cheap 24VAC power supply

[ngenius]

Hi,

I am trying to make a power supply circuit, which has to supply 5V @ 50mA max, and 15V @ 1 mA max.
I was thinking of using a single phase rectifier with 1 diode, a reasonably sized cap of say 47µF, and behind that for the 15 V part, just a zener and a resistor, and for the 5 V part an 7805.
However, the 7805 can only take 35 V max, and also I do not have any space for large cooling or anything.
Is there a simple and cheap way of limiting the input voltage to my 7805, without having to just dissipate it through a series resistor ?

thx

 

[Hero999]

24√2 = 33.9V but you're right to be worried about exceeding the voltage rating of the LM7805 as the 24V could be 26V off-load.

Use an LM7815 for the 15V supply and connect the LM7805 to its output.

 

[ngenius]

hi,

would that be the cheapest solution? And what about power dissipation? With the 2 casaded voltage regulators, the power dissipation is just divided over 2 voltage regulators.
Isn't there a way of 'shaving' the peaks? I have seen a circuit somewhere where they use some kind of a transistor and a zener diode to clip the peaks. But I do not know what kind of a transistor, the package is too small to read the label from it.

Heat might be a problem in my application, hence I would prefer smaller dissipation...

 

[Hero999]

50mA isn't much current so the power dissipation will be low <1.7W in total so the regulators probably won't even need a heat sink.

The power dissipation in each regulator is proportional to the current and voltage drop. The 7815 will disipate more heat than the LM7805 because it has a higher voltage across it.

Using a full bridge rectifier my capacitor sizing speadsheet recommends a minimum of a 35:mu:F capacitor which will need to be doubled for a half wave rectifier, giving 70:mu:F so use 100:mu:F which is the next standard value up.

 

[Willbe]

Two 10v Zeners in series, cathode to cathode, will drop the peaks.

 

[ngenius]

with the zeners in series, do you mean in front of the bridge rectifier or after that? and in series with the rest of the circtuit or in parallel? and where does the power go? dissipated in the zener diodes?

 

[Willbe]

with the zeners in series, do you mean in front of the bridge rectifier or after that? and in series with the rest of the circtuit or in parallel? and where does the power go? dissipated in the zener diodes?

Two zeners upstream of the diodes will reduce the AC into the diodes.
One zener downstream of the diodes will reduce the DC.
The zeners dissipate 10v x I or 20v x I worth of watts, but the I is rms and these values for these waveforms are tricky to calculate.

 

[Hero999]

Sorry, I was wrong about the zeners, you do need them if you want to use the LM7815

I thought that the LM7815 was rated to 40V but it isn't, it's only rated to 35V.

Why not just use an LM317 for the 15V regulator and put the LM7805 on the output?

It needs a couple of resistors but they're cheaper than zeners.

**broken link removed**
For 15V
R1 = 220R
R2 = 2k4

 

[ngenius]

Hi,

If I do nothing with these regulators, even with an LM317 and an 7805, I still have to dissipate too much energy, unless I have a large cooling island on the PCB or something, for which I do not have the space.

However, I have seen a circuit (I have reverse engineered an existing product and drew a little schematic of it) which seems to be able to do it all with standard SMD components and the regulator even in an SOT8 package.

I am just not sure how this works, and how much current I could draw from it. The components on it however are really cheap, and it saves me from going to some switching regulator.

Anyone any insights in this?

 

[Hero999]

I don't understand why you haven't just used the LM317.

At 50mA the power dissipation in the LM317 would be about 950W.

The power in the LM7805 would only be 500mW.

If you worried about the LM317 overheating then connect a 1W 180R resistor in series with it so it will only need to dissipate 300mW.

The power still needs to be dissipated somewhere so those zeners will still need to be appropriately rated.

You can alway use a switching regulator which wouldn't need any heatsinking.
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