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what i did here is take current over the 4.7 kilo ohm and since there are no other voltage input other than the one then shouldn't the Vout be 0.2127 volts? thats not possible for a non-inverting amplifier right? do i just assume Vin = Vout in this case then?

enter image description here

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  • \$\begingroup\$ Where is that schematic from? I've got a few words for the author of that – If they are using the battery symbol where they actually mean a voltage source, they should at least realize that this symbol is polarized. \$\endgroup\$ – Marcus Müller Mar 3 at 18:09
  • \$\begingroup\$ No, that calculation is totally wrong. How did you end up with 0.2127 V? Can you explain? Because, then we can help you understand your misunderstanding \$\endgroup\$ – Marcus Müller Mar 3 at 18:10
  • \$\begingroup\$ i took the voltage over the 4.7 volt resistor so then i got 0.2127 mA and then i also considered the 1 kilo resistor and the current shouldn't have dissipated anywhere as so i thought it would be the same current over the 1k resistor getting 0.2127 volts, which is parallel to Vout? \$\endgroup\$ – HaidyE Mar 3 at 18:15
  • \$\begingroup\$ Why do you think that any current is flowing in the 4.7k resistor? What is the input impedance of an ideal opamp? \$\endgroup\$ – Dave Tweed Mar 3 at 18:17
  • \$\begingroup\$ i just assumed since we also have a Vin of 1 volt. Does that mean that Vin= Vout then? \$\endgroup\$ – HaidyE Mar 3 at 18:21
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That's just a unity gain amplifier. So gain=1, Vout=Vin. The 4.7k resistor doesn't matter, unless you get close to (or beyond) the supply voltages.

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