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My supply voltage is +10 V, collector current at my bias point is 5 mA, base current is 28.8 uA, amplification is 173.3.

enter image description here

I already calculated the base resistor to be $$R_1=\frac{V_{CC}-V_{BE}}{I_b}$$ and it seems to be right.

These are the correct values according to the solution:

  • \$R_1=322\text{ kΩ}\$
  • \$R_2=2140\text{ Ω}\$

I can't seem to find out how to get \$R_2\$.

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1 Answer 1

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Based on the numbers you have, and assuming you want the collector to rest at 50% of the supply voltage. then Ohm's Law:

E = I x R

R = E / I

R2 = (Supply/2) / 0.005 A

R2 = 5 / .005

R2 = 1 K

Note that the circuit will not actually work. Biasing the base with a single resistor is called dangle biasing, and is severely unstable. The gain of the transistor changes with temperature, with changes in collector current, and from part to part.

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  • \$\begingroup\$ why would you assume the collector should rest at 50% of the supply voltage, is that common scenario ? Also, the solution is not 1k but 2.14k. Any clues to that ? \$\endgroup\$
    – kellogs
    Mar 3, 2019 at 18:50
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    \$\begingroup\$ Given the large lack of information from the OP, assumptions happen. It is common for a linear amplifier to be biased to the power rails midpoint to maximize the available peak-to-peak voltage swing. 10 V Vcc and 5 mA Ic are stated, so the solution for the operating point I mentioned is 5 mA. How did you get 2.14 K? Note that 2.14 K is not an E96 standard value. \$\endgroup\$
    – AnalogKid
    Mar 3, 2019 at 19:38
  • \$\begingroup\$ Ah, I see. 2.14k is just the stated correct solution, probably picked up from an exercise book. Anyway, I was trying to to use one such NPN transistor not in its active quadrant but at saturation. How would R2 be determined in that case ? Say, for 5V supply, my transistor can withstand 50V across the CE junctions; do I even need an R2 in this scenario ? \$\endgroup\$
    – kellogs
    Mar 3, 2019 at 19:53
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    \$\begingroup\$ R2 limits the collector current to a max value at saturation. Without it the transistor will attempt to dead short the power supply and fail. \$\endgroup\$
    – AnalogKid
    Mar 3, 2019 at 21:18
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    \$\begingroup\$ Using Ohm's Law, the voltage drop across 2140 ohms at 5 mA is 10.7 V. This is not possible in the circuit. \$\endgroup\$
    – AnalogKid
    Mar 3, 2019 at 21:19

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