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I tried to build a simple stabilised voltage supply for basic use to operate a maximum of 2 PC fans (0.15A and 0.45A) and also a 12V DC LED (0.5A) eg: http://www.led-slt.com/Spotlight/139.html.

Using parts I had around, I built a simple circuit based on an ac transformer (WDB4109 with o/p: 16.7Vac), a schotty rectifier STPS30 (giving 25V dc); an 18v zenner to drop the voltage to 18V which are fed to a voltage regulator L78S12, hoping to have a stabilized 12V dc. I've supplied a circuit diagram. Without load, I have 18.2V after the zenner and a nice 12V output. When I connected the 0.13A fan, it rotated very very slowly as if there is no power.

Why I have very limited power in this circuit?

For your info, the transformer circuit is as follows: enter image description here

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  • \$\begingroup\$ The schematic shows a half-wave rectifier... is that really how it's wired? \$\endgroup\$ – Brian Drummond Oct 13 '15 at 18:44
  • \$\begingroup\$ Indeed. The secondary needs a center tap, and a 36V rating. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 13 '15 at 18:45
  • \$\begingroup\$ What is the value of R? I am guessing it is quite high. \$\endgroup\$ – HandyHowie Oct 13 '15 at 19:17
  • \$\begingroup\$ In the absenceof a centre tapped transformer, a bridge rectifier (4 of those diodes) would work. \$\endgroup\$ – Brian Drummond Oct 13 '15 at 19:20
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    \$\begingroup\$ OH! it is L78S12. From the data sheet, I can see that it can tolerate up to 40V input so I may remove the Zenner and see what happens : alldatasheet.com/datasheet-pdf/pdf/22628/STMICROELECTRONICS/… \$\endgroup\$ – MarzTronix Oct 13 '15 at 19:39
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As already noted, if your drawing is accurate then the bottom diode of the STPS30 isn't being used at all and all you're doing is half-wave rectifying and filtering the output of the secondary.

With no load on the regulator, (assuming it's a 7812 clone) its output voltage will look like 12 volts, but as soon as a load starts drawing substantial current out of the regulator the drop across R will increase, making the Zener invisible and decreasing the voltage available into, and out of, the regulator.

An easy fix would be to get rid of "R" and the Zener, and to let the regulator do the dirty work, but in order to do that, we need to known what the transformer's secondary looks like

EDIT:

Now that we know what the transformer looks like, you're left with a couple of basic choices: Use either pins 5 and 6 or 7 and 9 as the inputs to a full-wave bridge rectifier, then smooth and regulate the output of the bridge to get the 12 volts you need to drive your loads.

If you use the center-tapped 11 volt RMS winding as the input to the bridge, you'll lose about 1.5 volts through the diodes so, if your filter cap is big enough, you'll wind up with about 14 volts into the 7812 clone.

Typically, a 7812 needs about 2 volts of headroom to work properly so, depending on the \$I^2R\$ losses in the transformer, the actual drop across the bridge's diodes and the ESR of the reservoir cap, you could wind up with significant ripple into your load.

On the other hand, if you use the 16 volt winding you could wind up ripple-free, but with something like 21 volts into the 7812 and with the 1.1 ampere load from the fans and the LED, the 7812 will have to drop the difference between 21 volts and 12 volts at 1.1 ampere, which means it'll have to dissipate about:

$$ P = IE = 1.1A \times (21V - 12V) \approx 10 \text{ watts}$$

That's a lot for a 7812 in a TO-220 package to dissipate, and one way to keep from having to use a large heat sink to do it is to use a power resistor in front of the 7812 in order to drop the 21 volts to something the 7812 can more reasonably handle.

NASTY SURPRISE:

Becoming suspicious because of the roughly 23VA capacity required of the transformer (13 watts for the load and 10 watts wasted in the regulator) I did a cursory search for the transformer and found that the best it can do is less than 10VA!

I lost the link, but you can probably find it if you Google WDB4109 or GAL4118E-WDB-01, which is the same transformer.

