I was looking at a 12 volt regulated power supply at Dick Smith Electronics cat. number M9935 @
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for a regen radio I'm planning on building
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1-12af6.htm) but the AU$24 is a bit too high for my liking. I was thinking of using a center tapped transformer, 6.3-0-6.3V @ 500mA, in series with a 1.3 watt zener diode, I need help with obtaining 12.6 volts from a center tapped transformer, I never had to do it before. I thought that the two positive primary leads were connected, I don't want to blow the transformer if this isn't the case.
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
There were several power supply designs offered in response to your post on a voltage doubler. The same principles apply to your 12 VDC circuit. It would be good for you to become familiar with LTspice where you can try different configurations and do experiments without blowing something up or hurting yourself.
For the components you have, you will need a full wave bridge made from four 1N400x diodes, a capacitor of about 1000 uF 25 VDC, and a resistor that will drop about 5 volts at the current you need to draw, plus the zener current. With a 1.3 watt 12 V zener, its maximum current is about 100 mA, so the maximum current draw for your supply should be about that much.
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.
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:
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.
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.
I like that 7812 idea. The tube (12AF6) draws 150 mA, then there's the current draw of the 12 volt B+, a zener won't handle all that. A
5600 MFD 40 volt capacitor is AU$7.30, I don't like paying that for low quality Asian capacitors, so I'm paralleling 3 2200 MFD 25 volt caps in series to give a total 6600 MFD. Each of these caps cost $0.97
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You can\'t "parallel caps in series".
According to the tube specs the plate current will be about 1mA, so
the total draw, with 175mA of filament current, will be 176 mA. With
that current, the capacitance needed to get 1 volt of ripple will be:
Idt 0.176A * 0.01s
C = ----- = ---------------- = 1.76E-3F = 1760µF
dV 1V
If your 2200µF caps have a tolerance of +/-10% then the lowest their
capacitance will be is 1980µF, so you\'ll only need to use one.
Matter of fact, even if they\'re +/-20% that comes out to 1760µF, which
is right on the money, so you could still get away with just one!
JF
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