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

[BenBa]

Isn't that a poor Qpoint because it is near the edge, so any amount of swing in one direction of input will cause the transistor to hit the rail. I was told that good transistors have Qpoints in the center of the line for maximal symmetric swing.

EDIT: or is it a perfect Q point because it means that when the voltage in is positive only one transistor is on and vice versa for negative voltage?

 

[gneill]

Isn't that a poor Qpoint because it is near the edge, so any amount of swing in one direction of input will cause the transistor to hit the rail. I was told that good transistors have Qpoints in the center of the line for maximal symmetric swing

That is a concern for single transistor amplifiers. This is a push-pull configuration. When one transistor "hits the rail", the other is just beginning to turn on and take over. Each transistor handles half of the signal (assume a sinusoidal input for simplicity). As a result, theoretically they can both access their entire load lines to handle their half of the signal. A single transistor configuration must handle the entire signal on its single load line.

 

[BenBa]

Awesome that makes so much sense!

Okay so now to get the open loop voltage gain. We just need to ignore the feedback resistor and capacitor and put in a signal and find out what happens to it.

Here is what i think, the input will either be positive or negative (or zero) so only one transistor will be on at a time, so we just need to find the gain from the push pull amplifier from one transistor.

But wait, if it's open loop and the inputs of the op amp are pinned at zero (because its ideal V+ = V-) then shouldn't all of Vin be lost over the first resistor? What does this mean for the output of the op amp?

 

[gneill]

If you cut the feedback loop, the mechanism that keeps the op-amp inputs potential difference at (essentially) zero is removed too...

 

[BenBa]

I thought that the mechanism that kept the op amp inputs both at zero was a property of ideal op amps. Meaning that the inputs are both still held at zero because the plus terminal is at ground. Doesn't this imply that the voltage must drop to zero from Vin over just R1? How does that effect the output of the op amp?

 

[gneill]

I thought that the mechanism that kept the op amp inputs both at zero was a property of ideal op amps. Meaning that the inputs are both still held at zero because the plus terminal is at ground. Doesn't this imply that the voltage must drop to zero from Vin over just R1? How does that effect the output of the op amp?

The mechanism is the very high gain of the op-amp coupled with the feedback loop. Without the feedback loop the ideal op-amp is just a differential amplifier with infinite gain... A "real" op-amp is a differential amplifier with very, very high gain whose output is limited by the supply rails.

 

[BenBa]

The mechanism is the very high gain of the op-amp coupled with the feedback loop. Without the feedback loop the ideal op-amp is just a differential amplifier with infinite gain... A "real" op-amp is a differential amplifier with very, very high gain whose output is limited by the supply rails.

So without the feedback loop in there the output of the op amp is it's rail voltage? Would that be +15 volts? Or -15Volts? Or does it depend on the input at all?

 

[gneill]

So without the feedback loop in there the output of the op amp is it's rail voltage? Would that be +15 volts? Or -15Volts? Or does it depend on the input at all?

It'll be pinned to one rail or the other depending upon the polarity of the input signal, and it can be a very minute signal. In a perfect circuit you might expect the op-amp output to be zero if you grounded the input. But in the real world op-amps aren't perfect and suffer noise and offset currents and voltages. Random noise could set the amplifier bouncing between the rails if the input was open or grounded, or perhaps it would be pinned to one rail in particular due to these imperfections. It's tricky to design practical op-amp circuits lacking feedback.

 

[BenBa]

So for the purposes of getting gain let's assume that the op amp pins the voltage at +15, that means the Q1 is active, so how do we get the gain of the push pull amplifier? we know Vb is 15 volts except it has to drop across R3...

Or is the gain of the circuit infinite since it takes infinitesmally small amount of voltage to get a gain of something clearly large?

 

[gneill]

So for the purposes of getting gain let's assume that the op amp pins the voltage at +15, that means the Q1 is active, so how do we get the gain of the push pull amplifier? we know Vb is 15 volts except it has to drop across R3...

