I want to know why there is a minus sign in the output voltage.
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2 Answers
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From the model viewpoint, the voltage-controlled current source is pointing in the direction that an increase in input voltage leads to a decrease in output voltage.
From a practical point of view, more voltage on the base means more collector current, which decreases the output (collector) voltage in an NPN common-emitter configuration.
If your output was taken across the load resistor (from Vcc to collector) rather than from the collector to ground, then the sign would be positive.
answered Jul 24, 2014 at 16:09
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\$\begingroup\$ In addition to the above explanation: You should realize that in the small-signal equivalent ac circuit diagram the ground connection of the load RL must be interpreted as a connection to the positive supply voltage (which has a zero ac resistance to ground). This explains the direction of the output current source. \$\endgroup\$– LvWJul 24, 2014 at 16:36
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\$\begingroup\$ I must be missing something. Regardless of whether you take the voltage across the load resistor (Rc at this link) or from collector to ground it should be positive, right? See CE Amplifier. How could the collector ever become negative as suggested in the OP? \$\endgroup\$ Jul 24, 2014 at 18:44
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\$\begingroup\$ Perfect, thank you! Except how did you get -228mV? Is Vbe in this case the magnitude of change in Vbe? If it were the change then the output should be -2.28mV, right? If it were the magnitude then it would actually be -139mV. \$\endgroup\$ Jul 24, 2014 at 19:06
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\$\begingroup\$ @sherrellbc Should be 2.28V .. gm is Ic/Vt for a BJT, with Vt ~= 26mV at room temperature. We know RL = 6K, Ic = 1mA, so gm is 38mS. gmVbeRl = 0.038 * 0.01 * 6000 = 2.28V. \$\endgroup\$ Jul 24, 2014 at 19:14
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1\$\begingroup\$ @sherrellbc It's a small-signal model. So imagine the bias point with power supply 12V, collector at 6V, base at 6.6. In the small signal model, Vout is 0V at the bias point. Then you change Vbe by +10mV (to 6.61V), the collector will go down by gmVbeRL (so less than 6V). Say RL = 6K, so 1mA bias, gm might be 38mS at room temperature, so Vout will be -2.28V (small signal). The actual voltage at the collector (large signal) will be 6V-2.28V = 3.72V \$\endgroup\$ Jul 24, 2014 at 19:15
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Recall that, for a resistor, the resistor current enters the more positive terminal thus, the resistor current exits the more negative terminal.
In the referenced circuit, the current \$g_mv_{be}\$ exits the resistor terminal labelled \$v_{out}\$ thus, this terminal is the more negative terminal.
But, the other resistor terminal is connected to the zero volt reference node (AKA 'ground') thus, for positive \$g_mv_{be}\$, the output voltage is more negative than zero.
answered Jul 24, 2014 at 18:09