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
  • \$\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


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.


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