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The Full-Wave Rectifier
Consider the following junction diode circuit:
Note that we are using a transformer in this circuit. The job of this transformer is to step-down the large voltage on our power line (120 V rms) to some smaller magnitude (typically 20-70 V rms).
Note the secondary winding has a center tap that is grounded. Thus, the secondary voltage is distributed symmetrically on either side of this center tap.
For example , ifv S = 10 V, the anode ofD 1 will be 10V above
ground potential, while the anode ofD 2 will be 10V below
ground potential (i.e., -10V):
v s(t)
vO (t)
R
Power Line
v s(t)
D 1
D 2
Conversely, ifvS=-10 V, the anode ofD 1 will be 10V below
ground potential (i.e., -10V), while the anode ofD 2 will be 10V
above ground potential:
v S=10 V
vO (t)
R
Power Line
v S=10V
D 1
D 2
v S=-10 V
vO (t)
R
Power Line
v S=-10V
D 1
D 2
Note that the secondary voltages at either end of this circuit
are the same , but have opposite polarity. As a result, ifvS=10,
then the anode of diodeD 1 will be 10 V above ground, and the
anode at diodeD 2 will be 10V below groundโjust like before!
Now, letโs attempt to determine the transfer function v O = f v ( (^) S) of this circuit.
First, we will replace the junction diodes with CVD models.
Then letโs ASSUMED 1 is forward biased andD 2 is reverse
biased, thus ENFORCE vD i 1 = 0 and iD i 2 = 0. Thus ANALYZE:
v S=
+
_
vO (t)
vS=10^ R
+
_
D 1 D 2
v S
+
_
vO (t)
_
v S^ R
+
_
iD i 1 +v Di 1 = 0 โ - 0.7 + iD i 2 = 0
2
i
โ vD +
i
Note that we need to determine 3 things: the ideal diode
current iD^ i 1 , the^ ideal diode voltage^ vD^ i 2 , and the^ output
voltage v O. However, instead of finding numerical values for
these 3 quantities, we must express them in terms of source
voltage v S!
From KCL: i = iD i 1 + iD i 2 = iDi 1 + 0 =iDi 1
From KVL: vS โ vDi 1 โ 0.7 โ R iDi = 0
Thus the ideal diode current is:
1 i S 0. D
i v
R
=^ โ
Likewise, from KVL: vS โ vDi 1 โ 0.7 + 0.7 + vDi 2 + vS = 0
Thus, the ideal diode voltage is:
2 2 S
i
vD = โ v
And finally, from KVL: vS โ vDi 1 โ 0.7=vO
Thus, the output voltage is:
vO = vS โ0.
Thus, we have found that the following statement is true about this circuit:
vO = vS โ 0.7 V when vS >0.7 V
Note that this statement does not constitute a function (what
about vS < 0.7?), so we must continue with our analysis!
Say we now ASSUME thatD 1 is reverse biased andD 2 is
forward biased, so we ENFORCE iD i 1 = 0 and vD i 2 = 0. Thus, we
ANALYZE this circuit:
Using the same proceedure as before, we find that
vO = โv S โ 0.7, and both our assumptions are true when
vS < โ 0.7 V. In other words:
vO = โv S โ 0.7 V when vS < โ0.7 V
Note we are still not done! We still do not have a complete
transfer function (what happens when โ0.7 V < vS < 0.7 V?).
v S
+
_
vO (t)
_
v S^ R
+
_
iD i 1 = 0 - 0.7 + iDi 2
1 i
+ vD โ
i
โ vD = +
i
Finally then, we ASSUME that both ideal diodes are reverse
biased, so we ENFORCE iD^ i 1 =^0 and^ iD^ i 2 =^0. Thus ANALYZE:
Following the same proceedures as before, we find that
vS = 0 , and both assumptions are true when โ0 7. < vS < 0 7..
In other words:
vS = 0 when โ 0 7. < vS <0 7.
Now we have a function! The transfer function of this circuit is:
0 7 V 0 7 V
0 0 7 0 7 V
0 7 V 0 7 V
S S
O S
S S
v. for v.
v V for. v.
v. for v.
โง^ โ^ >
โชโฉโ^ โ^ < โ
Plotting this function:
v S
+
_
vO (t)
_
v S^ R
+
_
iD i 1 = 0 - 0.7 + iD i 2 = 0
1 i
+ vD โ
2 i
โ vD +
i
Likewise, compare the output of this junction diode full-wave rectifier to the output of an ideal full-wave rectifier:
Again we see that the junction diode full-wave rectifier output is very close to ideal. In fact, if A>>0.7 V, the DC component of this junction diode full wave rectifier is approximately: (^2) 0 7 V O
V A.
Just 700 mV less than the ideal full-wave rectifier DC component!
t
A
v (^) S(t ) -A
v v (^) O (t )
