0
\$\begingroup\$

Find the operating point of the transistors T1 and T2.

I've been given this problem and when I try to apply Thevenin's theorem on resistors R2, R3 and then try to calculate \$I_{C1}\$, I get \$I_{C1} = 0\$ and have no idea why.

schematic

simulate this circuit – Schematic created using CircuitLab

Sorry for the not so good explanation I'm new in this community and new in the electronics domain, it isn't really my strength. Any help will be appreciated.

Applying thevenin's on R2, R3, we get:

schematic

simulate this circuit

If we take the loop with V1, R1, T1, V2, using K2:

$$Vcc - 0 = U_{EB}+I_{B1}*R1+ V1$$

$$I_{c1}/\beta_0 = (Vcc - V1)/R01.$$

With \$\beta_0 = 100\$ and \$U_{EB} = 0.6V\$ and approximating V1 to 9.6 we get I_c1 = 0. And it's really weird.

New contributor
C. Cristi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
\$\endgroup\$
  • \$\begingroup\$ Show all of your work, please. \$\endgroup\$ – Elliot Alderson 2 days ago
  • \$\begingroup\$ @ElliotAlderson Ok, I'll try to edit the question \$\endgroup\$ – C. Cristi 2 days ago
  • \$\begingroup\$ And please draw the ground connections towards the bottom (Rotate V1 180 degrees) \$\endgroup\$ – Oldfart 2 days ago
  • \$\begingroup\$ FYI, we use \$ to start and end inline math and $$ for display math. \$\endgroup\$ – The Photon 2 days ago
  • 1
    \$\begingroup\$ Am i the only one who is having a hard time figuring out how Q2 is supposed to work? If i'm not mistaken, in that configuration, Q2 would never be in a conductive state since current needs to flow from collector to base for it to be active yet for Vb to be lower than Vc, current needs to flow into the base. or am i missing the question completely? \$\endgroup\$ – Francois landry 2 days ago
3
\$\begingroup\$

This problem isn't a good candidate for hand solution.

Because the resistor divider formed by R2 and R3 is going to have about \$(10\ V)\frac{0.6}{9.3+0.6}\approx 0.6\ V\$ across R2, and thus there is about 0.6 V across the b-e junction of T1.

That means that T1 is only very weakly turned on, and our basic rules for hand calculations don't handle this situation well. If we have a rule that the b-e junction must be -0.7 V (negative because it's a PNP) to turn the transistor on, then we think this transistor is in cut off mode. Or if we think the b-e junction needs -0.6 V to turn on then we see that the transistor is just barely turned on, but there's nothing to tell us how much current is flowing out of the base.

Even if we use the more accurate 9.393 V you calculated for the Thevenin equivalent voltage of the 10 V source and resistor divider, there's nothing to say that the b-e voltage of the transistor will be exactly -0.6 or -0.7 under these bias conditions. It might be -0.65 V or -0.55 V, and given the small difference between the Thevenin source output and the base voltage in the circuit, this will cause substantial errors in the calculation result.

The best way to solve this circuit would be to plug it in to a simulator (with reasonably chosen transistor models). Or to adjust the resistor values to be sure the transistor is either fully in active mode or fully cut off.

\$\endgroup\$
  • \$\begingroup\$ Why negative because it's pnp? \$\endgroup\$ – C. Cristi 2 days ago
  • \$\begingroup\$ Because a PNP BJT enters active operation when \$V_{be} < -0.6\ \rm V\$ or so. \$\endgroup\$ – The Photon 2 days ago
  • \$\begingroup\$ can you tell me more how did you get that resistor divider? and why it isn't \$ \frac {9.3}{9.3+0.6}(10 V)\$? \$\endgroup\$ – C. Cristi 2 days ago
  • \$\begingroup\$ I was calculating the voltage across R2, instead of across R3, since that's what's relevant of finding the b-e voltage of T1. If I did it more accurately, instead of finding the voltage was 9.393 V above ground, I'd have found that it was 0.607 V below the 10 V supply node. \$\endgroup\$ – The Photon 2 days ago
  • \$\begingroup\$ Can you explain more why it's wrong to have a voltage of 0.6 V across R2 while the b-e junction is also 0.6 V? I don't really get it \$\endgroup\$ – C. Cristi 2 days ago

Your Answer

C. Cristi is a new contributor. Be nice, and check out our Code of Conduct.

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.