This question came up in my exam and is goes like this:

The two following montages were dimensioned to work as signal amplifiers but none of them works correctly.
Explain why and say what signal we get on the output when the input is a sinusoidal signal with 20 mV of amplitude.

The first one seems like an subtractor but i have no ideia why its wrong

About the second one it has positive feedback so i think it can only be a comparator and never an amplifier.
Usually the expression for the second one is something like Vi=-R2/R1*V1 but since we have a resistor on the non inverting input it messes up the things a little bit..
Also is seems more like a subtractor circuit but it can not be so since it has positive feedback....

Thanks

 

What is the DC (no input signal) output level of the opamp in the first one?

 

What is the DC (no input signal) output level of the opamp in the first one?

I would say something like the paralel of the resistors attached to the non-inverting input...but since the inverting input is grounded,so i would say the OpAmp saturates and Vo=+VCC,or possibly and most likely zero since the capacitor behaves like an open-circuit for DC signals

 

I would say something like the paralel of the resistor attached to the non-inverting input...but since the inverting input is grounded,so i would say the OpAmp saturates and Vo=+VCC,or possibly and most likely zero since the capacitor behaves like an open-circuit for DC signals

I have no idea what "the parallel of the resistor attached to the non-inverting input" could possibly mean. It's gibberish. Please try to use complete thoughts and sentences.

If the opamp saturates, why is it saturated? What is the DC (or bias) voltage at the non-inverting input? Assuming the opamp was powered by sufficiently broad rails so that it didn't saturate, what would the output voltage be?

 

I have no idea what "the parallel of the resistors attached to the non-inverting input" could possibly mean. It's gibberish. Please try to use complete thoughts and sentences.

If the opamp saturates, why is it saturated? What is the DC (or bias) voltage at the non-inverting input? Assuming the opamp was powered by sufficiently broad rails so that it didn't saturate, what would the output voltage be?

V+=V-

The circuit seems like a subtractor montage but since we do not have two reference voltages its confusing me

If i had two reference voltages on the + an - inputs i would say

V-=V2*(1k)/(1k+100k)

V+=47k//5.1k/(47k//5.1+7.5k)V1+47k//5.1k/(47k//5.1+0.75k)V2

This configuration is strange to me...

 

The problem is that you are trying to say, "What circuit diagram does this remind me of and what are the magical mystery equations that I memorized for that circuit?". Instead, what you need to learn to do is to analyze circuits based on the fundamentals.

So let's take it one step at a time.

Start from the left of the circuit. If the input voltage is not changing and if you ignore the base current (assume it is negligibly small), what is the voltage at the base of the transistor?

 

The problem is that you are trying to say, "What circuit diagram does this remind me of and what are the magical mystery equations that I memorized for that circuit?". Instead, what you need to learn to do is to analyze circuits based on the fundamentals.

So let's take it one step at a time.

Start from the left of the circuit. If the input voltage is not changing and if you ignore the base current (assume it is negligibly small), what is the voltage at the base of the transistor?

So Ib aprox =0 A,so this means that the transistor is like a short-circuit.

Therefore Vb=Vcc*(7.5k+0.5k)/(47k*5.1k) so Vb=0.5V

 

So Ib aprox =0 A,so this means that the transistor is like a short-circuit.

If something has zero current, then it is like an open circuit. A short circuit is when something has zero votlage across it (which is why we talk about a "virtual short" between the input terminals of an opamp when it is in the active region).

Therefore Vb=Vcc*(7.5k+0.5k)/(47k*5.1k) so Vb=0.5V

[/QUOTE]

Where is this coming from?

Thank you for using units or, more precisely, showing them. You still aren't *using* them, but this is a major improvement.

Look at the units of your expression. You have Vb on the left side, which means you are looking for a voltage. On the right side, you have Vcc, a voltage, multiplied by the ratio of two things. The numerator is the sum of two resistances, so the numerator has units of resistance. The denominator is the product of two resistances, so it has units of resistance-squared. The ratio of resistance to resistance-squared is one-over-resistance. So the voltage is divided by resistance, yielding an answer that has units of current. But you want voltage. The units don't work out and so you KNOW the answer is wrong.

If no current flows in the base of the transistor (imagine cutting the wire going to it), what is the voltage at the junction of the 47kΩ and 5.1kΩ resistors?

 

If something has zero current, then it is like an open circuit. A short circuit is when something has zero votlage across it (which is why we talk about a "virtual short" between the input terminals of an opamp when it is in the active region).




Where is this coming from?

Thank you for using units or, more precisely, showing them. You still aren't *using* them, but this is a major improvement.

Look at the units of your expression. You have Vb on the left side, which means you are looking for a voltage. On the right side, you have Vcc, a voltage, multiplied by the ratio of two things. The numerator is the sum of two resistances, so the numerator has units of resistance. The denominator is the product of two resistances, so it has units of resistance-squared. The ratio of resistance to resistance-squared is one-over-resistance. So the voltage is divided by resistance, yielding an answer that has units of current. But you want voltage. The units don't work out and so you KNOW the answer is wrong.

If no current flows in the base of the transistor (imagine cutting the wire going to it), what is the voltage at the junction of the 47kΩ and 5.1kΩ resistors?

I was seing the transistor like a short instead of an open circuit.
IF we have no current flowing on the base of the transistor then the tension on the base of the transistor is equal to 0 Volts.

