First, note that R1 is not 1kΩ, it's 87kΩ

When the input to the circuit is open, the bias current through R1 leads to a voltage drop of -160 nA * 87 KΩ

This is then the input to the circuit (with the input open), and is in series with Vos and then multiplied by the gain of the circuit.

If you're not sure of the polarity of Vos, you might need to make two calculations using both possibilities.

 

When the input to the circuit is open, the bias current through R1 leads to a voltage drop of -160 nA * 87 KΩ

This is then the input to the circuit (with the input open), and is in series with Vos and then multiplied by the gain of the circuit.

V01 = ((R3 + R2) /R2) * Vos = ((70KΩ + 1KΩ) /1KΩ) * 1,8mV = 127,8 mV

V+ = V-
V+ = - IB-x * R1 = -160 nA * 87 KΩ = -13,92 V

(Vx - V-) / 1KΩ = IB-x
Vx = IB-x * 1 KΩ + V- = 160 nA * 1 KΩ - 13,92 V = -13,76

((Vx - Vo) / R3) + IB-x + ((Vx - 0) / R2) = 0

((-13,76 - Vo) / 70KΩ) + 160nA + ((-13,76 - 0) / 1KΩ) = 0 , and then isolate Vo and multiplicate it with V01.

 

V01 = ((R3 + R2) /R2) * Vos = ((70KΩ + 1KΩ) /1KΩ) * 1,8mV = 127,8 mV

V+ = V-
V+ = - IB-x * R1 = -160 nA * 87 KΩ = -13,92 mV should be -0,01392 V

(Vx - V-) / 1KΩ = IB-x
Vx = IB-x * 1 KΩ + V- = 160 nA * 1 KΩ - 13,92 V = -13,76

((Vx - Vo) / R3) + IB-x + ((Vx - 0) / R2) = 0

((-13,76 - Vo) / 70KΩ) + 160nA + ((-13,76 - 0) / 1KΩ) = 0 , and then isolate Vo and multiplicate it with V01.

Pay attention to orders of magnitude. Redo all these calculations.

For the approximate results you are seeking you can calculate Voutdc for the given offset, and for the given bias current, separately, and then add them. Superposition applies here.

 

Yes, I got 70,97 as you now

No, I don't know how to calculate the output peak amplitude. And what about question 1? I can see that my teacher made those mistakes. He meant, that it was a non inverting amplifier.

Hi,

We'll work on question 3.

STEP 1:
To calculate the peak output voltage Vpk, input the given peak voltage (100mv) and given the gain of the circuit calculate the peak output voltage Vpk.
Next, find the slope (dv/dt) of the output voltage at t=0 for a sine with arbitrary frequency F with a peak voltage of Vpk calculated above.
Compare the slope you find there with the slew rate, and calculate the frequency F=F1 where that slope equals the slew rate. That tells you the max frequency before distortion.

STEP 2:
To calculate the max frequency given the bandwidth of the op amp, divide the bandwidth by the gain. That's the max frequency F2 that you can amplify provided F1 is higher than that.

STEP 3:
If F1 is lower than F2, then F1 is the highest frequency that can be used, but if F2 is lower than F1, then F2 is the highest frequency that can be used with this circuit.

Recap:
1. Calculate slope of output voltage at t=0, from that and the slew rate determine max frequency F1.
2. Calculate max frequency F2 from bandwidth and gain.
3. Compare F1 and F2, make the decision which one to use as the max for this circuit.

 

Pay attention to orders of magnitude. Redo all these calculations.

For the approximate results you are seeking you can calculate Voutdc for the given offset, and for the given bias current, separately, and then add them. Superposition applies here.

V01 = ((R3 + R2) /R2) * Vos = ((70KΩ + 1KΩ) /1KΩ) * 1,8mV = 127,8 mV

V+ = V-
V+ = - IB-x * R1 = -0,00016 * 87 KΩ = -0,01392 V

(Vx - V-) / 1KΩ = IB-x
Vx = IB-x * 1 KΩ + V- = 0,00016 * 1 KΩ - 0,01392 V = -0,01376

((Vx - Vo) / R3) + IB-x + ((Vx - 0) / R2) = 0

((-0,01376 - Vo) / 70KΩ) + 1,6*10^-7 + ((-0,01376 - 0) / 1KΩ) = 0
Vo = [(-0,01376 / 70KΩ) - (0,01376 / 1KΩ) + 1,6*10^-7] * 70 = 0,977

 

Hi,

We'll work on question 3.

