I'm getting more confused.

If B = (R2)/(R1+R2), then what do you mean by "Vinverting is (0.5vcc)(1+R1/R2 or B)"? I take that to mean

Vinverting = (0.5vcc)(1+R1/R2) = (0.5vcc)(1+R1/R2)

But this implies that B = 1 + (R1/R2) = (R1+R2)/(R2)

I also don't know what "the bypass capacitor is doing Verror =0" means.

 

Vinverting= Vout(R2)(R1+R2) as the bypass capacitor is open Vinverting=(infinity)(Vout)/(R1+infinity), Vinverting=1*Vout

Okay, now you are getting someplace, though cancelling out the "infinity" from the numerator and denominator is not valid. There are, well, an infinite number of cases where infinity/infinity is not 1.

But the conclusion here is correct and, at DC, Vinverting = Vout. So if Vnoninverting = Vcc/2 and if Vinverting=Vnoninverting (due to negative feedback), what is Vout at DC?

 

Vout/vnoninverting=R2/R1+R2 as R2 approaches infinity, Vout/vnoninverting=1, given vnoninverting= 0.5Vcc, vout=0.5Vcc and the output remains stable 0.5VCC at DC input(bypass capacitor open)
Please can you correct my statement please or otherwise confirm
Thanks

 

Okay, now you are getting someplace, though cancelling out the "infinity" from the numerator and denominator is not valid. There are, well, an infinite number of cases where infinity/infinity is not 1.

But the conclusion here is correct and, at DC, Vinverting = Vout. So if Vnoninverting = Vcc/2 and if Vinverting=Vnoninverting (due to negative feedback), what is Vout at DC?

Let say when the capacitor is open at dc ,let say R2=1000000ohms, therefore 1000000/(R1+1000000)~=1 but R2 is much bigger than R1
do you get this I think

 

sorry the last post I did a mistake I corrected it

 

You are basically there, but let's clean things up and get in the habit of thinking with a bit more rigor.

First, R2/R1+R2 is NOT the same as R2/(R1+R2). Remember order of operations!

The statement

\(
\frac{Vout}{Vin} \; = \; \frac{R_2}{R_1+R_2}
\)

is incorrect since it ignores the reactance of Cby.

And R2 is R2 -- it's not going to infinity at all.

What you have is

\(
\frac{Vout}{Vin} \; = \; \frac{Z_2}{R_1+Z_2}
\)

where Z2 is the impedance of the series combination of R2 and Cby

To show that at DC this reduces they way you are trying to show, you need to avoid making claims like infinity/infinity equals one. You do this through a chain of reasoning as follows:

\(
\frac{Vout}{Vin} \; = \; \frac{Z_2}{R_1+Z_2}
\;
\frac{Vout}{Vin} \; = \; \frac{1}{\(\frac{R_1+Z_2}{Z_2}\)}
\;
\frac{Vout}{Vin} \; = \; \frac{1}{\(\frac{R_1}{Z_2}+1\)}
\)

As f -> 0, Zs >> R1, therefore R1/Z2 -> 0

\(
\(\frac{Vout}{Vin}\)_{DC} \; \approx \; 1
\)

Now, I'm not saying that you need to write things out so explicitly, but your thinking should follow these lines and it seems like your thinking is a bit too sloppy right now. That will bite you at some point.

 

Let say when the capacitor is open at dc ,let say R2=1000000ohms, therefore 1000000/(R1+1000000)~=1 but R2 is much bigger than R1
do you get this I think

On what basis do you say that R2=1MΩ? What if R2=100Ω? And even if R2=1MΩ, what if R1=10MΩ?

R2 has nothing to do with the DC level. It's all about the capacitor at DC.

 

You are basically there, but let's clean things up and get in the habit of thinking with a bit more rigor.

First, R2/R1+R2 is NOT the same as R2/(R1+R2). Remember order of operations!

The statement

\(
\frac{Vout}{Vin} \; = \; \frac{R_2}{R_1+R_2}
\)

is incorrect since it ignores the reactance of Cby.

And R2 is R2 -- it's not going to infinity at all.

What you have is

\(
\frac{Vout}{Vin} \; = \; \frac{Z_2}{R_1+Z_2}
\)

where Z2 is the impedance of the series combination of R2 and Cby

To show that at DC this reduces they way you are trying to show, you need to avoid making claims like infinity/infinity equals one. You do this through a chain of reasoning as follows:

\(
\frac{Vout}{Vin} \; = \; \frac{Z_2}{R_1+Z_2}
\;
\frac{Vout}{Vin} \; = \; \frac{1}{\(\frac{R_1+Z_2}{Z_2}\)}
\;
\frac{Vout}{Vin} \; = \; \frac{1}{\(\frac{R_1}{Z_2}+1\)}
\)

As f -> 0, Zs >> R1, therefore R1/Z2 -> 0

\(
\(\frac{Vout}{Vin}\)_{DC} \; \approx \; 1
\)

Now, I'm not saying that you need to write things out so explicitly, but your thinking should follow these lines and it seems like your thinking is a bit too sloppy right now. That will bite you at some point.

