# Op-amps circuit with diodes I am a beginner at electronics and I am trying to solve my first problems with diodes and opamps. I have to find and draw the output voltage uο related to the input voltage ui knowing that the voltage drop on the diodes is 0.7V when they conduct. It is also known that opamps are ideal so I guess that no currents flow in or out of the inputs and the difference between the inverting and non-inverting terminals can be considered to be zero.

However, I am bit confused with the diodes. How do I start solving a problem like this? Do I have to assume first that ui is positive and then that ui is negative? Any help is appreciated! Thanks in advance.

• Yes, make some assumptions, like ui is +ve, -ve and zero, Make some additional assumptions like any particular diode is conducting or not. When you chase through what all those assumed conditions mean, some sets of assumptions will be inconsistent, some consistent. Reject the inconsistent ones. Apr 14 '18 at 14:20

You can analyse it in two conditions of input ui.

1. When ui > 0

The op-amp output is positive and hence D1 will be forward biased and D2 will be reverse biased. Assuming ideal diodes and op-amps, we can draw the circuit like this. Op-amps are ideal and have negative feed back. So $V_B = V_A = ui$

At the node B, apply KCL/nodal analysis for three branches.

Input currents to op-amps are zero. So:

$$\frac{(V_B - uo)}{R} + 0 + 0 = 0$$ $$\implies V_B = uo$$ $$\therefore uo = ui$$

1. When ui <0

The op-amp output is negative and hence D1 will be reverse biased and D2 will be forward biased. Assuming ideal diodes and op-amps, we can draw the circuit like this. A little more neatly. It is simply a buffer driving an inverting amplifier. The input to the inverting amplifier is ui itself. I can therefore write the traditional input output relation of inverting amplifier like:

$$uo = -ui(\frac{R}{R})$$ $$\therefore uo = -ui$$

Now you can plot a sine waveform for ui as $V_{in}$ and u0 as $V_{out}$ to see that it would just act as a full wave rectifier. --EDIT--

Even if diode drops are considered, the same output is obtained at the end. This is an example of Full Wave Precision Rectifier. It acts as a rectifier with ideal diodes.

• Thanks a lot! You were really helpful! Just one question. In the exercise, its is mentioned, that only op-amps are ideal and the voltage drop on diodes is 0,7V when they conduct. Doesn't that change our analysis? I mean we have assumed that diodes are ideal and the voltage drop on them is 0V.
– MJ13
Apr 15 '18 at 8:21
• Concepts are same. They will just appear as offsets as + 0.7 or - 0.7 at places in equations. You can try it out yourself. If you understood this properly. Apr 15 '18 at 8:23
• yeah i see but will i have to make again the same assumptions?I mean if i assume that ui>0 then the output of the first op-amp is negative but how do i know which of D1 and D2 are forward or reversed biased?
– MJ13
Apr 15 '18 at 8:35
• Biasing of diodes is same still. Only 0.7 drop comes. Actually you will find out that diode drops are not relevant at all in the end.... Apr 15 '18 at 8:37
• Thanks a lot man for your time!It's a bit difficult to understand everything at first sight cause i am a beginner. I will try to solve the problem the way we discussed. Thanks again!
– MJ13
Apr 15 '18 at 9:18

A transfer function of a circuit is the output-to-input ratio of two voltages which must have the same shape (sinusoidal) - independent on the signal amplitudes! Otherwise, such a ratio cannot be found. What does this mean?

This means - for computing the transfer function - the circuit must be considered as LINEAR. With other words - all signals must be small enough to fulfill this restriction. In the given circuit, this can be accomplished for very small signals only which do NOT open the diodes considerably.

Question: Perhaps it is the task to find the output voltage as a timely function of the input signal?

• Οk i will tell you exactly how the exercise is given so that you maybe get the answer: "In the following circuit, the diodes have a voltage drop of 0.7Volts when they conduct and the operational amplifiers are considered ideal. Draw the output voltage as a function of input voltage. What is the use of the circuit?"
– MJ13
Apr 14 '18 at 15:21
• OK - this sounds a bit different. No mentioning of the term "transfer function".
– LvW
Apr 14 '18 at 15:35
• Yeah you are right. I just thought that transfer function was needed. So how can i approach this problem? Do i have to assume anything about the input voltage?
– MJ13
Apr 14 '18 at 15:40