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JRE
JRE
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if u rIf you are using the same circuit and same input range, the maximum output voltage uyou will get is ((2*vin) + 5)/3\$\frac {2 \times V_{in} +5}{3}\$ (use superposition theorem). So, for vin=2) For \$V_{in}\$ =2.5v5 V (peak), the vo=\$V_o\$ = 3.34v34V (peak) and ov0V (min)minimum.)

if u r using the same circuit and same input range, the maximum output voltage u will get is ((2*vin) + 5)/3 (use superposition theorem). So, for vin=2.5v (peak), the vo= 3.34v (peak) and ov (min).

If you are using the same circuit and same input range, the maximum output voltage you will get is \$\frac {2 \times V_{in} +5}{3}\$ (use superposition theorem.) For \$V_{in}\$ =2.5 V (peak), the \$V_o\$ = 3.34V (peak) and 0V (minimum.)

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pavan g
pavan g
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if u r using the same circuit and same input range, the maximum output voltage u will get is ((2*vin) + 5)/3 (use superposition theorem). So, for vin=2.5v (peak), the vo= 3.34v (peak) and ov (min).