 

It is obvious (a little bit of simple arithmatic) that the collector of the transistor is about +7.3V.
The opamp it feeds has a voltage gain (a little more simple arithmatic) of 101 times so its output will try to be at +737V which is impossible so it will be saturated as high as it can go.

The second opamp is obviously not an amplifier because it has positive feedback instead of negative feedback. The resistor in series with its (-) input does absolutely NOTHING and can be removed.

Did you notice that in both circuits the opamps are not powered?

 

I was seing the transistor like a short instead of an open circuit.
IF we have no current flowing on the base of the transistor then the tension on the base of the transistor is equal to 0 Volts.

Why do you say that?

And, I specifically didn't ask for the voltage on the base in the last post (when I said to imagine cutting the wire to it). I want to know the voltage at the junction of the two resistors that previously were connected to it.

See the attachment.

 

Why do you say that?

And, I specifically didn't ask for the voltage on the base in the last post (when I said to imagine cutting the wire to it). I want to know the voltage at the junction of the two resistors that previously were connected to it.

See the attachment.

v_=7.32313423

 

V would be equal to 20V minus the voltage drop in the first resistor of 47K(the current that passes through the resistor is very low) so would be some thing like V=14.99V

If this is the level of your understanding, then you really, really, really are simply not prepared to be doing anything with transistors or opamps. You need to go back to the very first circuits class where all that existed was batteries and resistors and learn things like Ohm's law and Kirchhoff's Laws and fundamental circuit analysis because you have a really weak understanding of these fundamental.

I know that sounds harsh, but I'm afraid that you are faced with a pretty harsh reality that, sooner or later, you are going to have to deal with. It may or may not be your fault that you are in that situation -- many schools are more interested in keeping students enrolled by passing them along in order to keep getting paid than actually making sure they learn anything along the way. The students, in that situation, really can't be faulted either for not learning anything or for not knowing that they haven't learned anything because the school's been telling them that they are doing fine.

But at the end of the day it doesn't matter -- it is YOU that has the weak fundamentals and it is YOU that have to address them, some way, some how, because things will only go downhill from here. Each class you take you will be even less prepared to understand the concepts and, eventually, you will either quit out of frustration, get kicked out because of poor performance that can no longer be overlooked or, after getting milked for everything you've got, get a worthless degree and dumped into a job market that will have nothing to do with you.

But it doesn't have to end up like that! Take charge and do what is necessary to overcome your deficiencies and strengthen your fundamentals. The first thing to do is consider whether the problem is because you haven't been a good student or because you haven't been attending a school that is teaching anything. If the latter, go someplace else! If the former, then reflect and determine if you've been a poor student because of things beyond your control that aren't going to change, or things that you do have control over and have the determination to change. Perhaps you are working long hours to support a big family and simply don't have the time to be a good student. Perhaps you just aren't cut out for this kind of stuff and you really don't see yourself doing much better. If that's the case, find some other field that is better suited to your abilities -- you'll be happier and more successfull in the long run. If the causes are things that you do have the ability and determination to change, then make up a game plan to do so and then execute the plan.

 

V would be equal to 20V minus the voltage drop in the first resistor of 47K(the current that passes through the resistor is very low) so would be some thing like V=14.99V

Absolutely NOT because the supply is 15V, not 20V.

Ignoring the very low base current of the transistor, the simple arithmatic for the transistor is:
1) Current in the base voltage divider is 15V/47k + 5.1k)= 288uA
2) The base voltage is 288uA x 5.1k= 1.47V
3) The emitter voltage is about 1.47V - 0.7V= 0.77V
4) The emitter current is about 0.77V/750= 1.03mA.
5) The collector current is also 1.03mA then the collector resistor has a voltage drop of 1.03mA x 7.5k= 7.7V.
6) Then the collector voltage is 15V - 7.7V= 7.3V

The opamp has a voltage gain of 1+ (100k/1k)= 101

 

Absolutely NOT because the supply is 15V, not 20V.

Ignoring the very low base current of the transistor, the simple arithmatic for the transistor is:
1) Current in the base voltage divider is 15V/47k + 5.1k)= 288uA
2) The base voltage is 288uA x 5.1k= 1.47V
3) The emitter voltage is about 1.47V - 0.7V= 0.77V
4) The emitter current is about 0.77V/750= 1.03mA.
5) The collector current is also 1.03mA then the collector resistor has a voltage drop of 1.03mA x 7.5k= 7.7V.
6) Then the collector voltage is 15V - 7.7V= 7.3V

The opamp has a voltage gain of 1+ (100k/1k)= 101

So i have a colector voltage of 7.3V.What does that tell me?That V+>V-?

 

So i have a colector voltage of 7.3V.What does that tell me?

It tells you that it is biased correctly so the collector is able to swing from about 1.4V to 15V when it is amplifying an input signal.

 

So i have a colector voltage of 7.3V.What does that tell me?That V+>V-?

Recall that, in the active region, the voltage on the two inputs of an opamp are virtually identical, So if V+ is 7.3V, what is V-? If V- is 7.3V what is the current in the resistor going from V- to ground? Given that current, what is the current in the feedback resistor? Given the current in the feedback resistor, what is the voltage across the feedback resistor? Given the voltage at V- and the voltage across the feedback resistor, what is the votlage at the output? Is this voltage reasonable?