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Assuming that the op amp and the diode are ideal (for op amp: infinite input impedance and 0 output impedance, for diode: 0 voltage drop):

D1 is open <=> $$V_i \frac{R2}{R1}>E$$

  • case 1 (D1 closed):

$$V_i < E\frac{R1}{R2}$$

$$V_o = -V_i \frac{R2}{R1}$$ (It's a simple inverter)

  • case 2 (D1 open):

$$V_i > E \frac{R1}{R2}$$

$$Vo = -V_i \frac{R2||R3}{R1} -E \frac{R2||R3}{R3}$$

How do you get the last relation (case2, Vo=...) ?

where $$R1||R2 = \frac{R1 \cdot R2}{R1+R2}$$

  • 3
    \$\begingroup\$ Gotta love those "ideal" 741 opamps. \$\endgroup\$ Commented Jan 30, 2013 at 18:12
  • 1
    \$\begingroup\$ Those "ideal" diodes are pretty sweet, too. \$\endgroup\$
    – HikeOnPast
    Commented Jan 30, 2013 at 18:29
  • 2
    \$\begingroup\$ They are limited edition... \$\endgroup\$
    – Cristi
    Commented Jan 30, 2013 at 20:07

1 Answer 1


Is this homework problem from your superposition chapter? Short E, solve for Vo in terms of Vin. Short Vin, solve for Vo in terms of E. The final answer you have there could be simplified..


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