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I'm trying to analyze a common-emitter amplifier circuit, but am not sure why \$R_1\$ and \$R_2\$ make sense for making a quiescent Volt of \$5 V\$. In this, \$\beta=100\$. I've calculated \$I_c = I_e = 0.5\text{ mA}\$, \$I_b = 5 \mu \text{A}\$, so the voltage drop over \$R_e = 0.5\text{ V}\$, which means the voltage at the base is \$0.5\text{ V}\$ and \$0.6\text{ V}\$, from the voltage drop of the transistor, which equals \$1.1\text{ V}\$. But I'm not sure why this matters for \$R_1\$ and \$R_2\$. Is it having to do with the need to ensure the transistor is not broken by too large of a voltage input?

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    \$\begingroup\$ Please fix up your schematic. It has no ground, and it also appears that you did not connect R1 and R2 to the base of the transistor. The reason Vout is 5V (or around 5V) is because of the drop across Rc. 10V - Ic*Rc = 5V. I am assuming ground is the bottom horizontal wire. The voltage at the base sets the voltage at Re, and Re, therefore, sets the current. Assuming Ic=Ie, the drop across Rc determines Vout. \$\endgroup\$ – mkeith Oct 22 '15 at 16:20
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With an amplifier like this, you want to bias the collector to more or less mid-rail, this maximises the +/- swing it can make from the quiescent state before hitting the top rail, or running out of voltage drop across the transistor. Choosing 5v, as half the rail voltage, is about right, though it could be a tad higher.

Once we have chosen the output voltage, we choose a collector resistor. The range of sensible values is quite wide at this stage. Unless we are asked to make a certain output impedance, then 1k to 10k sort of range is reasonable. They have chosen 10k. With 5v on the collector, this means we need a collector current of 500uA.

Now we need to choose, again fairly arbitrarily, the voltage across the emitter resistor. It needs to be enough to swamp variations in transistor Vbe with batch variation and temperature variation. It needs to be small enough to not use up all of the available rail volts. I tend to go for about 10% of rail, as I'm quite conservative. This circuit author has gone for 5% of rail, which is still OK. This is 0.5v, so now we can compute the value of the emitter resistor as 1k.

Now we have the emitter voltage, we need to add Vbe to find the base voltage. I tend to call Vbe 0.7v, ending up at 1.2v. But the beauty of this type of biassing is, it can be a bit off, and still work well. Once you've built it, and see what Vbe you get with your transistor at your current, then you can adjust slightly.

Finally, choose R1 and R2 as a potential divider to give your base voltage. The base voltage gives you their ratio, but you have a further choice, over a relatively wide range, of what the actual values are. Too small, and they load the input excessively. Too big, and variations in the base current will alter the base voltage. You can design for a nominal base current, but a good design will tolerate variations in beta over at least a 2:1 range. As the emitter current is fixed by Re, beta variations will alter the base current the transistor draws.

With an output impedance of 10k, the choice of 12k and 100k for R1/R2 seems a bit low, but then I guess this is just an exercise question. A higher base voltage would allow a larger R2, and therefore higher input impedance.

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  • \$\begingroup\$ does the power source being a 50 mV AC take any effect into this? \$\endgroup\$ – TheStrangeQuark Oct 22 '15 at 17:23
  • \$\begingroup\$ Also, I'm still confused by your last two paragraphs. 100/12 is way higher than 1.2 V \$\endgroup\$ – TheStrangeQuark Oct 22 '15 at 17:31
  • \$\begingroup\$ @TheStrangeQuark How so? The pot-down ratio of 12k and 100k is 12/(12+100) = 0.107, which powered from 10v gives an open circuit voltage of 1.07v. The R1/R2 divider has an output impedance of 12k in parallel with 100k = 10.7k, which means for every 1uA the base draws, the base voltage will drop another 10.7mV. \$\endgroup\$ – Neil_UK Oct 22 '15 at 20:40

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