Experiments Power Supply

I have been playing around with the experiments on here and have successfully built the power supply with a transformer robbed from an old turntable. I have also completed the rectifier projects. I would like to build a permanent power supply using the bridge rectifier and filter capacitor experiment but thought it might be beneficial to be able to switch between 6 and 12 volts as well as AC or DC. I have attached a drawing. My initial concern is the use of a filter capactior in conjunction with center tap rectification. Can this be done? Also, will the bridge rectifier work in this capacity (center tap rectification)? The main purpose of this power supply will be the later projects which call for 6 and 12 volt batteries.

You can use a centre tap transformer to output 6V and 12v simultaneously using 2 full bridges and filter capacitors for each. Alternatively, you can also add a cheap variable voltage supply by using an LM317 chip to vary your voltage very precisely from 1.25v to about 12.8v. If you intend to build this circuit to test something for a few seconds at 6v, the LM317 would be fine. But if you are going to soak test or charge a battery at 6v for hours, better settle for the first idea to minimize heat loss.

To obtain 6v and 12v simultaneousely, connect wires 1 & 2 of the transformer to the bridge to get 6v. and wires 1 and 3 to another bridge to get 12v. Or if you only need one output at a time, use one bridge and switch the inputs to the bridge from wire 2 and 3 to get 6v and 12v. When you do like this use atleast a 1000uF 25V electrolytic cap as filter.

Thanks for the reply. I don't know if you were able to look at my drawing. Sorry about the format, I don't have drawing software right now and when I tried converting to pdf everything got distorted. Anyway, I don't require simultaneous outputs, just switching capability. Your last recomendation is similar to what I have drawn except that instead of switching wires two and three to one side of the bridge, wires one and three always attach to either side of the bridge and I switch the output negative between the bridge negative (for 12V) and the center tap (for 6V). I am basically switching from full bridge rectification to full center tap rectification. What I am not certain about is whether or not a capacitor can be used to filter the center tap rectifier and also if the bridge rectifier will work since I am only using half of it. Intuition tells me it should work just fine. But why trust intuition when I have you guys to back me up? I have most of the components now, and they will all function for either design, its simply a matter of which wires I switch. Thanks again.

A 12.6V transformer might be 14V with a low current load. Its peak voltage is 19.8v which is reduced by the recifiers to 18.4VDC which is much higher than 12VDC.

When the transformer is fully loaded its voltage is 12.6V which is a peak of 17.8V. The rectifiers will reduce it to 15.8V and since the filter capacitor is too small the ripple voltage at the output will reduce the pulsing DC voltage a few volts.

I didn't check your drawing, but Im attaching the circuit of what you should do.

The switch used is a SPDT type to switch between voltages. Any 2way switch will also work. The transformer must be 6-0-6 to achieve what you are trying to do. If its 12-0-12 please post it before attempting to do this. Diodes are rated according to ampereage of transformer. 1N4001 will mostly be sufficient for you. With the above components you can switch between 12v and 6v on the fly with a fully rectified bridge and pure DC output. 12v transformers give 12x1.4=16.8VDC which you must pass through a LM7812 for safe usage.

P.S. You dont need any drawing software to draw circuits. MS Paint is enough. I have about 4-5 drawing softwares, but MS Paint gives the most appealing output, provided you know how to illustrate. Make a copy of all the components and save it as a Bitmap file and copy the components whenever needed.

I have completed the power supply as drawn and checked its operation under no load with and without the capacitor. Without it I see around 12.7V with my meter set to DC. Switched to 6V I see around 6.3V on DC.

The addition of the capacitor to the circuit really surprised me though. The voltages have jumped to over 18V and 9V, right around Audioguru's prediction. So my understanding is that the capacitor is charging to the peak voltage while I am really only reading a sort of RMS without the capacitor. I never really thought about that so its kind of cool to see.

