# OpAmp circuit with diodes: conduction conditions

Consider the circuit below, where $\ v_2$ is an arbitrary input voltage and not necessarily equal to 3V as in the image. Both diodes and the OpAmp are ideal.

If only diode 1 is conducting or if both are, the output voltage, $\ v_0$, would be given by $$v_0=v_2 +2V$$ If only diode 2 is conducting the output would be $$v_0=2v_2$$ those two results where obtained by application of the standard rules (negative feedback, infinite input impedance...).

What I can't figure out is for which values of $\ v_2$ each situation is verified. According to the solutions the first situation is verified for $\ v_2\leq 2$ and the second for $\ v_2\geq 2$.

• which source are you talking about? is it V1=2V or V2=3V.. you have mixed them up, creating confusion... – nkg2743 Jun 11 '16 at 5:51
• @nkg2743 , sorry. I've already corrected it. – delta_omega Jun 11 '16 at 10:20

Lets analyse this circuit by Kirschoffs law:

as long as your (perfect) OPAMP is not saturated it tries to regulate the output like like that the diferential input is 0V.

M1: $U_{R1} + u_{diff} - V_2 = 0 => U_{R1 }= V_2$

M2: $-U_{R1} + U_{D1} - V_1 + U1_{OUT} = 0 => U1_{OUT} = V_2 - U_{D1} + V_1$

M3: $-U_{R1} - U_{D2} - U_{R2} + U1_{OUT} = 0 => U1_{OUT} = V_2 + U_{D2} + U_{R2}$

Lets say both diodes are ideal and conduct @0.5V And lets also say the part D1 is not assembled then you get:

$U1_{OUT} = 2*V_2+0.5V$ (D2 is always conducting a $U1_{OUT}$ always bigger than $V_2$).

With this knowledge we add now D1 and get:

$2*V_2+0.5V = V_2 - U_{D1} + V_1$

$V_2 + 0.5V = -U_{D1} + V_1$

$V_2 = -2*0.5V + V_1$ => D1 is conduction between $V_2=0$ and $V_2 = 1V$

So between $V_2 = -VDD..1V$ voltage $U1_{OUT}$ is determined by:

$U1_{OUT} = V_2 + V_1 - U_{D1} = V2 + 1.5V$

Above $V_2 = 1V..VSS$ its determined by:

$U1_{OUT}=2*V_2+V_{D2}$

• I have one doubt D2 is always conducting a U1OUTU1OUT always bigger than V2 is this valid for both the case inverting and non-inverting amplification depending upon the value of V2 as it is arbitrary.. for some value of V2, D2 will conduct & for value of V2, D2 will not as mentioned by OP. – nkg2743 Jun 11 '16 at 4:29
• Nope, even when V2 is getting negative, the D1 pass will keep U1OUT 1.5V above V2. D2 will still conduct, but with lower current (VR2 = 1.5V-0.5V = 1V => IR2 = 100uA) – KarlKarlsom Jun 11 '16 at 5:21