Timeline for Why does this transistor amplifier fail if Rc is greater than 50 ohms?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 18, 2024 at 3:13 | comment | added | periblepsis | @SRobertJames So, for example, I might guess 1.5 V for Ve, add 0.9 V for Vbe to get 2.4 V for the thevenin base voltage. I'd also set Rth*Ib = Vcc/beta, for example. A solution I get from those two is R3=1912 Ohms and R4=478 Ohms. I'd use R3=1800 and R4=470. For example. Then R2 might be 150 or 180? | |
| Jul 18, 2024 at 2:55 | comment | added | periblepsis |
@SRobertJames As a stand-alone circuit that isn't part of something much larger, you should want the drop across the Thevenin resistance to be small with respect to the voltage supply magnitude. A factor of about 100 (scales with \$\beta\$.) So with 12 V, I'd probably expect to see 120 mV across the Thevenin resistance for biasing purposes. That's not a requirement. But it gets me a ballpark guess of what to expect.
|
|
| Jul 18, 2024 at 2:45 | comment | added | periblepsis | @SRobertJames By "very tiny biasing-pair resistor current" I mean the current from the 12 V supply through those two resistors, which is only 180 uA. Normally, when I see a 50 Ohm resistor at the emitter, I might imagine a collector current of about a volt or two divided by that resistance. On the order of tens of mA. The base current of that will then be about the same magnitude as the earlier 180 uA, about 200 uA? So signal variations of base current are likely to significantly alter the drop across the Thevenin source resistance. Which means the biasing point is no longer true. Etc. | |
| Jul 18, 2024 at 2:39 | comment | added | SRobertJames | As your model predicts, lowering the bias voltage allows a much higher Rc, and thus much higher gain! And, in this case, since we're dealing with such a small signal, there's no reason the bias voltage can't be close to ground. Now, you also write "Using a very tiny biasing-pair resistor current, as a ratio of the expected collector current." Mathematically, I don't see how you concluded this, since even with a low Ib, beta*Ib is still sufficiently large. And indeed LTspice shows nothing improved by lowering R3 and R4 by a constant factor. | |
| Jul 18, 2024 at 2:37 | history | edited | periblepsis | CC BY-SA 4.0 |
added 14 characters in body
|
| Jul 18, 2024 at 2:28 | comment | added | periblepsis | @SRobertJames Thanks for letting me know it helped out! :) | |
| Jul 18, 2024 at 2:28 | history | edited | periblepsis | CC BY-SA 4.0 |
added 14 characters in body
|
| Jul 18, 2024 at 2:23 | comment | added | SRobertJames | Very helpful! Thank you for the clear and thorough mathematical explanation. | |
| Jul 18, 2024 at 2:22 | vote | accept | SRobertJames | ||
| Jul 18, 2024 at 0:16 | history | edited | periblepsis | CC BY-SA 4.0 |
edited body
|
| Jul 18, 2024 at 0:05 | history | answered | periblepsis | CC BY-SA 4.0 |