Find Vp, Vn and Vo and please make it easier.
i am asking on answers because my teacher sucks at teaching
http://img844.imageshack.us/img844/4999/…
i am asking on answers because my teacher sucks at teaching
http://img844.imageshack.us/img844/4999/…
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This was already asked and answered.
What an opamp does is force the voltage between the inputs to zero.
(all currents in mA, resistances in k)
Assume a voltage Vo at the output. Current in 1k is (Vo–Vn)/1, current in 3k os Vn/3. Currents in that node are (Vo–Vn) – (Vn/3) = 1
current in 4k is (Vo–Vp)/4, current in 2k is Vp/2. Currents in that node are
(Vo–Vp)/4 – Vp/2 = –1
Vp = Vn, simplifying those two equations
Vo – Vn – Vn/3 = 1
Vo – (4/3)Vn = 1
3Vo – 4Vn = 3
(Vo–Vn)/4 – Vn/2 = –1
(1/4)Vo – (1/4)Vn – (1/2)Vn = –1
(1/4)Vo – (3/4)Vn = –1
Vo – 3Vn = –4
two equations in 2 unknowns
3Vo – 4Vn = 3
Vo – 3Vn = –4
3Vo – 4Vn = 3
–3Vo + 9Vn = 12
5Vn = 15
Vn = 3v = Vp
Vo = 3Vn – 4
Vo = 5v
checking, Vo = 5v, inputs are 3v
current in ik = 2mA
current in 3k = 1 mA
adds up to 1 ma, which is the current source
current in 4k = 2/4 = 0.5 mA
Current in 2k = 3/2 = 1.5 ma
difference is 1ma, the current source
What an opamp does is force the voltage between the inputs to zero.
(all currents in mA, resistances in k)
Assume a voltage Vo at the output. Current in 1k is (Vo–Vn)/1, current in 3k os Vn/3. Currents in that node are (Vo–Vn) – (Vn/3) = 1
current in 4k is (Vo–Vp)/4, current in 2k is Vp/2. Currents in that node are
(Vo–Vp)/4 – Vp/2 = –1
Vp = Vn, simplifying those two equations
Vo – Vn – Vn/3 = 1
Vo – (4/3)Vn = 1
3Vo – 4Vn = 3
(Vo–Vn)/4 – Vn/2 = –1
(1/4)Vo – (1/4)Vn – (1/2)Vn = –1
(1/4)Vo – (3/4)Vn = –1
Vo – 3Vn = –4
two equations in 2 unknowns
3Vo – 4Vn = 3
Vo – 3Vn = –4
3Vo – 4Vn = 3
–3Vo + 9Vn = 12
5Vn = 15
Vn = 3v = Vp
Vo = 3Vn – 4
Vo = 5v
checking, Vo = 5v, inputs are 3v
current in ik = 2mA
current in 3k = 1 mA
adds up to 1 ma, which is the current source
current in 4k = 2/4 = 0.5 mA
Current in 2k = 3/2 = 1.5 ma
difference is 1ma, the current source