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Transistor help!

hantto

Member
Please help me. I get so confused when i try to read the BC547 and BD140 datasheets! I want to know what's the max current and voltage that one can feed into transistors BASE whitout them braking!
 

kinjalgp

Active Member
For BD140 Max. base emitter volatge is 5V and for BC547 it is 6V.
Hfe min. for BD140 is 40 at bias current of 150mA and
Hfe min. for BC547 is 110 at bias current of 2mA.
 

hantto

Member
OK so the resistor for the BC547 will be (let's say i use 11,3V suply) 11,3-2=9,3 (i want to give 2V to the base) and then the 2mA current 9,3/0,002=4650

So the resistor before the base will be 4150ohm?

(this is just an example to help ME understad a bit more)

And thx kinjal!
 

kinjalgp

Active Member
If you want to give 2V to base, a potential divider should be designed such that it will produce 2V at transistors base as well as it is capable of supplying at least 2mA of current to the base. Lets take an example where Vcc=12V (please use '.' instead of ',') Assuming Emitter is directly grounded,
Applying KVL to base loop
VBB - IB*RBB - Vbe = 0
Here,
VBB = 2V
IB = 2mA
Vbe = 0.7V (Silicon)
So, 2 - 2m*RBB - 0.7 = 0
or, RBB = 650 Ohms

RBB is the parallel combination of R1 and R2 in potential divider. Thus to supply 2mA to base, equivalent resistance of R1 and R2 in parallel should be 650 Ohms.

Now let us assume R2 = 820 Ohms.
Using potential divider formula,
VBB = (R2 * Vcc)/(R1 + R2)
2 = (820 * 12)/(R1 + 820)
Solving above equation, R1 = 4.1k ~= 3.9k (Standard value)

So now we have R1 = 3.9K, R2=820 Ohms
Lets see if parallel combination of both these match 650 Ohms which is our requirement.
RBB = (R1 * R2) / (R1 + R2) = (3.9k * 820) / (3.9k + 820) = 677.54 Ohms which is nearly equal to 650 Ohms.

Thus we have designed potential divider bias circuit for BC547 for Vbe = 2V and Ib = 2mA.

Hope you got it.
 

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