hi guyz!how r u doin?i've a little problem with an assignment i've got
regarding voltage regulator.
ASSIGNMENT:-
my assignment is to design a voltage regulator which gives an output
voltage of
15 volts & a current of 350mA(mili-ampere).the components used should
be a step-down transformer, a fullwave rectifier bridge,2
resistors(Rin,Rload ),a capacitor & a zener diode.
PROBLEM:-
the problem is that i am not able 2 adjust the zener diode
accordingly.when i use an ideal
one the current & voltage are not exact."can any one of you tell me
the specifications of the
zener diode and the transformer.'ll be very thankful.plz help.
---
I'll help you, but I'd appreciate it if you didn't use chat room
slang and if you bottom posted.
First, here's a circuit, which you can view in Courier, which shows
how your system should be wired up
15V
/
+---+---[R1]--+-------+
+------+ | | | |
MAINS>---P||S--|~ +|--+ |+ |+ |
R||E | | [BFC] [ZENER] [R2]
MAINS>---I||C--|~ -|--+ |C1 | |
+------+ | | | |
+---+---------+-------+
With this kind of problem you have to work backwards, so your load
resistor has to allow 350mA through it with a voltage drop of 15
volts across it. That means the value of the resistor, R2, has to
be:
E 12V
R = --- = ------- ~ 34.3 ohms
I 0.35A
and the power it must dissipate will be:
P = IE = 12V * 0.35A = 4.2 watts
Now, to select the Zener you have to determine what the load's
dynamic characteristics are going to be like. That is, what will
the change of current through the load look like as a function of
time?
In this case, there's no problem since the load is just a resistor
and its resistance shouldn't change much when it gets hot.
Now, the deal with Zeners is that when you reverse bias them so that
what's called the "test current", Izt, passes through them, the
voltage across them is guaranteed to be within a certain range,
usually 5 or 10% of the nominal voltage, so your job is to make sure
that that test current will be going through the Zener while the
load current is going through the load.
Now you have to choose a Zener. Since your load isn't going to vary
much, Izt is only going to vary because of changes in the mains
voltage and because of the ripple in C1. For a first cut, let's try
a 1N5352B, which is a 15V 5W Zener with an IZT of 75mA. The data
sheet is here:
http://www.onsemi.com/PowerSolutions/product.do?id=1N5352B
Since we have 350mA in the load and 75mA through the Zener, that
current is all having to go through R1, but now, before we can
figure out what value of R1 to choose, we have to consider the
"front end" of the supply. That is, the transformer, rectifier
bridge, and filter.
To start, let's say that we never want the voltage on the C1 end of
R1 to go below 16V. That'll give us a volt of headroom with the load
current and Izt through R1, so for a first cut,
Ein - Vz 16V - 15V
R1 = ---------- = ---------------- ~ 38 ohms
IL + Iz 0.35A + 0.075A
Now, looking at the capacitor, we know that since:
IdT
C = -----
dV
Where C is the capacitance in farads,
I is the steady-state load current in amperes,
dT is the reciprocal of the rectified mains frequency, and
dV is the desired ripple voltage,
if we don't want dV to go below 16V when the capacitor is
discharging into the load and the mains voltage cycle is below the
point where it's high enough to charge the cap and feed the load,
then we must choose the capacitance of the capacitor so that its
output voltage never falls below 16V during that mains "dead time."
Assuming we allow one volt of ripple, then the capacitor will charge
to a peak of 17V and discharge to 16V before the next mains cycle
comes along and starts charging it up again for the dead time after
the present peak. if we plug some numbers into the equation, we'll
wind up with:
IdT 0.425A * 8.33E-3s
C = ----- = ------------------ = 3.54E-3F = 3540µF. Not bad...
dV 1V
But now we've got to get that 17V from somewhere. The bridge. If
you want to use a full-wave bridge, then there will always be two
rectifiers in series in front of the reservoir cap, and assuming a
forward voltage drop of 0.7V per rectifier, that means that if you
want 17V out of the bridge you've got to put:
Vin = Vout + 2Vf = 17V + 1.4V = 18.4V into it.
But now we've got to get that 18.4V from somewhere. That'll be the
transformer, and we've got to make sure that we can get it when the
mains are 10% under their nominal 120V rating, so that's at 108V.
So, we need a transformer than can put out 18.4 volts, peak, with
108 volts, RMS, on the primary. Since transformers are rated for RMS
in and out, we'll have to convert that 18.4VPK into RMS, and we do
that by dividing it by the square root of 2:
VP 18.4
VRMS = --------- = ------- ~ 13VRMS
sqrt(2) 1.414
Transformers are also rated with 120V on the primary, so since that
13V is with 108V on the primary, what we need to do is up the
secondary voltage by the ratio of nominal mains to low mains:
Vmains nom 120V
Vsnom = Vslow ------------ = 13V ------ = 14.4 volts, RMS
Vmains low 108V
So we need a transformer with a 120V primary and a 14.4V secondary,
and it'll need to be able to supply about 1.8 times the current into
the load and the Zener because it also has to charge up the cap so,
since the load needs 350mA and the Zener needs 75mA, the transformer
secondary needs to be rated for:
Is = 1.8* (Il + Izt) = 0.765 ampere
What we need then, is a transformer with a 120V primary and a
secondary rated to supply 0.765 amps at 14.4 volts.
Unfortunately, a rare beast to find.
Fortunately, there are lots of standard transformers out there which
are pretty close, among them a Triad FP30-400:
http://www.triadmagnetics.com/subcategory/ptfp.html
which can be found at Digi-Key for $14.00. That's a little high and
you can do better if you shop around for different vendors and
different manufacturers.
In any case, the transformer is rated to put out 15VRMS at 800mA
with a 120V input, so now we can work out the value of R1.
With low mains, the transformer will put out 13.5VRMS which, after
the 1.4V lost in the bridge, will charge the cap up to about 17.7V
so,with the Zener end of R1 at 15V and the cap end at 16.7V (the low
voltage point of the ripple waveform) the resistor will need to drop
1.7V at 475mA, so its value needs to be:
E 16.7V -15V
R = --- = ------------ ~ 3.6 ohms
I 0.475A
and it'll be dissipating roughly:
P = IE = 0.475A * 1.7V ~ 0.8 watt.
So, finally, we have:
15V
R1 /
+---+--[3R6]--+-------+
+------+ | | | |
MAINS>---P||S--|~ +|--+ |+ |+ |
R||E | | [4700] [1N5352] [34R3]
MAINS>---I||C--|~ -|--+ |C1 | |R2
+------+ | | | |
+---+---------+-------+
Now we have to check that nothing's going to fry at high line.
With high mains, the voltage into the transformer will be 132V, so
the secondary will rise to 16.5VRMS, which is 23.3 peak, so after
the diode loss of 1.4V through the bridge the cap will charge up to
about 22 volts, so C1 should be 4700µF at 35V. It's 4700µF instead
of the 3540 we originally figured because of the 20% tolerance, and
4700µF is the closest standard value that'll assure us at least
3540µF if the cap is 20% low.
Now, with 22V on the cap end of R1 and 15V on the Zener end, the
current through the resistor will be:
E 22V - 15V
I = --- = ----------- ~ 1.94A
R 3.6R
of which 350mA will be going through the load and the remainder
through the Zener. That's about 1.6 amps through the Zener, which
means it'll be dissipating about 24 watts, so the scheme is flawed
if it needs to run from low mains to high mains.
However, if your assignment only deals with a single mains voltage,
use 120VRMS and the procedure outlined above to find the proper
value for R1.
