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?

  • \$\begingroup\$ Do you know the value of beta for this transistor? Usually, \$\frac{I_c}{I_b}\$ is constant, and you can use the constant to fill in the missing equations. \$\endgroup\$
    – Greg d'Eon
    Apr 28 '15 at 0:20
  • \$\begingroup\$ @Greg I'm afraid not. I actually need to find beta, and that's why I need \$I_c\$ and \$I_b\$ \$\endgroup\$ Apr 28 '15 at 0:22
  • \$\begingroup\$ You can write an equation for beta in terms of the unknown Ic and the Ib, which you should be able to calculate. \$\endgroup\$ Apr 28 '15 at 0:28
  • \$\begingroup\$ @SpehroPefhany My question is how to do the latter \$\endgroup\$ Apr 28 '15 at 0:30
  • \$\begingroup\$ Assume Vbe is 0.7V and proceed from there. Most transistors will not be in saturation so I think you should treat Ic as a variable. \$\endgroup\$ Apr 28 '15 at 0:58

Unless you are good at iterative math, you'll have to make some simplifying assumptions:

  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\$.

  • \$\begingroup\$ What will happen if I do not connect 330 ohm resistor? \$\endgroup\$
    – Fennekin
    Nov 17 '16 at 4:36
  • \$\begingroup\$ In that case, you will simply have \$I_C=\beta I_B\$. We know that \$I_B=38\rm{\mu A}\$; \$\beta\$ is a bit tougher though, as it's generally a strong function of selected transistor, process variation, temperature, and current level. Because of this, it would take some digging in the datasheet, or possibly just building the circuit, to figure out what the resulting \$I_C\$ will be. \$\endgroup\$
    – Zulu
    Jan 18 '17 at 22:32

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