Download Some Electric Circuits

Some Standard Electric Circuits
Ang Man Shun
2012-3-3
Some Resistor Network
Some OpAmp Circuits
The Virtual Short Concept for ± pins in an OpAmp
The pin − and + are virtual shorted : V− = V+
( This is because of finite output and infinite gain of ideal OpAmp vout = A(v+ − v− ) )
Thus, when one of pin is grounded, then both pins are grounded.
1
1
The Resistor Network
1.1
Averager
The current of R1 is
V1
R1
Thus at the output point, all the current sum together : Iout =
(
Since the 3 resistor is in parallel, so the equivalent resistance is
(
Thus the output voltage is Iout Req =
V1
V2
V3
+
+
R1 R2 R3
)(
V2
V3
V1
+
+
R1 R2 R3
1
1
1
+
+
R1 R2 R3
1
1
1
+
+
R1 R2 R3
)−1
)−1
=
R1 R2 R3
R1 R2 + R2 R3 + R3 R1
V1
V2
V3
+
+
R
R2 R3
= 1
1
1
1
+
+
R1 R2 R3
When resistor are equal
Req =
R
3
Iout =
V1 + V2 + V3
R
Vout =
V1 + V2 + V3
3
Generalized, for n-resistor averager
The equivalent resistance is
)−1
(
R1 R2 ...Rn
1
1
1
=
+
+ ... +
Req (n) =
R1 R2
Rn
R1 R2 ...Rn−1 + R1 R2 ...Rn−2 Rn + ... + R2 R3 ...Rn
)−1
( n
∏n
∑ 1
Ri
Sum of R taken n at a time
= ∑n ∏i=1
Req (n) =
=
n
Ri
Sum of R taken n − 1at a time
R
i=1
i=1
j=1 i
(Analogy to Algebra of Polynomial )
j ̸= i
The output current is Iout =
V1
Vn
+ ... +
R1
Rn
And the ouput voltage is Vout
1.2
V1
+ ... +
R1
=
1
+ ... +
R1
Vn
Rn
1
Rn
Birdge
The path R1 R4 is parallel to path R2 R3 , thus voltage across R1 + R4 is equal to R2 + R3
R4
R3
By potential divider, Va = Vin
, Vb = Vin
( R1 + R4
)R2 + R3
R3
R4
−
Thus, Vab = Va − Vb = Vin
R1 + R4 R2 + R3
When Vab = 0 , then Va = Vb , and thus
R3
R4
=
R1 + R4
R2 + R3
⇒
R3 1/R3
R4 1/R4
=
R1 + R4 1/R4
R2 + R3 1/R3
=⇒
⇒
R1
R2
=
R4
R3
2
1
R1
+1
R4
=
1
R2
+1
R3
⇒
R2
R1
+1 =
+1
R3
R4
2
OpAmp Circuits
2.1
Inverting Amplifier
Vin
Rin
Since the ±pin is virtual grounded (Va = 0 ), so the all current go into Rf
Rf
Thus the voltage of Rf is VRf = Iin Rf = Vin
Rin
Rf
By KVL : Va = VRf + Vb ⇐⇒ 0 = Vin
+ Vout
Rin
Vout
Rf
Thus, the gain is
=−
Vin
Rin
When Rf = Rin , the OpAmp become a phase invertor , Vout = −Vin
Input current is Iin =
2.2
Non-inverting Amplifier
Input voltage Vin
By potential divider at point C : VC = Vout
By virtual short of ±pin, VC = Vin
R1 + R2
R2
Vout
=
=1+
So the gain is
Vin
R1
R1
2.3
R1
R1 + R2
Voltage Follower
The input voltage Vin equal to V−
The output voltage Vout equal to V+
By virtual short in ±pin, V− = V+
Thus Vout = Vin
Vout
=1
So the gain is
Vin
2.4
Weighted Summer
Notice that the resistor network is an averager.
Vin,k
The input current of resistor k : Ik =
Rk
∑
By KCL, the total current in node a will be Ia = all k Ik
Since + pin is grounded, so by virtual short in ±pin, − pin is also grounded (Va = 0)
Thus the voltage of Rf is
(
)
∑
V2
Vn
V1
Rf
Rf
Rf
= If Rf = Ia Rf = ( Ik ) Rf =
+
+ ... +
Rf = V1
+ V2
+ ... + Vn
R1 R2
Rn
R1
R2
Rn
all current go into Rf
VRf
( Rf
Rf
Rf )
V1
+ Vout
+ V2
+ ... + Vn
R1
R2
Rn
(
)
Rf
Rf
Rf
Thus Vout = − V1
+ V2
+ ... + Vn
R1
R2
Rn
Vout
Rf
Rf
Rf
If all the input voltages are equal, the gain is
=
+
+ ... +
Vin
R1
R2
Rn
By KVL, Va = VRf + Vb
⇐⇒ 0 =
3
2.5
Differential Amplifier
Virtual short in ±pins, so Va = Vc , but Va , Vc ̸= 0
If V1 = 0 , this amplifier become non-inverting amplifier , At point C , Vc = V2
Rg
, this Vc is
R2 + Rg
the Vin in non-inverting amplifer
If V2 = 0 , this amplifier become inverting amplifier, and
Rf
Vout1 = −V1
R1
Thus, Vout2 = Vc
Rg R1 + Rf
R1 + Rf
= V2
R1
R2 + Rg R1
The total output is Vout = Vout1 + Vout2
Rg R1 + Rf
Rf
+ V2
= −V1
R1
R2 + Rg R1
For R1 = R2 , Rg = Rf : Vout = (V2 − V1 )
between 2 inputs
2.6
Rf
, thus, the output is proportional to the difference
R1
Integrator
Vin
R
Since + pin is GND and virtual short of ±pin,
Va = 0, so all current goes into the capacitor
The input current is Iin =
Capacitor charge-capacitance-voltage relation is : Qcap = Vcap C
dVcap
dQcap
=C
Thus Icap =
dt
dt
dVcap
1 ´t
Vin
=C
⇐⇒ Vcap =
Iin = Icap
⇐⇒
Vin dt , assume Vcap (t = 0) = 0
R
dt
RC 0
By KVL : Va = VCap + Vb
⇐⇒ 0 = Vcap + Vout
⇐⇒ Vout = −Vcap
−1 ´ t
Thus Vout =
Vin dt
RC 0
2.7
Differentiator
The capacitor equation : Qcap = CVcap
dVcap
dVin
dQcap
=C
=C
dt
dt
dt
By +pin GND & ±pin virtual short, IR = Icap
dVin
=⇒ VR = CR
dt
dVin
+ Vout
By KCL : Va = VR + Vb ⇐⇒ 0 = RC
dt
dVin
Vout = RC
dt
Remark. If the input is A sin ωt , then output is ωA cos ωt , thus the ouput will have a larger amplitude
for high frequency component. Thus differentiator will also amplify the high frequency noise.
=⇒ Icap =
−EN D−
4