I am having troubles understanding a circuit where to base of a transistor is connected to the ground.

As far as I know a transistor needs a certain voltage on the base so electricity can flow from the collector to the emitter. However, if the base is connected to the ground this can never happen.

In the image you can see, that the base of T2 is connected to the ground and as Ue = 0, the base of T1 is connected to the ground as well. U+ = 5V (don't know if this is important).

Should an explanation be too much work, please point me in the right direction or post links to further resources that might help me.

Thank you very much in advance!

enter image description here

  • 1
    \$\begingroup\$ Not 'a certain voltage on the base', the requirement for forward bias of a transistor is on the base-emitter voltage difference. \$\endgroup\$ – Whit3rd Mar 29 '18 at 18:40

Take a look at this circuit:


simulate this circuit – Schematic created using CircuitLab

Even though the base is grounded, there is still a base current flowing because of the negative power supply in the emitter loop, this is similar to what you have.

The loop equation would look like:

$$0-I_BR_B-V_{BE}-I_ER_E+V_S=0 $$

Which results in \$I_B>0\$, so current does flow in this case (for proper \$-V_S\$).

$$ I_B=\dfrac{V_S-I_ER_E-V_{BE}}{R_B}$$

As long as the base-emitter is forward biased (\$V_{BE}>0\$), then current will flow.

  • \$\begingroup\$ Thank you very much for the detailed explanation. However I would like to ask for clarification regarding the -Vs. So if the ground = 0 and Vs = -10, then there is a difference of 10V and therefore current can flow. Did I get that right? \$\endgroup\$ – pls_help Mar 29 '18 at 18:51
  • \$\begingroup\$ @pls_help: the -Vs is simply a negative power supply, just as +Vs is a positive power supply. \$\endgroup\$ – Peter Bennett Mar 29 '18 at 19:00
  • \$\begingroup\$ @pls_help correct. If you connect one end of a resistor (A) to 0V and the other (B) to -10V, then VAB=+10V. \$\endgroup\$ – Big6 Mar 29 '18 at 19:10

For an NPN BJT to be in the active region the base must be 0.6-0.7V more positive than the emitter. Since base is grounded the emitter should be around -0.6-0.7V. And that is possible. The emitters can go as much negative as \$U_-+V_{ce}(sat)+I_o*R_e\$. If, for example, \$U_-=-5V\$, \$I_o\$=1mA, \$R_e\$=1k ohms, and \$V_{ce}(sat)\$=0.2V, then it can be minimum -5+0.2+1=-3.8V.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.