In any case, assuming a decent transformer, here's what your power supply should look like:

enter image description here

and here's the LTspice circuit list if you want to play around with the circuit:

Version 4
SHEET 1 880 852
WIRE -256 48 -368 48
WIRE -160 48 -256 48
WIRE -96 48 -160 48
WIRE 224 48 160 48
WIRE -368 80 -368 48
WIRE -256 80 -256 48
WIRE -704 176 -928 176
WIRE -256 176 -256 144
WIRE -256 176 -624 176
WIRE -928 224 -928 176
WIRE -704 224 -704 176
WIRE -624 224 -624 176
WIRE -160 224 -160 48
WIRE 224 224 224 48
WIRE -672 320 -672 208
WIRE -656 320 -656 208
WIRE -928 352 -928 304
WIRE -736 352 -928 352
WIRE -704 352 -704 304
WIRE -704 352 -736 352
WIRE -624 352 -624 304
WIRE -592 352 -624 352
WIRE -368 352 -368 144
WIRE -368 352 -592 352
WIRE -368 384 -368 352
WIRE -256 384 -256 176
WIRE -736 448 -736 352
WIRE -704 448 -736 448
WIRE -592 448 -592 352
WIRE -592 448 -624 448
WIRE -368 480 -368 448
WIRE -256 480 -256 448
WIRE -256 480 -368 480
WIRE -160 480 -160 288
WIRE -160 480 -256 480
WIRE 32 480 32 144
WIRE 32 480 -160 480
WIRE 224 480 224 304
WIRE 224 480 32 480
WIRE -368 544 -368 480
FLAG -368 544 0
SYMBOL schottky -240 144 R180
WINDOW 3 -63 1 Left 2
WINDOW 0 -38 30 Left 2
SYMATTR Value B520C
SYMATTR InstName D3
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL ind2 -720 208 R0
WINDOW 0 -43 40 Left 2
WINDOW 3 -43 75 Left 2
SYMATTR InstName L1
SYMATTR Value 50
SYMATTR Type ind
SYMATTR SpiceLine Rser=,1
SYMBOL voltage -928 208 R0
WINDOW 3 24 104 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR Value SINE(0 339 50)
SYMATTR InstName V1
SYMBOL ind2 -608 208 M0
WINDOW 0 -46 46 Left 2
WINDOW 3 -62 79 Left 2
SYMATTR InstName L2
SYMATTR Value .229
SYMATTR Type ind
SYMBOL res 208 208 R0
SYMATTR InstName RL
SYMATTR Value 8
SYMBOL res -608 432 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName Rimg
SYMATTR Value 1G
SYMBOL PowerProducts\\LT1085-12 32 48 R0
SYMATTR InstName U2
SYMBOL schottky -352 448 R180
WINDOW 3 24 0 Left 2
WINDOW 0 42 32 Left 2
SYMATTR Value B520C
SYMATTR InstName D2
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL schottky -352 144 R180
WINDOW 3 24 0 Left 2
WINDOW 0 42 31 Left 2
SYMATTR Value B520C
SYMATTR InstName D1
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL schottky -240 448 R180
WINDOW 3 -65 0 Left 2
WINDOW 0 -37 31 Left 2
SYMATTR Value B520C
SYMATTR InstName D4
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL polcap -176 224 R0
SYMATTR InstName C1
SYMATTR Value 1500µ
TEXT -352 512 Left 2 !.tran 1
TEXT -720 144 Left 2 !K1 L1 L2  1
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  • \$\begingroup\$ You are right: I think the main issue is the STP30S connecting to a non-centred tap transformer. Still, would such setup (half wave rect.) produce only 0.15Amps??? google.com/… \$\endgroup\$ – MarzTronix Oct 13 '15 at 19:57
  • \$\begingroup\$ @MarzTronix: try to scrounge 4 diodes and make a full-bridge rectifier for your 16VAC output. Alternatively, try connecting the STPS30 as a full-wave rectifier on the two 5.4VAC secondaries; you should be getting about 15V DC, which will be marginal for the L78 as input (needs about 3V headroom), but worth a try. Refer to hammondmfg.com/pdf/5c007.pdf for schematics. Also, how big is your filter capacitor? \$\endgroup\$ – Fizz Oct 13 '15 at 20:05
  • \$\begingroup\$ Update: I removed the 50Ohms resistor, hence connected the output of the STPS30 to the input of the Volt. regulator and I managed to get a 0.3A output. The light was flickering - very fast but noticable. \$\endgroup\$ – MarzTronix Oct 13 '15 at 20:25
  • \$\begingroup\$ Filter capacitor is currently 330uF (25V). \$\endgroup\$ – MarzTronix Oct 13 '15 at 20:27
  • \$\begingroup\$ So if at present it seems I can have a max of 0.3A, with a full-wave rectification, should I get twice as much or more? \$\endgroup\$ – MarzTronix Oct 13 '15 at 20:28
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Considering the available transformer taps the best way with the given components would be to use the output from the transformer tappings labelled as 5 and 6. According to the image this would give 16.33 V. Connecting this directly to the linear regulator would still allow a 12 V Output.
However depending on the current your circuit draws the linear regulator could get hot and require additional cooling.

