Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
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
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.