Op-Amp circuit with a gain of 4

Thread Starter

TL314

Joined Mar 29, 2015
64
So would the value of capacitor not even matter since the current would be zero in this resonant tank?
 

Thread Starter

TL314

Joined Mar 29, 2015
64
Resonance the parallel LC tank circuit acts like an open circuit with the circuit current being determined by the resistor, R only. So the total impedance of a parallel resonance circuit at resonance becomes just the value of the resistance in the circuit and Z = R as shown.
Frequency 40kHz?
 

WBahn

Joined Mar 31, 2012
30,078
The LC tank is at parallel resonance at a particular frequency. That frequency may or may not be 40kHz, but at whatever frequency it is it has an infinite impedance at that frequency.

In order to get a gain of -2 your circuit, because of the two R values in that circuit, whatever is in parallel with the feedback resistance has to have infinite impedance at the frequency you want it to have a gain of -2 at.

What does that tell you about what the resonant frequency of the LC tank needs to be for this circuit?
 

WBahn

Joined Mar 31, 2012
30,078
that the C value has to make the LC tank have infinite impedance?
But at what frequency does it have to make the LC tank have infinite impedance? I can choose a value of C that makes it have infinite impedance at 1MHz. Or I can choose a value of C that makes it have infinite impedance at 1kHz. Get in the habit of stating your conditions in terms that you could turn the problem over to someone that is a bit behind you in your studies and that they could then do it.
 

Thread Starter

TL314

Joined Mar 29, 2015
64
since it tells us a gain of -2 at 40kHz wouldnt I need to find the value of C that creates the LC tank to have infinite impedance? like C=1.58314e^-8
 

WBahn

Joined Mar 31, 2012
30,078
since it tells us a gain of -2 at 40kHz wouldnt I need to find the value of C that creates the LC tank to have infinite impedance? like C=1.58314e^-8
I'm not sure that you are still grasping things (maybe you are and I'm just not getting that warm fuzzy).

If it asked for a gain of -1.5 at 40kHz, you would NOT be looking for the value of C that causes the LC tank to have infinite impedance at 40kHz just because it tells use to find a certain gain at 40kHz.

You want to find the value of C the causes the LC tank circuit to have infinite impedance at 40kHz because you need the LC tank circuit to have infinite impedance at the frequency in order for the gain of the circuit to be -2 at that frequency.
 

Thread Starter

TL314

Joined Mar 29, 2015
64
I'm not sure that you are still grasping things (maybe you are and I'm just not getting that warm fuzzy).

If it asked for a gain of -1.5 at 40kHz, you would NOT be looking for the value of C that causes the LC tank to have infinite impedance at 40kHz just because it tells use to find a certain gain at 40kHz.

You want to find the value of C the causes the LC tank circuit to have infinite impedance at 40kHz because you need the LC tank circuit to have infinite impedance at the frequency in order for the gain of the circuit to be -2 at that frequency.
im not grasping it, so the 40kHz is irrelevant in looking for a value of C that will cause the LC tank to have infinite impedance?
 

Thread Starter

TL314

Joined Mar 29, 2015
64
the resonant frequency = 1/(2 pi sqrt(LC))
so any value of C gives a different resonant frequency when compared to L=.001H
so what resonant frequency would I be looking for to find the value of C that corresponds with it I guess would be my question


earlier you said "In order to get a gain of -2 your circuit, because of the two R values in that circuit, whatever is in parallel with the feedback resistance has to have infinite impedance at the frequency you want it to have a gain of -2 at."

the question asked "what capacitor will this circuit have a gain of -2 at 40 kHz. "

so 40kHz is the frequency I need, so my C value would be 1.58314e^-8
that would give me a resonant frequency of 40kHz

right?
 
Last edited:

Thread Starter

TL314

Joined Mar 29, 2015
64
oh but then you stated "If it asked for a gain of -1.5 at 40kHz, you would NOT be looking for the value of C that causes the LC tank to have infinite impedance at 40kHz just because it tells use to find a certain gain at 40kHz."

so those seem to be conflicting statements so im confused again :/
 

WBahn

Joined Mar 31, 2012
30,078
the resonant frequency = 1/(2 pi sqrt(LC))
so any value of C gives a different resonant frequency when compared to L=.001H
so what resonant frequency would I be looking for to find the value of C that corresponds with it I guess would be my question


earlier you said "In order to get a gain of -2 your circuit, because of the two R values in that circuit, whatever is in parallel with the feedback resistance has to have infinite impedance at the frequency you want it to have a gain of -2 at."

the question asked "what capacitor will this circuit have a gain of -2 at 40 kHz. "

so 40kHz is the frequency I need, so my C value would be 1.58314e^-8
that would give me a resonant frequency of 40kHz

right?
This is right, except that C can't have a value of 1.58314e^-8 because that is just a number, not a capacitance. That's like saying that my height is both 72 and 6. That's meaningless. A physical quantity is inherently tied to the units with which it is measured. I am 72 inches tall and I am 6 feet tall. (72 inches) is equal to (6 feet), but 72 is NOT equal to 6.

So you got a value for C that is 1.58314e-8 F (or farads).

Next, you are not justified in expressing this to six sig figs. In fact, since the inductance is only given to one sig fig you technically aren't justified to more than one or two sig figs. In engineering, the convention is to use three sig figs unless the situation clearly justifies more. The convention is also to use engineering prefixes, so you would have C=15.8 nF (or even just 15 nF).

Now, putting all of that aside, look at the reasoning presented here. We are finding the value of C that results in resonance at 40kHz not because 40kHz happens to be the frequency we are operating at. That is the point that I'm not sure that you are catching.
 

WBahn

Joined Mar 31, 2012
30,078
oh but then you stated "If it asked for a gain of -1.5 at 40kHz, you would NOT be looking for the value of C that causes the LC tank to have infinite impedance at 40kHz just because it tells use to find a certain gain at 40kHz."

so those seem to be conflicting statements so im confused again :/
Okay, so let's change the problem slightly and make that feedback resistor 2000Ω instead of 1000Ω. Everything else is the same, namely you are trying to find the value of C that makes the gain of the circuit -2 at a frequency of 40kHz.

How would you proceed?
 

Thread Starter

TL314

Joined Mar 29, 2015
64
This is right, except that C can't have a value of 1.58314e^-8 because that is just a number, not a capacitance. That's like saying that my height is both 72 and 6. That's meaningless. A physical quantity is inherently tied to the units with which it is measured. I am 72 inches tall and I am 6 feet tall. (72 inches) is equal to (6 feet), but 72 is NOT equal to 6.

So you got a value for C that is 1.58314e-8 F (or farads).

Next, you are not justified in expressing this to six sig figs. In fact, since the inductance is only given to one sig fig you technically aren't justified to more than one or two sig figs. In engineering, the convention is to use three sig figs unless the situation clearly justifies more. The convention is also to use engineering prefixes, so you would have C=15.8 nF (or even just 15 nF).

Now, putting all of that aside, look at the reasoning presented here. We are finding the value of C that results in resonance at 40kHz not because 40kHz happens to be the frequency we are operating at. That is the point that I'm not sure that you are catching.
oh yes im sorry, I wouldnt be given full credit without the units thanks you.

Okay, so let's change the problem slightly and make that feedback resistor 2000Ω instead of 1000Ω. Everything else is the same, namely you are trying to find the value of C that makes the gain of the circuit -2 at a frequency of 40kHz.

How would you proceed?
I am kind of understanding it but what if the frequency and value of C wasnt given, how would I find the resonance and value for C? or in your case your question. if my fb resistor was 2000 then my gain would be -4 if I kept everything the same right?
 
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