), then you want ~1-2 mA to flow through your input resistor (call it R1). Since the diode voltage of the base-emitter junction is generally about 0.7 V, you then want R1 < (4.3V / 2 mA) => R1 < 2.15 k

Since 'on' is 100 mA and a saturated transistor is effectively a short, your R2 resistor would then be R2 = 12V / 100 mA = 120 ohms.

 

As a follow up, I would note that when using a transistor as a switch, I think about the resistors involved as _limiting_ the current. If you don't care whether the load you're driving is sinking 200 mA or 300 mA, and it's only important that it doesn't exceed a maximum of 300 mA, just choose R2 such that no more than 300 mA can flow through it. Then you can saturate the transistor however much you want (e.g. inject 5 mA into the base, which will attempt to cause between 500 and 1000 mA to flow throw the collector) and you won't have to worry about burning up your transistor.

 

mmm.. for example shouldn't it work with 4.7k as R2 and 56 as a R1??
Or even as I tried from beginning 4.7k for both? Shouldn't Ib be AT LEAST 1/100 of Ic?
so if R1=R2 it should be more than ok!

I'm burning my brain.. what am i doing wrong?

 

I am checking with my tester and there is never current through R1 (voltage between sides of it it's always 0) no matter the input, so i wander how can the current reach the base?? I'm starting to think that this bjt is burned.. how can i check it?

 

Without a schematic, "R1" and "R2" are meaningless labels. Is "R1" the base resistor? The collector resistor? Something else?

 

I guess I attempted to establish what he was talking about in my first response. This is how I envisioned the circuit:
http://i258.photobucket.com/albums/hh276/verita5/transistor.jpg

 

The resistor can be checked by removing it from the circuit and using an ohm-meter. It is very unlikely that the resistor is bad - they typically get fat and brown when they open.

The transistor can also be checked by removing it from the circuit and using an ohm-meter: http://www.allaboutcircuits.com/vol_3/chpt_4/3.html A bad transistor seldom gives visual clue that it is bad.

 

Similar to what veritas has explained:

The C2383 xstr has a minimum gain of 60.

If you were going to use a 12V relay with a 180 ohm coil, that would result in
12V / 180 ohm = 67mA collector current

67mA / 60 hfe = 1.1mA base current.

4.3V / 1.1mA = 3.9K this is the highest value base resistor to use with the above described relay and collector current (lower base resistor values will work provided the source can provide the needed amount of current, for instance 2.2K = ~2mA, or recalculate accordingly if there is a lower source voltage <5V).

 

hFE is used for a transistor operating as a linear amplifier with plenty of collector to emitter voltage. Most transistors have their max saturation voltage (when it is used as a switch) spec'd with the base current 1/10th the collector voltage.

We don't know if the 5V signal is still 5V when loaded. It might come from a low current CD4xxx Cmos IC. Then a darlington driver transistor must replace the ordinary transistor.

 

I stand by my assesment of Shocwaver's requirements for a switching transistor (also as in another thread on a similar subject).

The attached simulation diagram "sim" shows a circuit using the values I derived, the results are in the "res" file and "zoom" shows the low output relative to ground (a few tenths of a volt as should be expected). With 0V-4V drive @ 1KHz.

You don't have to drive the base with 1/10th the collector current, that is usually overkill and it places unnecessary stress/requirements on the source driver.