Need help analyzing and audio amplifier [final exam study guide]

[BenBa]

AH! So I_c = \beta I_b
I_b + (\beta)I_b = I_e
15V-I_bR_3-(\beta+1)I_bR_4 -V_{be} = 0
Using this we solve for Ib and then use that to find the Vb and then use that to find Ve. Now that we have the voltage out how to we use it to find the feedback fraction?

 

[gneill]

AH! So I_c = \beta I_b
I_b + (\beta)I_b = I_e
15V-I_bR_3-(\beta+1)I_bR_4 -V_{be} = 0
Using this we solve for Ib and then use that to find the Vb and then use that to find Ve.

Sure. That or solve for $I_e$ and then the voltage across R4 as $I_e R4$.

Now that we have the voltage out how to we use it to find the feedback fraction?

The feedback fraction is determined by the components in the feedback loop; the old -R2/R1 for the gain stuff. That's another part of the question. This calculation concerns the open loop gain. You've got the output voltage given that the op-amp is "just" pinned to a rail, and you previously determined the input voltage that would "just" pin the op-amp output to that rail... so what's the open loop gain?

 

[BenBa]

Okay so if 150microVolts is the maxmimum voltage that gets pinned to 15volts, but the output of Vout is actually 15V-I_bR_3-V_{be} then the open loop gain of the circuit is \frac{15V-I_bR_3-V_{be}}{150\mu V}
Correct?

 

[gneill]

Sure, but you can solve for Ib and provide a numerical figure for the open loop gain.

 

[BenBa]

Yes, so i calculate Ib to be 1.4mA, which makes the top of that fraction 13.46 (note that this doesn't match when i calculate this voltage from (\beta+1)I_b*100 which comes out to be 14.14, why is this?

If we take the first value to be correct, then the gain is 13.46/1.4e-6 which is approximately 89000, does this seem reasonable?

EDIT: I did my math wrong, the do match (well they are differing by 0.02 volts). So the Vout is 14.14 V/150microVolts. Thus the gain is 94266.7, is this reasonable? It is close the Open Loop gain of the op amp

 

[gneill]

Yes, so i calculate Ib to be 1.4mA, which makes the top of that fraction 13.46 (note that this doesn't match when i calculate this voltage from (\beta+1)I_b*100 which comes out to be 14.14, why is this?

If we take the first value to be correct, then the gain is 13.46/1.4e-6 which is approximately 89000, does this seem reasonable?

I think $I_e$ should turn out to be about 142 mA, so Vout will be about 14.2V. $I_b$ will be about 1.4 mA as you've found.

A note about real components. In reality, many op-amps can't pin their outputs to the rails. There's often a diode-drop or three difference. Some specially made op-amps can do better. I imagine that this circuit, in "real life" would have an op-amp "pinned" output voltage of just over 13 V typically.

Also, transistors driven with large base currents will experience a larger than 0.7 V V_BE, which will reduce the base current somewhat. I imagine V_BE will be closer to ≈0.8 V with a large base current.

I think that given the number of approximations and assumptions required, your value for the open loop gain is fine. Anything around -85,000 ± several thousand should be acceptable. Note the sign on the gain!

 

[BenBa]

Awesome! Thats great! So we are almost done! I just need feeback fraction and the open loop gain.

I remember seeing somewhere that the feedback fraction was something like \frac{R_1}{R_1+/R_2} where R1 is the input resistor and R2 is the resistance of the feedback loop. Is this still correct if R2 is R2||C1 thus giving an imainary feedback fraction?

For closed-loop gain do we simply use the same equation G=\frac{R_2}{R_1} except replace R2 on top with the complex impedance of R2||C1?

 

[gneill]

I'm not absolutely certain about the "feedback fraction". It's been a long while since I've encountered it. I thought the definition had it as inverse of the closed-loop gain, making it related to a ratio of the components involved in setting the gain of the circuit.

Assuming that is true then the impedance R2||C2 will play a role. But you should confirm what effect C1 is having on the gain! 470 pF is not that large, so it may be there only to compensate for op-amp and transistor gain roll-off at higher frequencies. Check the magnitude of the gain across the frequencies of interest.

 

[BenBa]

To check the gain across multiple frequencies can i use that the gain is (R2||C1)/R1?

For high frequencies the R2||C1 just acts as a short, so that puts Vout at zero volts (because the positive terminal of the op amp is at zero). So does that make this a low pass filter?

 

[BenBa]

I believe I have solved it after re-reading some sections in Horowitz and Hill. Thanks for all the help! Now i need just need to solve the cascode amplifier problem...