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I have a problem I am working on where I have to find the voltage gain of the following circuit with a dc collector current of 0.3mA.

My attempt at a solution involved using the formula A=-(RcIc)/Vt = -(10000*3*10^-4)/0.025 = -120. I am not confident in this solution as I feel like it was too simple and does not account for the "load" resistor. Any advice on where to go from here? I have received a hint to use Thevenin's theorem to transform the circuit but am lost on how that should look.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ What section of that circuit would you feel confident converting to a Thevenin equivalent? \$\endgroup\$ – TimWescott Nov 6 '18 at 0:58
  • \$\begingroup\$ I believe that I am supposed to manipulate the circuit in a way that removes the 10k resistor from collector to ground and replaces it with a voltage source at the base but I am unsure of how to do this. \$\endgroup\$ – Burritos Nov 6 '18 at 3:34
  • \$\begingroup\$ Try this again. "Feel confident" is the key phrase. Put your thumb over the transistor. Does the two resistors with a 5V supply look like something you could reduce to a single source and one resistor? \$\endgroup\$ – TimWescott Nov 6 '18 at 16:50
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At 0.3 milliamp collector current, the derivative of (Vbe / Iout) which I personally label 'reac', is about 80 ohms.

Simply divide that into the collector load resistance, which is the parallel combination of the two 10 KOhm in parallel

5,000 / 80 = 63X

By the way, the derivative of (Vbe / Iout) is 1/gm.

Thus my 'reac' is just 1/gm.

At 1milliAmp Ic, the gm is 1/26 ohms and the reac is 26 ohms.

If you have an emitter resistance that is not bypassed, as in this schematic below, simply add the reac to the Re, then divide into the collector load.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thank you for the response, could you clarify what the variables g and m stand for in the react = 1/gm? If I am assuming correctly, at 0.3mA the gm = 80 and at 1mA gm=26? \$\endgroup\$ – Burritos Nov 6 '18 at 6:22
  • \$\begingroup\$ \$g_m\$ is the transconductance gain, and has units of Siemens (\$\mho\$). \$\endgroup\$ – TimWescott Nov 6 '18 at 16:48
  • \$\begingroup\$ The name of the variable is 'gm', and gm is the derivative of Icollecter versus Vbase_emitter. The 'reac' is just 1/gm. By knowing the 'reac' at 1milliAm is 26 ohms, its very easy to walk thru pure-bipolar designs and understand them. FETs have various degrees of freedom (Gate Spacing of Source to Drain, for example), and channel-doping, that prevent detailed analysis on a napkin. \$\endgroup\$ – analogsystemsrf Nov 7 '18 at 3:12
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If I remember well the transconductance of a bipolar junction transistor (BJT) in common emitter configuration (CE) is \$1/(40 I_c)\$. The fact that \$I_c\$ is given makes me follow the path of writing the small signal current transfer function (actually it's a transconductance: output current over input voltage) as \$i_c = 1000/12 \approx 83 \, u_{in}\$.
For small signals the two \$10 \, k\Omega \$ resistors are considered to be in parallel as the one from collector to \$+5 V\$ is also considered grounded for AC (small) signals.

This gives us the (small signal) output voltage \$u_{out} = 5 \, k\Omega * (1000/12) \, u_{in} \$.
The voltage gain therefore is \$ u_{out}/u_{in} = 415 \$.

If you look up Thevenin source transformation you'll find that the current source in the given circuit (the transistor in the small signal model) is paralleled with \$2\$ resistors of \$10 \, k\Omega \$ each and will form a voltage source with output voltage \$ (5/12)10^6 \, u_{in} \$ with a series resistance of \$ 5 \, k\Omega \$.

The Thevenin source transformation transforms a bunch of sources, in this case only 1 current source--the transistor, into one voltage source with a series impedance.

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