Or is the gain of the circuit infinite since it takes infinitesmally small amount of voltage to get a gain of something clearly large?

The gain will be very large since the op-amp will pin its output and cause the maximum drive the circuit is capable of to the transistors (either positive or negative) for very tiny input signals. You might determine an order of magnitude from the typical open-loop gain of the op-amp specified (look up the device characteristics). Also be sure to specify the sign of the amplification.

 

[BenBa]

The gain will be very large since the op-amp will pin its output and cause the maximum drive the circuit is capable of to the transistors (either positive or negative) for very tiny input signals. You might determine an order of magnitude from the typical open-loop gain of the op-amp specified (look up the device characteristics). Also be sure to specify the sign of the amplification.

Sorry i don't quite understand.
So the Open loop gain of the LF411 is just above 100,000. So clearly the input of the transistors is pinned at 15V (absolute value), but how does the push pull amplifier behave at this voltage? I assume it amplifies it even more, right? Or is the gain from that negligible compared to that of the op amp?

Also what exactly is feedback fraction? How can we have a feedback fraction if it's open loop, or is this asking us about the original circuit again?

 

[gneill]

Sorry i don't quite understand.
So the Open loop gain of the LF411 is just above 100,000. So clearly the input of the transistors is pinned at 15V (absolute value), but how does the push pull amplifier behave at this voltage? I assume it amplifies it even more, right? Or is the gain from that negligible compared to that of the op amp?

You'll have to determine whether or not it's negligible. With a tiny enough input signal the output of the op-amp can (theoretically) be kept between the rails, even open-loop. Given a gain of 100,000 for the op-amp, what input voltage would just bring the op-amp output to a rail? When the op-amp is at the rail, what is the output voltage of the push-pull section?

Also what exactly is feedback fraction? How can we have a feedback fraction if it's open loop, or is this asking us about the original circuit again?

If memory serves, for a closed-loop amplifier the feedback fraction is the reciprocal of the closed-loop gain when the open-loop gain is very large.

I'm not sure whether they are asking (B) about the open or closed loop case. If it's open-loop then no portion of the output is fed back as you have surmised, in which case the answer is trivial. If it's closed-loop then that's a different matter. I wonder if they're still talking about the DC conditions, too. The feedback fraction would change with frequency (thanks to the capacitor in the feedback path).

 

[BenBa]

You'll have to determine whether or not it's negligible. With a tiny enough input signal the output of the op-amp can (theoretically) be kept between the rails, even open-loop. Given a gain of 100,000 for the op-amp, what input voltage would just bring the op-amp output to a rail? When the op-amp is at the rail, what is the output voltage of the push-pull section?

I see, just from some quick math if the input voltage(absolute value) is less than 150 microVolts it will be between the rails. Noise is approximately on that level, i think its fair to assume for this problem that it hits rail voltage.

OH duh! Push pull amplifiers only amplify current not voltage!

If memory serves, for a closed-loop amplifier the feedback fraction is the reciprocal of the closed-loop gain when the open-loop gain is very large.

I'm not sure whether they are asking (B) about the open or closed loop case. If it's open-loop then no portion of the output is fed back as you have surmised, in which case the answer is trivial. If it's closed-loop then that's a different matter. I wonder if they're still talking about the DC conditions, too. The feedback fraction would change with frequency (thanks to the capacitor in the feedback path).

I believe they must be asking about the closed loop case, or else it would be trivial. If the feedback fraction is just the reciprocal of the closed-loop gain when open loop gain is very large (which it seems to be) doesn't that mean it does not vary with frequency?

And how do we find out the closed-loop gain? I know that since the push-pull only amplifies current we are just looking for the gain of the op amp and normally inverting op amps have a gain of -R2/R1 but does this hold when the feedback loop doesn't occur until the all the way after the push-pull amp, and how do we account for the capacitor?

 

[gneill]

I see, just from some quick math if the input voltage(absolute value) is less than 150 microVolts it will be between the rails. Noise is approximately on that level, i think its fair to assume for this problem that it hits rail voltage.

