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Half‐wave rectifier with capacitive filter
Elettronica IIProf. Paolo Colantonio 2 | 9
Breadboard
Elettronica IIProf. Paolo Colantonio 3 | 9
Source
Elettronica IIProf. Paolo Colantonio 4 | 9
Oscilloscope
Elettronica IIProf. Paolo Colantonio 5 | 9
VM
t
vo
t1 t2 t3
o i ONv v V
t1 < t < t2When vi > VON, the diode is ON and the current flows toward R and C. C is charging with a low
t2 < t < t3The diode turns OFF and the capacitance discharge with law
t
RCo m ONv V V e
• Let us consider a half‐wave rectifier with a capacitive load, used to produce asteadier output
RLv(t)i v (t)o
D
C
Elettronica IIProf. Paolo Colantonio 6 | 9
t
Assuming an high time constant RC>>T (period), then the exponential behaviour can be approximated with the tangent
* *
* 1tRC
B P Ptv t V e VRC
In the time interval t2‐t3, we can write
t2 t3
vp
Assuming t2‐t3 T, we can write
V
1 Pp P
V TTV V VRC RC
VP
From which
P
V TRIPPLEV RC
Elettronica IIProf. Paolo Colantonio 7 | 9
B
DR
AiD
B BD C R
dv t v ti t i t i t C
dt R
t1 < t < t2 Diode ON
t2 < t < t3 Diode OFF
iCiR
vA
t
t3
The maximum current occurs at the circuit start up, when the capacitor C is discharged.
vB
t1 t2
iD
,00
sin 2BD MAX p p
tt
dv t dI C C V t f C Vdt dt
C
Elettronica IIProf. Paolo Colantonio 8 | 9
vA
t
vBiD
V
VP
PV TV
RC
TRIPPLE
RC
, 2D MAX pI f C V
Dimensioning of RC to control the Ripple
Dimensioning of C to reduce the maximum current in the diode
Elettronica IIProf. Paolo Colantonio 9 | 9
SetupMount the circuit and apply a voltage signal Vpp=10V and f = 1kHz
Work• Determine the values for R and C in order to assure in B a Ripple < 10% of
maxximum output voltage
D1N4001
R C
BA Datasheet
Diode 1N4001
Conducting voltage V=0.65V
Max current Imax=30A
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