Many thanks I understood I appreciated much much Your effort to teach me!

 

Last question at DC the output of the circuit is 0.5Vcc before the coupling capacitor(Cout), and after Cout I think the output is 0V? am I right?
Sorry for my last question!
thanks

 

What is the voltage on Cout? How did it get that way? Do you really have any way of saying what the voltage on that capacitor is for the circuit as given? The DC voltage on the output side of Cout will be dictated by whatever other circuit is connected to it and, without information about that circuit, that voltage is indeterminate.

 

Hello
Sorry but I am not so proffeciant in English I am trying to describe what happen when Vin=0V,kindly can you say that I am basically there or not?
a)Suppose that Vin=0V ,the input is 0.5Vcc(because of the potential divider) and at dc Vout/Vin=1, therefore Vout(before Cout with respect to ground)=0.5Vcc(D.C.) and finally Vout(Cout is open at dc)=0V
Thanks
sM

 

Hello
Sorry but I am not so proffeciant in English I am trying to describe what happen when Vin=0V,kindly can you say that I am basically there or not?
a)Suppose that Vin=0V ,the input is 0.5Vcc(because of the potential divider) and at dc Vout/Vin=1, therefore Vout(before Cout with respect to ground)=0.5Vcc(D.C.) and finally Vout(Cout is open at dc)=0V
Thanks
sM

You are basically there but still missing an important point.

1) Suppose that Vin=0V ,the input is 0.5Vcc(because of the potential divider) and at dc -- TRUE
2) Vo/Vin=1 (AT DC), therefore Vo (before Cout with respect to ground)=0.5Vcc(D.C.) -- TRUE
3) and finally Vout(Cout is open at dc)=0V -- FALSE

Note that I tweaked the terminology a bit. The output of the op amp is Vo (not labeled) and the output of the circuit is Vout (labeled in the schematic).

In order for Vo to be 0.5Vcc (relative to ground) and Vout to be 0V (relative to ground), the voltage on Cout has to be 0.5Vcc. How did it get charged to just that level? Where did the electrons that are on the negative plate come from?

If this capacitor started off uncharged, then it is still uncharged because there is no way for charge to enter/leave the negative side of it. So, IF this capacitor started off uncharged, what will the value of Vout be if Vo = 0.5Vcc?

 

Thanks for your help
when the capacitor is charged I think no current will flow through the capacitor and vout is zero I think

 

Thanks for your help
when the capacitor is charged I think no current will flow through the capacitor and vout is zero I think

Again, HOW is this capacitor getting charged AT ALL?

If the initial charge on Cout is 0 C, then what will be the charge on Cout once the system is powered up, with no input signal, and the system reaches DC steady state?

If I know that no current is flowing in a capacitor, then that ONLY tells me that the voltage across the capacitor isn't changing, it does NOT tell me anything about what the voltage across the capacitor actually IS.

 

a) is getting charged if there is a Load
b) Cout=0.5Vcc once charged

 

a) is getting charged if there is a Load
b) Cout=0.5Vcc once charged

But, as given, there IS no load. And you don't know what that load might be even if you assume that there is one. What if the load is a circuit that has a DC bias at the input that is 0.5Vcc? Or what if load is a pull-up resistor? Or what if it is a Zener circuit that is biased to 4.7V relative to ground? You don't know.

 

If you find sometime please read this:
http://forum.allaboutcircuits.com/t...ns-but-im-troubled-by-them.64696/#post-444315
http://forum.allaboutcircuits.com/t...single-supply-op-amp-setup.67075/#post-463133
http://forum.allaboutcircuits.com/t...e-if-my-theory-is-correct.104247/#post-789916

 

Again, HOW is this capacitor getting charged AT ALL?

If the initial charge on Cout is 0 C, then what will be the charge on Cout once the system is powered up, with no input signal, and the system reaches DC steady state?.

Hello
I tried a little experiment with a capacitor in series with a load resistor, when the capacitor is charged 0.5Vcc the voltage across resistor is 0V(please see image attached) I am trying to say If I am correct that when the input voltage(Vin) is 0V (in previous opamp circuit) for some time(the capacitor to be fully charged), Vout is becomes zero across RL. Kindly can you say if am I there?
Thanks very much

 

Again, by putting a load resistor on there you are working with a circuit that is NOT the circuit in the diagram.

And just saying that you have a capacitor in series with a load resistor is not enough. What is the other end of that resistor connected to? There is nothing special about the arbitrarily chosen and labeled GND node. What do you get if you put that resistor to Vcc instead of GND? What do you get if you put it to the tap of a resistive voltage divider similar to the one used to bias the noninverting input of the amp?

 

Ok I understood in some way