OH duh! Push pull amplifiers only amplify current not voltage!

Push-pull amps are great for achieving large output currents, but there will still be an output voltage.

Open loop, only R4 is in the emitter paths of the transistors and the output voltage will be developed across it. You should be able to estimate the output voltage that occurs when the voltage driving the push-pull stage is ± 15V.

I believe they must be asking about the closed loop case, or else it would be trivial. If the feedback fraction is just the reciprocal of the closed-loop gain when open loop gain is very large (which it seems to be) doesn't that mean it does not vary with frequency?

The closed-loop gain will vary with frequency due to the capacitor in the feedback path (But by how much?). It would be helpful to know what effect the capacitor has on the gain over the given frequency range.

This question can be either relatively simple to answer or rather more complicated... you'll have to be the judge based on the material in the course section that this problem comes from.

Unless your text defines some particular conditions for determining the gain of an audio amplifier (input signal voltage and frequency, amplifier load), my suggestion would be to look at the low frequency (f → DC) gain to answer the problem. That would be the simple version.

And how do we find out the closed-loop gain? I know that since the push-pull only amplifies current we are just looking for the gain of the op amp and normally inverting op amps have a gain of -R2/R1 but does this hold when the feedback loop doesn't occur until the all the way after the push-pull amp, and how do we account for the capacitor?

It holds for the whole amplifier, too. The op-amp will drive the push-pull stage with whatever amount of voltage required so that feedback will null the voltage difference between its inputs. Effectively, the push pull stage becomes the new output stage of the op-amp. (you could draw a bigger triangle and include the push-pull stage inside and call that your op-amp!)

The feedback loop is a filter thanks to the capacitor in the path, so the gain will vary with frequency. The only question is by how much? Check to see what effect it will have at either extreme of the signal frequency range (20 Hz - 20 kHz). How does the capacitor impedance compare to R2 for those frequencies?

 

[BenBa]

Push-pull amps are great for achieving large output currents, but there will still be an output voltage.

Open loop, only R4 is in the emitter paths of the transistors and the output voltage will be developed across it. You should be able to estimate the output voltage that occurs when the voltage driving the push-pull stage is ± 15V.

Are we to again assume that no current flows in the base (meaning there is no drop of voltage across the base resistor? If so it is not a hard calculation because it is 15-0.7. I'm not sure how to calculate it if we can't assume the above.

The closed-loop gain will vary with frequency due to the capacitor in the feedback path (But by how much?). It would be helpful to know what effect the capacitor has on the gain over the given frequency range.

This question can be either relatively simple to answer or rather more complicated... you'll have to be the judge based on the material in the course section that this problem comes from.

Unless your text defines some particular conditions for determining the gain of an audio amplifier (input signal voltage and frequency, amplifier load), my suggestion would be to look at the low frequency (f → DC) gain to answer the problem. That would be the simple version.

This is an introductory course and the professor very barely went over feedback. So i think it's a fair assumption to take the easier route.

So when frequency goes to DC the capacitor acts as an open circuit and we simply have feedback through R2. I know that for an inverting op amp the closed loop feedback gain is -R2/R1 but that is when the feedback loop is taken right after the output of the op amp, in this scenario we have an output resistor, a push-pull amplifier, and a resistor to ground all before the feedback loop starts, so how do i factor those into the feedback?

It holds for the whole amplifier, too. The op-amp will drive the push-pull stage with whatever amount of voltage required so that feedback will null the voltage difference between its inputs. Effectively, the push pull stage becomes the new output stage of the op-amp. (you could draw a bigger triangle and include the push-pull stage inside and call that your op-amp!)

Does this mean i don't need to take R3 into account for the feedback loop of the op amp? But i still have to deal with R4 affecting the feedback loop, right?

The feedback loop is a filter thanks to the capacitor in the path, so the gain will vary with frequency. The only question is by how much? Check to see what effect it will have at either extreme of the signal frequency range (20 Hz - 20 kHz). How does the capacitor impedance compare to R2 for those frequencies?