One thing I don't get is Audioguru's statement about the capacitor being too small. The design in the experiments calls for a 1000μF 50WVDC capacitor. I have used a 2200μF 50WVDC unit. Is this not "bigger"? What would be big enough?

I will have to go get a 12V lamp for a load since all I have on hand is a wimpy little motor right now. I am curious to see the drop in voltage due to the discharge of the capacitor. How high the voltages remain under load may prompt my next project; a voltage regulator.

Which circuit have you connected??

The addition of the capacitor to the circuit really surprised me though. The voltages have jumped to over 18V and 9V, right around Audioguru's prediction.

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It isn't prediction, its theory.12.6VAC = 17.95VDC (12.6x1.414)

What you get after rectification is a clipped wave of which you read its RMS. When you add a capacitor, it flattens the pulses. and filters the AC component.

I put it together as shown in my drawing, switching from full bridge to full center tap rectification. The transformer I used is the same one recommended in the experiments section. Its a Radio Shack 1511 I think. The voltage is 6 from the center tap to the outer conductors and 12 between the outers. Is this what you mean by 6-0-6?

I do understand what you mean by the theory, its just always nice when reality reflects that theory BTW, I didn't use individual rectifier diodes, I used a pre-made bridge rated for 4amps.

So I have done a little more testing and took some pictures for anyone interested. This is the "12VDC" side, filtered, under no load:

Here it is powering a 35 watt 12 volt bulb:

I am still not completely familiar with the oscilloscope, so I cannot define the magnitude here; I simply adjusted the voltage multiplier to display the waveform. Under this load I calculate a current of around 2.7amps. Oddly enough, while the 3amp transformer and switch remained completely cool to the touch, the 4amp rectifier bridge was extremely hot. Next, switched to "6VDC", no load:

Note that this value is greater than half of the 12 volt operation. I think this is attributable to only using half of the rectifier bridge resulting in half the diode voltage drop. And, finally, powering a small 12VDC motor robbed from a turntable:

I guess my next step is to decide whether I want to use this to power heavier loads and then design an appropriate method of voltage regulation. Given the obvious limitations of the rectifier bridge, I was thinking I might try using one of the voltage follower circuits listed in the experiments section in conjunction with a zener diode regulated circuit. Haven't really studied it yet, but I think that would work at low power. Thanks for a great website and the input.

BTW, here is a photo of the wiring diagram:

Might be interesting to see the model number of the bridge.
Does it look like it should be mounted to a chassis or heat sink?

Its a Radio Shack 276-1146. It has no mounting hole in it but I suppose it would probably perform much better with the aid of a heat sink. I should have used the part recommended in the experiment section since it is rated for 25amps. But surely I don't need that capacity with only a three amp transformer right? Live and learn. I may upgrade if needed.

The voltage is 6 from the center tap to the outer conductors and 12 between the outers. Is this what you mean by 6-0-6?

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Yes i do. If you take 6 and 6's only, you get 12.

I do understand what you mean by the theory, its just always nice when reality reflects that theory BTW, I didn't use individual rectifier diodes, I used a pre-made bridge rated for 4amps.

Click to expand...

That is true. And bridges are better than diodes, provided you get them cheaper.


Its better if you use a variable voltage regulator chip like LM117K at 3A. It gives good filtering and regulated output. LM317 is more widely used at 1.5A

Also you dont need to take photos of your screen, just press the PRT SCR on keyboard and paste it in paint.

So I have been using this power supply as is (no regulation) and working my way through the experiments section. This presents no problems since these experiments have little to no load and are also not picky about specific voltage (I have been applying 9 volts to most of them to good effect).

But I got to the op-amp experiment: http://www.allaboutcircuits.com/vol_6/chpt_5/17.html
and this was how I THOUGHT I could regulate my voltage (at least at really low current). So here's what I did; I placed the op-amp in voltage follower configuration by applying full output feedback to the inverting "terminal" (I think that's the correct lingo). Next, I applied the simple voltage regulator experiment to the noninverting terminal in place of the potentiometer, drawing the signal from in between the 12V Zener diode and the 10kΩ resistor. And viola, I had 11.48VDC in and out (not a particularly accurate Zener diode).