STEP 1:
To calculate the peak output voltage Vpk, input the given peak voltage (100mv) and given the gain of the circuit calculate the peak output voltage Vpk.
Next, find the slope (dv/dt) of the output voltage at t=0 for a sine with arbitrary frequency F with a peak voltage of Vpk calculated above.
Compare the slope you find there with the slew rate, and calculate the frequency F=F1 where that slope equals the slew rate. That tells you the max frequency before distortion.

STEP 2:
To calculate the max frequency given the bandwidth of the op amp, divide the bandwidth by the gain. That's the max frequency F2 that you can amplify provided F1 is higher than that.

STEP 3:
If F1 is lower than F2, then F1 is the highest frequency that can be used, but if F2 is lower than F1, then F2 is the highest frequency that can be used with this circuit.

Recap:
1. Calculate slope of output voltage at t=0, from that and the slew rate determine max frequency F1.
2. Calculate max frequency F2 from bandwidth and gain.
3. Compare F1 and F2, make the decision which one to use as the max for this circuit.

Thanks for your time and help, but the above is too complicated for me for calculate.

 

V01 = ((R3 + R2) /R2) * Vos = ((70KΩ + 1KΩ) /1KΩ) * 1,8mV = 127,8 mV

V+ = V-
V+ = - IB-x * R1 = -0,00016 * 87 KΩ = -0,01392 V

(Vx - V-) / 1KΩ = IB-x
Vx = IB-x * 1 KΩ + V- = 0,00016 * 1 KΩ - 0,01392 V = -0,01376

((Vx - Vo) / R3) + IB-x + ((Vx - 0) / R2) = 0 You have units of current here

((-0,01376 - Vo) / 70KΩ) + 1,6*10^-7 + ((-0,01376 - 0) / 1KΩ) = 0
Vo = [(-0,01376 / 70KΩ) - (0,01376 / 1KΩ) + 1,6*10^-7] * 70 = 0,977 Units of current times a dimensionless number
can't give you a voltage.

Your calculation for Voutdc due to Vos is ok, but your calculation for the effect of the bias current on Voutdc has a problem.

For an approximate result all you need to do is calculate the voltage across R1 due to the bias current, then multiply that voltage times the gain and add the 127,8 mV contribution to Voutdc due to Vos. This is a good approximation because at the V- terminal of the opamp the effective resistance the bias current flows in is less than 1k, so the voltage developed there is much less than what is caused at the V+ terminal with its 87 k.

 

For an approximate result all you need to do is calculate the voltage across R1 due to the bias current, then multiply that voltage times the gain and add the 127,8 mV contribution to Voutdc due to Vos.

Like this?:

-0,01392 * 70,97 + 127,8 mV = ?

 

Like this?:

-0,01392 * 70,97 + 127,8 mV = ?

Yes, I think that's a reasonable approximation. Don't forget the possibility that the 127,8mV term might be -127,8 mV.

To get a better result requires a more complicated calculation and that result is not much different.

 

Yes, I think that's a reasonable approximation. Don't forget the possibility that the 127,8mV term might be -127,8 mV.

To get a better result requires a more complicated calculation and that result is not much different.

Thanks, buddy. So VoutDC is -0,01392 * 70,97 + 127,8 mV = 126,8 V? And how can I know, that 127,8mV can might be -127,8mV?

 

Thanks, buddy. So VoutDC is -0,01392 * 70,97 + 127,8 mV = 126,8 V? And how can I know, that 127,8mV can might be -127,8mV?

Watch your orders of magnitude.

-0,01392 * 70,97 + 127,8 mV = 126,8 V should be:

-0,01392 * 70,97 + 0,1278 V = -0,8601 V

The only way to know about the polarity of Vos is whatever your text book, or your instructor, shows as their assumption.

