Electronic – Can the diode be ignored in this Op amp circuit

circuit analysisdiodesoperational-amplifier

Consider the circuit below:

enter image description here

The diode is ideal and \$R=1000\Omega\$.

What is \$v_o\$ when \$v_{in}=3V\$?

What is \$v_o\$ when \$v_{in}=-3V\$?

So my idea was just to make node equations to solve these to find \$v_o\$, but when I do that I completely ignore the diode. I do like this.

For this op amp we have: \$v_-=v_+=0V\$

When \$v_{in}=3V\$:

\$\frac{v_–3V}{R}+\frac{v_–v_o}{R}=0\$

Inserting \$v_-=0V\$ and solving this equation we find that: \$v_o=-3V\$

When \$v_{in}=-3V\$:

\$\frac{v_-+3V}{R}+\frac{v_–v_o}{R}=0\$

Inserting \$v_-=0V\$ and solving this equation we find that: \$v_o=3V\$

As you can see, I'm not really thinking about the diode at all. Am I solving the problem correctly?

Best Answer

schematic

simulate this circuit – Schematic created using CircuitLab

  • When Vi > 0 the diode conducts and shorts out R2. Stability will be reached when Vo = -0.7 V (for a real Si diode) and at 0 V for the ideal diode in the question.
  • When Vi < 0 the diode is reverse biased and is effectively out of the circuit. \$ V_O = - V_I \$.
  • The circuit when created with a real diode is an imperfect (due to the -0.7 V output) inverting half-wave rectifier.

From the comments:

Doesn't an op-amp force the voltage on both inputs to be the same?

Yes, in negative feedback configuration.

  • In Figure 1b we have a current flowing through R3 tending to raise the voltage on the inverting input. This will cause the output to swing negative.
  • When the output swings to just below zero (or just below -0.7 V with a real diode) the input will have been pulled down to 0 V and will match the voltage at the non-inverting input. The output will stabilise at that point.
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