So then I wanted to power something. I figured those transisters couldn't take too much, so I hooked up the low intensity LED from the other experiments and do you know it barely lit? The voltage dropped clear down to around 2 volts while I still had 11.48 at the input. <snip>

So now I am looking more closely at how these individual circuits work because I am not sure why it couldn't handle such a light load (pardon the pun). It was my impression that the purpose of the voltage follower circuit was to take an input voltage signal with a really small current and output that same voltage at an amplified current. Which sounded really simple, but now that I look at the specifics, I find that I don't have the knowledge to calculate this gain.

So I am sort of stalled right now. I have purchased an LN317T at the recomendation of JJ and others, but I don't want to apply it without have some understanding of what is going on. I could use some direction in analyzing these results. But just hints. Don't give it away completely.

Okay, not sure what I did wrong, but I was fairly certain this should work. So I tore everything apart and built the circuit again. And now it does. I have 11.48V in, 11.48V out, about three after a series resistance, and the lil LED lights. Haven't measured current yet though. Still don't know what I did wrong.

If you need to know how to work with Lm317, search this forum and you'll get hundreds of threads. Ckts and explainations are available online and also on Lm317's datasheet

I finally got a chance to record some voltage drops and calculate currents through this op-amp regulator. Here is the schematic with results:

There is indeed a significant current gain; 14.16 if my calculations are correct. But I still don't understand why. Based on the explanation in the experiment for the op-amp, here is my understanding: The current mirror formed by the two NPN's determines total current through the whole of the differential amp. The current mirror made up of PNP's divides the current equally between the differential segments. Since my program circuit only has 175μA's, shouldn't there only be half that in each of the differential legs? And if that is so, shouldn't there only be, at the most, 87μA's available to the load since that can be all that is available through the top of the PNP current mirror?

I know for a fact that one or more of the above arguments is invalid due to the resulting numbers. So where have I gone wrong? Can someone explain this circuit's operation to me, or at least guide me in the correct direction?

Best Regards
Mark

I have been reviewing everything and something didn't seem quite right. So I inspected my breadboard and discovered a significant mistake. The resistior in series with the LED is a 10kΩ, not 1kΩ. This means a load current of 939μA, not 9.39mA. So gain becomes only 1.42. My apologies.

So I wanted to see what kind of power I could get out of this thing and I began putting in resistors as loads to see what I could get. The 10kΩ minus the LED was my first step; 11.41V/10k=1.14mA for a gain of 1140/663=1.72.

But when I parallel another 10kΩ for a measured 4.8kΩ, voltage dropped to only 8.55V for a current of 1.78mA. Throwing a series 1kΩ in there for a measured 5.89kΩ got me 9.1V for 1.54mA. A measured 7.84kΩ returned 10.79V for 1.38mA. So, by experimentation, I figure the maximum practical current to be around 1mA and I decided to stick with the 10kΩ load.

Next I wanted to see how much gain I could get. Since I couldn't increase output anymore, I began decreasing input. I started big by swapping in a 1MΩ resistor in series with the Zener. My output was 11.42 for 1.14mA while input was 6.49V for 6.49μA and a gain of 175. Doubling that got me similar voltages, half the input current and twice the gain. After that, output dropped to only 10.3V at 3MΩ, so I quit there. A gain of 350 to 1 with the 2MΩ seemed pretty adequate to me.

So that's the update for now. I am no closer to understanding what it SHOULD do and thinking about it makes my head hurt. But I do know a lot more about what it WILL do.

Best Regards
Mark

BTW, forgot to mention, the program resistor was decreased to 22kΩ to allow for the increased current. Looks like this right now: