3
\$\begingroup\$

From the following circuit:

enter image description here

I got the following results in comparison to my theorized calculations:

enter image description here

Why were my result different to my calculations?

Here is my calculations:

enter image description here

Equipment used:

  • Agilent E3630A
  • Agilent 33210a
  • Agilent MSO-X2012a
  • Agilent U3401A
\$\endgroup\$
9
  • \$\begingroup\$ It would help a lot if you provided the process and calculations you used to get your theorized values. \$\endgroup\$
    – jonk
    Nov 7, 2019 at 7:22
  • \$\begingroup\$ Note how your \$V_{ce}\$ is very small in the measurement. The transistor now operates in saturation mode. I suspect that you have an error in the circuit that you're measuring. To investigate, also compare the voltages at base, emitter and collector (\$V_b, V_e, V_c\$) \$\endgroup\$ Nov 7, 2019 at 7:33
  • \$\begingroup\$ I have provided my calculation work flow. \$\endgroup\$
    – Redsam121
    Nov 7, 2019 at 8:01
  • \$\begingroup\$ Do you assume that no voltage is applied to Vin? And are you calculating and measuring at DC steady state? \$\endgroup\$ Nov 7, 2019 at 8:14
  • 1
    \$\begingroup\$ Just one example: in your calculation you use \$\beta=200\$. Where do you get this value from? What reason do you have to assumne that it is the correct value for your real component? \$\endgroup\$
    – Curd
    Nov 7, 2019 at 8:33

1 Answer 1

17
\$\begingroup\$

Your measurements are impossible if the circuit shown is correct.

12V / 20kΩ = 0.6mA. Therefore there cannot be 8.03mA going into the Base. However the slightly high Base-Emitter voltage suggests there is, so the '20k' resistor isn't 20kΩ.

According to your measurements, the voltage at the Emitter should be 200Ω*(8.03mA+0.5114mA) = 1.708V, and voltage at the Base should be 1.708V+0.7745V = 2.48V. Therefore current through the 3.3k resistor should be 2.48V / 3.3kΩ = 0.75mA. (12V-2.48V)/(0.75mA+8.03mA) = 1.08kΩ. So the '20k' resistor is actually ~1.1kΩ.

Voltage at the Collector should be 12v-(10.83mV+1.708V) = 10.28V. 10.28V / 0.5114mA = 20.1kΩ. This suggests the '1.2k' resistor is actually 20kΩ.

In other words, the 20k and 1.2k resistor are swapped around.

\$\endgroup\$

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.