Schematic Link: http://i1244.photobucket.com/albums/gg572/amandafairchild/b307d96a.jpg
1. Describe in words what kind of circuit do you recognize in this schematic
My answer: 2 stage common emitter NPN BJT amplifier with bypass capacitor
2. Express the output voltage Vo based on Av1 and Av2 (it says no calculations are necessary???)
My answer: I must be over thinking this question because I immediately want to start doing calculations. I know that Av1(Av2) is the gain ratio but I don't see an obvious way of knowing that without some basic algebraic formulas.
3. If the input Vin comes from a source that has its own internal Rs, whould the answer from aboive change? Explain why or how.
My answer: Yes,every successive amplifier becomes the load for the previous amplifier in-series.
(did that make sense?)
Really appreciate the help everyone, extremely and truly grateful!
1. Describe in words what kind of circuit do you recognize in this schematic
My answer: 2 stage common emitter NPN BJT amplifier with bypass capacitor
2. Express the output voltage Vo based on Av1 and Av2 (it says no calculations are necessary???)
My answer: I must be over thinking this question because I immediately want to start doing calculations. I know that Av1(Av2) is the gain ratio but I don't see an obvious way of knowing that without some basic algebraic formulas.
3. If the input Vin comes from a source that has its own internal Rs, whould the answer from aboive change? Explain why or how.
My answer: Yes,every successive amplifier becomes the load for the previous amplifier in-series.
(did that make sense?)
Really appreciate the help everyone, extremely and truly grateful!
-
Those 1nF capacitors are supposed to be assumed as "dead shorts" for any analysis. (That's the usual assumption.) Little re=(k⋅T) ⁄ (q⋅Iq). So Iq is about 1.5mA (which you can compute after finding the Thevenin voltage at the base divider and dropping it by one Vbe before allowing it across the emitter resistor) and re ≈ 18Ω. The 1nF capactors then must be well below that... say 2Ω or so to be ignored. Which implies a frequency of 300Mhz. Not especially believable for your usual BJT amplifier stage like this. There is no inductace in the schematic and when 1nF is a dead short you can be pretty sure that a few nH (not uncommon for a short trace) start to look like serious impedance. They count. And should be in there. So the whole thing falls on its face. It's not credible. But assume that away. And assume the dead short caps, anyway. And assume a reasonable frequency of operation despite what you see there.
Then you are stuck with 5100Ω/18Ω or about a gain of 280 per stage. However, the 2nd stage's impedance does load things down, just as you say it does. There is the Thevenin of the biasing pair (about 5kΩ or so); and there is the β⋅re which with the given β=150 means 150⋅18 or about 2700Ohms, in parallel with the 5kΩ. So now you are talking about 1750Ω or so. That should roughly knock the signal down by a factor of 4. So the overall gain is 280⋅¼⋅280 or real close to 20,000. With an input of 25μV (peak, not peak-to-peak, I assume) this suggests about 500mV at Vo (or 1V p-p.) Unloaded Vo, of course.
If Vin has a source resistance then it will look like part of a divider with that 1750Ω input impedance of your 1st stage. If it is 5kΩ, that cuts it down, times ¼. So yeah, it has an impact.
(The little re value is important for you to know about in cases like these where a capacitor makes a dead short at the emitter to ground. Depending on little re is not smart unless the thing is encased in a thermally stable environment, such as a pace maker for a heart. [And even then one might question it.] You can also just use the value Av=40⋅Iq⋅Rc and get close to the right gain for the stage, too -- a bit high, is all. So 40⋅1.5mA⋅5100Ω comes out to about Av=300.)
I think you'll do okay. You sound like you have a handle on things.
Then you are stuck with 5100Ω/18Ω or about a gain of 280 per stage. However, the 2nd stage's impedance does load things down, just as you say it does. There is the Thevenin of the biasing pair (about 5kΩ or so); and there is the β⋅re which with the given β=150 means 150⋅18 or about 2700Ohms, in parallel with the 5kΩ. So now you are talking about 1750Ω or so. That should roughly knock the signal down by a factor of 4. So the overall gain is 280⋅¼⋅280 or real close to 20,000. With an input of 25μV (peak, not peak-to-peak, I assume) this suggests about 500mV at Vo (or 1V p-p.) Unloaded Vo, of course.
If Vin has a source resistance then it will look like part of a divider with that 1750Ω input impedance of your 1st stage. If it is 5kΩ, that cuts it down, times ¼. So yeah, it has an impact.
(The little re value is important for you to know about in cases like these where a capacitor makes a dead short at the emitter to ground. Depending on little re is not smart unless the thing is encased in a thermally stable environment, such as a pace maker for a heart. [And even then one might question it.] You can also just use the value Av=40⋅Iq⋅Rc and get close to the right gain for the stage, too -- a bit high, is all. So 40⋅1.5mA⋅5100Ω comes out to about Av=300.)
I think you'll do okay. You sound like you have a handle on things.
-
engineering makes me sick