# Electronic – How to calculate the base and collector current in this circuit

currenttransistors

I have the following circuit, using a BJT NPN transistor:

How can we calculate the current going into the base and collector?

I was thinking of using Kirchoff's current law:

\$I_1 = I_b + I_c\$ where \$I_1 \$ is the current coming out from the positive terminal.

Which gives us \$I_b = I_1 – I_c \$ and \$I_c = I_1 – I_b \$

I then however need to find \$I_1\$ and either \$I_b\$ or \$I_c\$ to calculate the last constant.

If this was an easier circuit, what I would do is add the resistors together with the formula

\$\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2}\$

and then treat them like one resistor. I could then use Ohm's law to find \$I_1 \$. However, the LED and the transistor confuse me. Does this method still work in this case? And if it does, would it just leave me with 1 resistor? Would I then ignore the other components?

1. The transistor is on, so its \$V_{BE}\$ is about 0.6V.
2. The transistor is in saturation, so that its \$V_{CE}\$ is about 0.2V.
3. Assuming the LED is white, its \$V_F\$ is about 3V.
From these, you can calculate the base current \$I_B=\frac{9\rm{V}-0.6\rm{V}}{220\rm{k\Omega}}\$ and collector current \$I_C=\frac{9\rm{V}-0.2\rm{V}-3\rm{V}}{330\Omega}\$, and then find \$\beta=\frac{I_C}{I_B}\$. If you find a transistor whose datasheet \$\beta_F\$ is larger than this value, then the circuit will work as advertised (once the transistor goes into saturation, its \$\beta\$ will drop to the calculated value). However, if you find a transistor whose datasheet \$\beta_F\$ is less, then it won't work as calculated -- and the collector current will be less, limited to \$I_C=\beta I_B\$.