Help calculating output voltage in operational amplifier using two reverse biased zener diodes

[DominikAntonioSajko]

Hi! I'm a 2nd year student of EE and need help in solving this assingment.
Operational amplifier uses two reverse biased Zener diodes, calculate output voltage (in the picture 'Uiz')

 

[rjenkinsgb]

The V+ input has 5.1V since no current flows thru R5. Its drop is
0 (zero).

5.0V in the unanswered problem, though that does not affect the principles.

 

[danadak]

I concur, my error, its a 5.0 V Zener in problem.

I will edit my other post.


Regards, Dana.

 

[Les Jones]

Your equation 12 = 24.2 x I is wrong. (Not just that 24.2 should be 24.2K).
What do you think the voltage on the inverting input of the op amp is ?)
The correct answer to that question should make you see what is wrong with the equation.

Les.

 

[DominikAntonioSajko]

Absolutely Les, I just want to learn and understand.
Huge thanks to everybody, including danadak Les Jones rjenkinsgb and crutschow .
I'm uploading solved question below, as well as Proteus scheme for anyone interested in learning the amazing world of electrical engineering.

 

[rjenkinsgb]

Just for information, another way of solving it - the way I did it in my head - is based on the fact that the gain of a simple inverting opamp circuit such as this is the ratio of input to feedback resistors.

You don't have to go through all the current calculations, if it's not classwork!

2k2 + 22k input to 22k feedback; 11:10

The input voltage is +7V, relative to the fixed positive input at +5V.

The output voltage will be +5V - (10/11 of 7V)

The approximation I used for for 10/11 is 7 - 0.7 + 0.07 = 6.37V

5 - 6.37 = -1.37

Calculated exact answer for 10/11 of 7V = 6.364 to three digits, so pretty close.

 

[Ramussons]

This is how it was given in exam:

As per this schematic, the output will be 0 if it is a Unipolar power supply, -Vcc if it is a Bipolar power supply.

 

[Ramussons]

And this is how it was given in auditory exercises with a solution:

Now something can be worked out since Z1 is reverse connected compared to the schematic given in the exam.

 

[Les Jones]

Ramussons makes a very good point. The op amp power rails should be shown in the question. I had just assumed that it had positive and negative power supply rails of sufficient voltage for the op amp to behave normally.
When I first looked at the question I did not notice that the ucc rail was only 12 volts so I started doing the calculation with 20 volts at the top of the 20 volt zener diode.

Les.

 

[DominikAntonioSajko]

Just for information, another way of solving it - the way I did it in my head - is based on the fact that the gain of a simple inverting opamp circuit such as this is the ratio of input to feedback resistors.

You don't have to go through all the current calculations, if it's not classwork!

2k2 + 22k input to 22k feedback; 11:10

The input voltage is +7V, relative to the fixed positive input at +5V.

The output voltage will be +5V - (10/11 of 7V)

The approximation I used for for 10/11 is 7 - 0.7 + 0.07 = 6.37V

5 - 6.37 = -1.37

Calculated exact answer for 10/11 of 7V = 6.364 to three digits, so pretty close.

rjenkinsgb Could you briefly explain the way you would have it written in formulas to calculate output voltage?

 

[danadak]

Classic superposition as shortcut -

NI = G = 1 + 2200/24200 so NIout = G x 5V = 1.909 x 5 = 9.545

INV = G = - 22000/24200 so INVout = -G x 12V = -.909 x 12 = - 10.908

NIout – INVout = -1.363


Regards, Dana.

 

[DominikAntonioSajko]

Thank you sir.

 

[danadak]

G's are closed loop gain.


Regards, Dana.

 

[Tony Stewart]

The simplest method can be done in your head.

What is the inverting gain ?Av(-)= -1 = -R4/R3
So what is the output from that inverting source? -0.7V

What is the Non-inverting gain? Av+= 1 + |Av-| = 2
What is the output from non-inverting source? 5.1 * 2 = 10.2V

That is the net total? -0.7 + 10.2V

ANy questions? R5 is irrelevant except for matching voltage drop if any significant input bias current. Otherwise neglect.