How would I install the zener diode? I thought they were used in
series with the voltage source much like a rectifier diode, but I just
read that they're used in parallel to the source. Excuse the dumb
question, I've only ever had experience with rectifier diodes like the
1N4007.
---
For starters: (View in courier)
The way to get 12V out of the transformer is to ignore the center tap
and use the the two ends of the secondary as your 12V source.
There are two ways to do that, one called half wave rectification and
the other full wave rectification.
Half wave rectification works like this:
MAINS>--+ +--[DIODE>]--+---->17.1VDC
P||S |+
R||E--CT [BFC]
I||C |
MAINS>--+ +------------+---->GND
When the end of the secondary connected to the anode of the diode goes
positive, charge is allowed to flow through the diode and accumulate
in the BFC where, when the capacitor is full will be the peak value of
the AC waveform minus the voltage dropped across the diode.
That is:
VDC = (VRMS * sqrt(2)) - Vf
= (12.VRMS * 1.414) - 0.7V
~ 17.1V
Full wave rectification works like this:
MAINS>-+ +----+-[CR1>]---+---+---->16.4VDC
| | | | |+
P||S +-[<CR2]-+ | [BFC]
R||E-CT | | |
I||C +-[<CR3]-+-|---+---->GND
| | | |
MAINS>-+ +----+-[CR4>]---+
This time, when the top of the secondary goes positive, charge will
flow through CR1 into the BFC and then through CR3 back into the
negative end of the secondary.
When the polarity reverses and the bottom of the secondary goes
positive, charge will flow through CR4 into the BFC and then back into
the negative end of the secondary through CR2.
The full wave rectifier has the advantage of using a capacitor with
half the capacitance required for the same ripple out of a half-wave
rectifier, plus some others, so lets assume that's the supply you'll
build.
OK, now for the Zener.
A Zener diode belongs to a class of devices called "shunt regulators"
because they shunt current that a load doesn't need to ground while
maintaining the voltage across the load constant even as the supply
voltage changes and the load current varies.
In your case since you have a 16.4V supply and you want to use that
12V Zener, we look at the data sheet for a 1N4742 and we find that IZt
is 21mA and IZmax is 76mA.
IZt is the Zener test current and is the current through the Zener
which will cause the voltage dropped it fall within the device's
specifications, and IZmax is the maximum current which can be pushed
through the diode without damaging it if the ambient temperature is
25C or lower.
Now, if we look at the regulator circuit we'll have:
16.4V>--[R]--+---------+
|+ |
[ZENER] [LOAD]
| |
GND>---------+---------+
Let's assume that your load draws 50mA and that we'll allow IZt to
flow through the Zener.
Then we'll have, for the value of R:
Vsupply - Vz 16.4V - 12V
R = -------------- = ---------------- = 61.9 ohms
IZt + Il 0.021A + 0.05A
which is pretty close to 62 ohms, a standard 5% value.
The power the resistor will dissipate will be:
P = (Vsupply - Vz) * (IZt + Il)
= (16.4V - 12V) *(0.021A + 0.05A)
= 0.312 watts
So a half-watt resistor would be fine.
Your Zener circuit now looks like this:
6.3VAC 16.4VDC 12VDC
/ / /
MAINS>-+ +----+-[CR1>]---+---+-[62R]-+--------+
| | | | |+ |K |
P||S +-[<CR2]-+ | [BFC] [1N4742] [LOAD]
R||E-CT | | | | |
I||C +-[<CR3]-+-|---+-------+--------+
| | | |
MAINS>-+ +----+-[CR4>]---+
\
6.3VAC
But...
There's a very much better way to go, and that's to use a series
regulator like a 7812 in place of a Zener, like this:
6.3VAC 16.4VDC 12VDC
/ / /
MAINS>-+ +----+-[CR1>]---+---+-[7812]--+
| | | | |+ | |
P||S +-[<CR2]-+ | [BFC] | [LOAD]
R||E-CT | | | | |
I||C +-[<CR3]-+-|---+----+----+
| | | |
MAINS>-+ +----+-[CR4>]---+
\
6.3VAC
Not only will this give you better regulation with line and load
changes, it'll allow you to use the full output capability of the
transformer (half an ampere) with a suitable heatsink on the 7812.
So, the last thing to figure out is how big the BFC has to be.
We start by looking at the droput voltage of the 7805, (that's the
voltage difference between the output and the input which will assure
the output stays in regulation) and it's 2.5V, so that means we can
never let the input fall below:
Vin = Vout + Vdo = 12V + 2.5V = 14.5 volts
Since the BFC is being charged with sinusoidal pulses and being
discharged by the load when the pulse voltage isn't high enough to
supply the load current, ripple will appear across the capacitor and
we have to make sure that ripple never falls below 14.5V or the 7812
will go out of regulation. to do that we can say:
Idt
C = -----
dV
where C is the capacitance in farads,
I is the load current in amperes,
dT is the ripple period in seconds, and
dV is the allowable ripple in volts.
Then:
Idt 0.5A * 0.01s
C = ----- = ------------- = 0.002F = 2000µF
dV 2.5V
Not bad, but that's cutting it pretty thin so let's lower the ripple
to one volt and see what we get.
Idt 0.5A * 0.01s
C = ----- = ------------- = 0.005F = 5000µF
dV 1V
Still not bad, and that'll make up for the capacitor's tolerance on
the low side.
So your final power supply will look like this:
MAINS>-+ +----+-[1N4001>]---+----+---[7812]--+---->+12VDC
| | | | |+ | |
P||S +-[<1N4001]-+ | [5100µF] | [100nF]
R||E-CT | | | | |
I||C +-[<1N4001]-+-|----+-----+-----+---->GND
| | | |
MAINS>-+ +----+-[1N4001>]---+
The 100nF cap on the output keeps everthing stable, and both caps
should be mounted as close to the 7812 as possible.
Also, there's nothing critcal about the BFC; a general purpose
aluminum electrolyticwill do fine and if you cant find 5100µF the
next largest size rated at around 25V will be even better.
JF