 

Thanks everyone. The deadline is over. But I will still try and solve question 3 tomorrow. I have nothing to lose. Goodnight.

 

Thanks for your time and help, but the above is too complicated for me for calculate.

Hi again,

Oh i guess you are in a different time zone than me. It's not midnight here yet, another four hours.

The output sine will have the form:
Vout=Vpk*sin(w*t)
To find the slope calculate the derivative of Vpk with repect to time t:
d(Vout)/dt=Vpk*w*cos(w*t)

We want the slope at t=0 because that is the steepest part, and the slew rate of the op amp can not keep up with that if it is too steep, so we get:
dv/dt=Vpk*w*cos(w*0)

which of course equals:
dv/dt=Vpk*2*pi*f

That's your maximum positive going slope for Vout. See if you can use that in combination with the actual peak output voltage and the slew rate of the op amp to find the max frequency before distortion starts to occur.

 

Oh i guess you are in a different time zone than me. It's not midnight here yet, another four hours.

Yes I live in Denmark

Hi,
We'll work on question 3.
STEP 1:
To calculate the peak output voltage Vpk, input the given peak voltage (100mv) and given the gain of the circuit calculate the peak output voltage Vpk.
Next, find the slope (dv/dt) of the output voltage at t=0 for a sine with arbitrary frequency F with a peak voltage of Vpk calculated above.
Compare the slope you find there with the slew rate, and calculate the frequency F=F1 where that slope equals the slew rate. That tells you the max frequency before distortion.

Is Vpk found this way?:
Vpk = 100 mV * 0,5 = 50 mV.

and then:
dv/dt = Vpk * 2 * pi * f = 50 mV * 2 * π * 1,79 MHz.

 

Yes I live in Denmark


Is Vpk found this way?:
Vpk = 100 mV * 0,5 = 50 mV.

and then:
dv/dt = Vpk * 2 * pi * f = 50 mV * 2 * π * 1,79 MHz.

Hi,

You are definitely on the right track here, however i have to ask why you used 0.5 as a multiplier when the gain of the amp is higher than 50 (more like 71 or something right?).
So if the gain was 50 we would have:
Vout=Vin*50
Vout=0.100*50=5.00

so if 0.100 was the peak voltage input, then 5.00 is the peak voltage output. You could then continue as you have done and calculate the highest frequency. Using the correct gain however then you will get the right result.

Oh BTW, you dont use the gain bandwidth frequency in this calculation, you leave the frequency as the variable to be solved for. The result will be much lower than the gain bandwidth product.
So given dv/dt=Vpk*2*pi*F and the gain A and the input 0.100 peak, calculate whatever is needed as the output peak and then solve for F:
Vpk*2*pi*F=SlewRateOfOpAmp
The op amp slew rate above is in units of volts per second.

 

I made a mistake with the gain. It is of course 70,97.

Is this correct?:

Vpk = 0,100 V * 70,97 = 7,097 V.

and then:
Vpk * 2 * π * f = 1,9 V/μS
7,097 * 2 * π * f = 1,9 V/μS
f = (1,9 V/μS) / (7,097 *2*π) = 0,043

 

I made a mistake with the gain. It is of course 70,97.

Is this correct?:

Vpk = 0,100 V * 70,97 = 7,097 V.

and then:
Vpk * 2 * π * f = 1,9 V/μS
7,097 * 2 * π * f = 1,9 V/μS
f = (1,9 V/μS) / (7,097 *2*π) = 0,043

Hi,

Wow that's close. The only thing wrong is the units used for the slew rate. The data sheets usually give it in volts per microsecond, but that should be converted into volts per second.
1.9v/us means we have 1900000 volts per second.

So if the peak is 70.97 then we have:
70.97*2*pi*F=1900000

and solve for F which is the highest frequency allowed before distortion appears, ideally. I also corrected a typo in my previous post the units for slew rate are of course not volts per volt they are volts per second.

Units check:
Vpk*K*F=volts/sec
volts*K*frequency=volts/sec
volts*K/sec=volts/sec
volts/sec=volts/sec
(OK)