I am not sure how to "plug in" different frequencies into the circuit to check what happens. I know that the capacitor has impedance 1/jωC so the impedance gets larger with lower frequencies, but how do we account for this in the feedback loop to get a gain?

 

[gneill]

Are we to again assume that no current flows in the base (meaning there is no drop of voltage across the base resistor? If so it is not a hard calculation because it is 15-0.7. I'm not sure how to calculate it if we can't assume the above.

KVL and the relationship between I_B and I_C or I_E via β. Assume β = 100, and V_BE = 0.7V, or look up values for the particular transistor. Solve for the emitter current.

This is an introductory course and the professor very barely went over feedback. So i think it's a fair assumption to take the easier route.

So when frequency goes to DC the capacitor acts as an open circuit and we simply have feedback through R2. I know that for an inverting op amp the closed loop feedback gain is -R2/R1 but that is when the feedback loop is taken right after the output of the op amp, in this scenario we have an output resistor, a push-pull amplifier, and a resistor to ground all before the feedback loop starts, so how do i factor those into the feedback?

You can "move" R4 out beyond the feedback connection point if you wish, it's all one node and where things connect to it don't matter. The added stage is encompassed by the feedback loop, so you can pretend it's part of the op-amp. The op-amp will drive that stage to accomplish its "mission".

Does this mean i don't need to take R3 into account for the feedback loop of the op amp? But i still have to deal with R4 affecting the feedback loop, right?

Yes, ignore R3 and everything else associated with anything "before" the feedback pick-off point. Ignore R4 too! The op-amp will DO ANYTHING to accomplish the requirements of the feedback in nulling the voltage difference at the inputs of the op-amp. This includes forcing the output stage to drive enough current to supply any load, just so long as the feedback requirements are met.

I am not sure how to "plug in" different frequencies into the circuit to check what happens. I know that the capacitor has impedance 1/jωC so the impedance gets larger with lower frequencies, but how do we account for this in the feedback loop to get a gain?

Compare the magnitudes of the capacitor impedance at the two frequency endpoints with the value of R2. Will it make a significant difference? If you wish you can solve for the parallel impedance of C1 || R2, call it Z2, then write the gain as -Z2/R1. Plot |Z2/R1| versus frequency.

EDIT: Fixed resistor number (red)

 

[BenBa]

I can't think of a good KVL loop to use because the op amp gets in the way of any loop i try to create.

I am having trouble comparin magnitudes of capacitor impedance because they are imaginary...

What is the purpose of plotting Z2 over R2 vs frequency? Did you mean Z2 over R1?

 

[gneill]

I can't think of a good KVL loop to use because the op amp gets in the way of any loop i try to create.

The op-amp looks like a 15V source if it's pinned at the top rail. So:

You might need a touch of KCL, too.

I am having trouble comparin magnitudes of capacitor impedance because they are imaginary...

Magnitudes are never imaginary. How do find the magnitude of a complex value?

What is the purpose of plotting Z2 over R2 vs frequency? Did you mean Z2 over R1?

Sorry, I meant R1 of course. It was a typo, and I've fixed it in my post now. Thanks for pointing it out. You want to see how the gain behaves across the frequency range of the amplifier.

 

[BenBa]

So we have 15V - I_3R_3 - V_{be} - I_{Q1e}R_4 = 0
and we have I_3 + I_{Q1c} = I_{Q1e}

But since I_{Q1c} = I_{Q1e} right? But i don't think we can assume that because that would imply I3 is zero.

 

[gneill]

So we have 15V - I_3R_3 - V_{be} - I_{Q1e}R_4 = 0
and we have I_3 + I_{Q1c} = I_{Q1e}

But since I_{Q1c} = I_{Q1e} right? But i don't think we can assume that because that would imply I3 is zero.

The emitter current is the sum of the base and collector currents (KCL). The collector current is β times the base current, or alternatively, the emitter current is (β+1) times the base current. A common assumption is β = 100.

 

[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...