For a more efficient circuit you could use a step-down regulator with a large capacitor in front for smoothing. For example a easy solution from the Traco TSR Series. Furthermore for high current draw a full wave rectification is advisable.

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  • \$\begingroup\$ I am going to replace the STPS30 with a full rectifier RS405. I hope the resulting voltage is not to high for the linear regulator. What smoothing capacitor you suggest after the output of the rectifier. Currently there is a 330uF / 25V but better put a higher V-rating capacitor. PS. In the original circuit I was trying to step down the output viltage by a 18V zenner, \$\endgroup\$ – MarzTronix Oct 13 '15 at 20:58
  • \$\begingroup\$ Question: I have difficulty were to connect the ground of the rectifier, because as shown in my circuit, the DC ground is common with the AC ground, and that would mean a direct short between pin 3 (one of the ac in) and pin 4 (dc -) of the IC \$\endgroup\$ – MarzTronix Oct 13 '15 at 21:00
  • \$\begingroup\$ If you use a bridge rectifier, your circuit ground will be the negative terminal of the bridge. Neither end of the transformer seconary should be connected to your circuit ground. \$\endgroup\$ – Peter Bennett Oct 14 '15 at 0:52
  • \$\begingroup\$ Thanks! I opened the track to isolate the grounds. I just need to buy a smoothing/filter capacitor 1000uF/50V and tell you how it went! \$\endgroup\$ – MarzTronix Oct 14 '15 at 4:57
  • \$\begingroup\$ That smoothing capacitor should improve the input to your linear regulator. Especially as most capacitor technologies have a lower capacitance then what they are labeled as, when they work close to their maximum voltage. The combination of resistor and zener can work, but it needs to be adjusted to the maximum current that could occur to ensure enough voltage at the input to the linear regulator. \$\endgroup\$ – Grebu Oct 14 '15 at 5:17
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This is how my circuit looks like: Lessons learnt (solution)

1: high resistor feeding zenner diode was limiting power output on load. When 500Ohms was changed to 50Ohms, there was a drastic improvement.

2: Full wave rectification gives further power output from half wave

3: Smoothing capacitor of 1000uF improves the sc stability

4: Read the specs! The LM7805(equiv) was found to be able and take up to 40V as input so the zenner was useless to step down the 25Vdc output from the rectifier

5: RS405 = Full fave rectifier, 4A

Basic regulated Power Supply for small applications

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Posting just to show you the final stages of the project. I have build an own terranium and wanted some heat regulation. I've ripped many components from an unused microwave, such as the heating elements, power supply, fan and light bulb. The main concern is to maintian a constant temp. The heating elements in their original stated where too hot, hence i have connected them to a triac regulator to lower their output once they heat the chamber to the given value.

One of the ac splits feeds the powersupply, which is used to give 4 switchable 12Vdc outputs to power more internal fans and lights. The vivarium has on ac fan, on ac heating element, one uv light, on cool white light, one bulb light (subtle heat), one dc light and two dc computer fans.

Main issue is to keep a constant temp all over the vivarium! [enter image description here][1]

Thanks for your help

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  • \$\begingroup\$ Glad we could help and nice to hear the application. Two final suggestions, as this site behaves different than a normal forum: 1) Select one answer as the accepted answer to show this question is solved. 2) Post updates like this in your initial question or if it fits better in your answer by editing those. Do not create multiple answers. An answer here is meant as a solution to the initial question. \$\endgroup\$ – Grebu Oct 22 '15 at 9:31

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