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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. enter image description here

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 \$.

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

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  • \$\begingroup\$ 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. \$\endgroup\$ – nkg2743 Jun 11 '16 at 4:29
  • \$\begingroup\$ 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) \$\endgroup\$ – KarlKarlsom Jun 11 '16 at 5:21

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