# Amplifier consisting three resistors and an OP-AMP

#### Rumination

Joined Mar 25, 2016
74 The above figure shows a simple amplifier consisting of three resistors and an op-amp type uA741. The amplifier is designed so as to strive for a gain of Af [V / V], using the classic design rule for a non-inverting amplifier stage Af = R3 / R2 +1.

1) Calculate VoutDC for the circuit with open input (assume that Ao = ∞).

2) With an ideal op-amp (Ao = ∞) the amplifier will precisely have a gain of Af [V / V]. Calculate the error of the gain when using a real op-amp with a Ao [V / V].

3) The setting is connected to a sinusoidal inputsignal at Vin 100 mVpeak. Calculate the highest frequency that can be reproduced without distortion and without a decrease in gain.

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The given data are:

R1 = 87 KΩ.
R2 = 1 KΩ.
R3 = 70 KΩ.
Ao = 179000 [V/V].
Gain Bandvidth Product = 1,79 MHz.
Slew Rate = 1,9 V/uS.
Vos = 1,8 mV.
Input Bias Current = 160 nA.

*The deadline is today before midnight*

Last edited:

#### absf

Joined Dec 29, 2010
1,963
The above figure shows a simple amplifier consisting of three resistors and an op-amp type uA741. The amplifier is designed so as to strive for a gain of Af [V / V], using the classic design rule for a non-inverting amplifier stage Af = R3 / R2 +1.
If this is a non-inverting amp, you have your 2 inputs drawn reversed.

See the tutorial here.

Allen

• Rumination

#### Rumination

Joined Mar 25, 2016
74
1) Calculate VoutDC for the circuit with open input (assume that Ao = ∞).

Can I solve it using superposition?

V+ = Vos
V- = Vx
Ao = ∞
----------------
V01 = ((R3 + R2) /R2) * Vos = ((70KΩ + 1KΩ) /1KΩ) * 1,8mV = 127,8 mV

V+ = V-
V+ = - IB-x * R1

Or am I completely wrong?

• ErnieM

#### dannyf

Joined Sep 13, 2015
2,197
The above figure shows a simple amplifier consisting of three resistors and
It is not an amplifier.

#### WBahn

Joined Mar 31, 2012
26,398
Due before midnight?

Main takeaway from the assignment -- start working on your assignments earlier!

I don't understand your notation. What do you mean, for instance, by

Ao = Ao-X

What is X?

How can Ao be equal to Ao-X unless X is identically zero?

Clearly you probably mean something different than this interpretation, but you need to explain what it is.

The offset voltages and bias currents are not fixed actual values, they are specifications that give a typical or maximum value that a parameter MIGHT be. It is usually less than that. In some cases it could also be either positive or negative and the spec only gives the magnitude. So asking what the output will be if the input is floating is a bit nonsensical as you don't know what the actual offsets and biases are for that particular amp. The best you can do is come up with bounds, which is often very useful.

#### Rumination

Joined Mar 25, 2016
74

#### Rumination

Joined Mar 25, 2016
74
Due before midnight?
Main takeaway from the assignment -- start working on your assignments earlier!
I only have 27 hours to complete my assignment, so I can't post or work on it earlier.

I don't understand your notation. What do you mean, for instance, by
Ao = Ao-X
What is X?
How can Ao be equal to Ao-X unless X is identically zero?
I have edited the text in #1, so it's easier to understand.

#### ErnieM

Joined Apr 24, 2011
8,196
According to my teacher, it is.
Remind your teacher people use negative feedback for good reason.

#1 looks good?

For #2 remember that Vo = Ao * ( V+ - V-)
V+ is your input. V- is the output thru the divider. So you have all you need.

Do note your input terminals are drawn backwards.

• Rumination

#### Rumination

Joined Mar 25, 2016
74
For #2 remember that Vo = Ao * ( V+ - V-)
V+ is your input. V- is the output thru the divider. So you have all you need.
V+ = -160nA * R1 = -160 nA * 87 KΩ = 13920
Vo = Ao * (V+ - V-) = 179000 * (13920 - V-)

#### ErnieM

Joined Apr 24, 2011
8,196
V- is related to Vo by the divider network of R2 and R3.

#### Rumination

Joined Mar 25, 2016
74
V- is related to Vo by the divider network of R2 and R3.
So V- is 127,8 mV?

#### Jony130

Joined Feb 17, 2009
5,251

#### MrAl

Joined Jun 17, 2014
8,554
Hi,

According to my teacher, it is.
Everyone is complaining because the circuit in post 1 is drawn as a COMPARATOR circuit, not an amplifier circuit. That is because that circuit will behave like a comparator with hysteresis.

If it was a non inverting amplifier, the input would go to the non inverting input and the feedback would go to the inverting input. That's the only way it would work out.

If it was a non inverting amplifier, then yes, the equation Vout/Vin=1+R3/R2 is right.
If the resistor ratio is 100 then the gain would be 101 for example.
However, since i see they give the Ao as well, that approximation may not be acceptable unless they allow infinite gain. In question #2 they want you to use the real life formula which takes into account the real internal gain of Ao.
To calculate that, assume that the amplifier is a voltage controlled voltage source with a gain of that given for Ao. You can then calculate any difference between that and the output that would occur with Ao being infinite.

The max frequency without distortion is based on the output amplitude and slew rate. Calculate the output peak amplitude, then try to figure out how the slew rate limits the max frequency at that amplitude. The key point is the rate of change of the output sine wave at the zero crossing.

If you hurry with your reply i'll try to hurry with any more of mine too if needed.

• Rumination

#### ErnieM

Joined Apr 24, 2011
8,196
So V- is 127,8 mV?
No. Write it in terms of Vo

V- = Vo * (something here to show the voltage divider or feedback network)

#### Rumination

Joined Mar 25, 2016
74
No. Write it in terms of Vo

V- = Vo * (something here to show the voltage divider or feedback network)
V- = Vo * (R2/R2 + R3)

#### Rumination

Joined Mar 25, 2016
74
Hi,
If it was a non inverting amplifier, then yes, the equation Vout/Vin=1+R3/R2 is right.
If the resistor ratio is 100 then the gain would be 101 for example.
However, since i see they give the Ao as well, that approximation may not be acceptable unless they allow infinite gain. In question #2 they want you to use the real life formula which takes into account the real internal gain of Ao.
To calculate that, assume that the amplifier is a voltage controlled voltage source with a gain of that given for Ao. You can then calculate any difference between that and the output that would occur with Ao being infinite.
β = R2 / (R2 + R3) = 1kΩ / (1KΩ + 70kΩ) = 0,014

Af = Ao / (1+β*Ao) = 179000 / (1+0,014 * 179000) = 179000 / 2507 = 71,40 [v/v]

#### MrAl

Joined Jun 17, 2014
8,554
Hi,

That looks good. When i use a higher precision for Beta i get closer to 70.97 however, and Beta then is 1/71 exactly (as understood by inspecting the equation for Beta with the two resistor values).

Did you try the bandwidth limiting question yet?

#### Rumination

Joined Mar 25, 2016
74
Hi,

That looks good. When i use a higher precision for Beta i get closer to 70.97 however, and Beta then is 1/71 exactly (as understood by inspecting the equation for Beta with the two resistor values).

Did you try the bandwidth limiting question yet?
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.

#### The Electrician

Joined Oct 9, 2007
2,866
For part 1 you can't ignore the input bias current. Part 1 says the input is open, which means the input bias current flows through R1 and through R2&R3. This current in the resistors leads to a voltage drop across the resistors, which affects Voutdc just as Vos affects Voutdc.

Also, in the real world Vos can have either polarity and on a data sheet it would say Vos = ±1,8 mV. Do you know what polarity to assume, if it's only in one direction?

#### Rumination

Joined Mar 25, 2016
74
For part 1 you can't ignore the input bias current. Part 1 says the input is open, which means the input bias current flows through R1 and through R2&R3. This current in the resistors leads to a voltage drop across the resistors, which affects Voutdc just as Vos affects Voutdc.
Can it be solved like this:

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. Last edited: