Dual Supply for Low Power but Sensitive Circuit

[MondyT]

Hi Mike

If your wiring is correct, first thing to check is that when the red and orange transformer leads are disconnected from the rectifier, see that you have 54v rms off load between the black and orange, and 27v rms from either of these leads to the black/yellow centre tap. If not try reversing *one* of the secondary connections.

Cheers
Ray

[pacificblue]

That transformer is too small to charge 2x4700 µF capacitors to ~29 V and may just be sufficient to maintain them charged with no load at the regulator output. To compensate for missing power it has too high voltage. That means you can reduce those 4700 µF to 100 µF and achieve a good regulated voltage. But the power dissipation will be much too high for the 78L/79L.

Your two options are
- replace the 78L/79L with the TO-220 version of the 78/79 regulators and add heatsinks, replace the 4700 µF capacitors with 100 µF and keep the transformer.
- replace the 78L/79L with the TO-220 version of the 78/79 regulators and add heatsinks, if the load draws more than 50 mA and/or the ambient temperature around the regulators is more than 25 °C, buy a transformer with 2x15 V secondary voltage and at least 30 VA to cope with the 2x4700 µF and 100 mA load at the output and keep the capacitors.

[MondyT]

Good point PacificBlue!

I doubt if it would drag the negative rail down to 0.7v so I would think there is still a problem somewhere.

Ray

[miallen]

Problem solved.

The VRs were in backwards. My make-shift diagram is a top-view but the pin layout in the datasheet is a bottom view and my brain was stuck on top-view mode so the pins ended up reversed. Lesson learned.

But there were other problems as well. It seems that after meddling with it so much, I fried at least one of the chips and probably a capacitor because I had no luck trying to resurrect the circuit. I just started over with a new circuit, with all new parts, tested frequently and it worked like a champ the first time.

I have not tried the existing 22V torriodal with the new circuit though. I have just been testing with two 9V batteries in series because I'm worried that the 33V coming out of the rectifier a little too close to the 35V limit of the larger caps. I'll just wait for the 15V torroidal. I'll use the 22V one for a little bench supply maybe.

Anyway this was all mostly just an exercise so I think I got what I was looking for out of this. Ultimately I want to see what sort of ripple I get from this 78L15AC/79L15AC circuit as opposed to an LM317T/LM337T supply. The LM317T/LM337T supply will probably be the one that I ultimately use for the device in question. I also have a single set of LT1962/LT1964 chips and some surfboards are on the way but I haven't tried SMT soldering yet so that will have to wait until stop making stupid mistakes.

Thanks to all who replied. If you haven't noticed, I'm not an EE so I appreciate the help.

[MondyT]

Well done Mike!

Remember, as PacificBlue said, either drop the value of the capacitors or up the VA of the transformer to say 15 or 30VA as the 7VA transformer will not cope with the circuit as it is.

Cheers
Ray

[miallen]

Originally posted by MondyT
Remember, as PacificBlue said, either drop the value of the capacitors or up the VA of the transformer to say 15 or 30VA as the 7VA transformer will not cope with the circuit as it is.

I did switch to 1000uF capacitors and I will be getting the 2x15V / 0.233A torroidal.

However, I must admit that I did not understand PacificBlue's comments. The power supply only needs to put out maybe 5mA. So why do I need all of that power?

[pacificblue]

The transformer needs to charge the capacitors. A discharged capacitor is very similar to a short-circuit. The inrush current would be several tens of amperes, with a bigger transformer. The 7 VA transformer cannot deliver that amount of current. The voltage will collapse, when you switch that circuit on. The capacitors charge to that lower voltage, their impedance rises, the transformer voltage does the same. If nothing else but the capacitors were in the circuit, the voltage would probably reach 29 V after sufficient waiting, during which the transformer is severely overloaded. In your circuit, however, the voltage regulators are discharging the capacitors, while the transformer is still trying to charge them. Therefore the full voltage will not be reached. One of the regulators is obviously faster to clamp the voltage, so the secondary voltages are unbalanced. That is a possible explanation, why you get the full voltage on one regulator and next to nothing on the other.

http://hyperphysics.phy-astr.gsu.edu...ic/capchg.html

[miallen]

I understand (mostly).

So given the size of a capacitor and a voltage, how do I roughly compute how much power will be required to charge and sustain that voltage?

For example, if I have a 15V torroidal (instead of 22V) and 2x1000uF capacitors, will 7VA be sufficient?

You stated I need 30VA for 2x4700uF caps with 100mA load. So if I use 1000uF that is a factor of 0.21 times 30 is 6.3VA and the load will only be ~10mA.

Regarding the TO-220 vs TO-92, is it also a problem that a larger capacitor can deliver too much current to a TO-92 and that's why I need a heat sink?

It seems to me if I do not draw that much power from the voltage regulator and provided that the regulator input voltage is not incorrectly high (forcing it to dissipate the extra power as heat), then should the TO-92 be ok?

I'm trying to get low ripple in less space but it seems that large filter capacitors (low ripple) without heat sinks (less space) are incompatible?

[AndrewT]

Originally posted by miallen


One thing that is strange is that when I measure the secondaries, I get 27.66 VAC and 27.72 VAC. The output of the rectifier is +33.95 DC and -33.08 DC.

So everything is a bit higher than I thought it would be. Why am I reading almost 28 VAC from the secondaries of a transformer that clearly shows on it's side that it has 22 VAC secondaries?

transformer regulation and mains voltage variation.

[pacificblue]

Originally posted by miallen
So given the size of a capacitor and a voltage, how do I roughly compute how much power will be required to charge and sustain that voltage?

You don't need a lot of power to charge it. The point is that you need it in a very short time span, usually in the first half of half a mains cycle. The formula is 1 F = 1 As/V, so to charge 1000µF to 20 V, you only need 0,02 As. The first half of half a mains cycle is 1/240 S, where you live, which means, the transformer would have to supply an average of 4,8 A during 1/240 s. It can probably not do that, so it takes several cycles at overload to achieve the task.

To sustain the capacitor voltage you need to refill, what is taken out by the load and the regulators. Also not much power. P=U*I, e. g. 30 V * 10 mA = 0,3 W.

The problem could be that the regulators are trying to build up a stable voltage and draw more current from the capacitors than the transformer can supply to fill them up. You could make two tests. 1) put a switch between the capacitors and the regulators. Measure the voltages. When (if) they reach ~30 V on both capacitors, switch the regulators on. Maybe it works like that. 2) Replace the capacitors with 100 µF and see, if it works without the switch.

Originally posted by miallen
For example, if I have a 15V torroidal (instead of 22V) and 2x1000uF capacitors, will 7VA be sufficient?

Probably.

Originally posted by miallen
Regarding the TO-220 vs TO-92, is it also a problem that a larger capacitor can deliver too much current to a TO-92 and that's why I need a heat sink?

The current is determined by the load. The heat dissipation is a result of (Vin-Vout)/Iin, where Iin is Iload+Iregulator. If you assume 30 V in, 12 V out and 100 mA the regulator has to dissipate 1,8 W of heat. With a thermal resistance of 140 K/W that means the TO-92 will be ~250 K above ambient, while the maximum temperature is defined as 150 °C. The TO-220 has 50 K/W -> 90 K above ambient, which could work up to 60°C ambient. If that circuit lives inside a case, 60°C can easily be reached and surpassed, which means heatsinks are needed.

If your assumption about 10 mA maximum current draw is correct, we have to add 6 mA quiescent current of the regulator to that and only get a disspation of 0,288 W. That means 40 K above ambient, in which case TO-92 is okay with no additional heatsinking.

Originally posted by miallen
I'm trying to get low ripple in less space but it seems that large filter capacitors (low ripple) without heat sinks (less space) are incompatible?

The low ripple is provided by the regulators with the small capacitor at the output. The capacitors at the regulator input are there to maintain the voltage high enough, so that the regulator can work, i. e. above 12 V + what the regulator needs to maintain a stable output. Therefore you only need 100 µF, when you have ~30 V DC, but 4700 µF, when the voltage is below 18 V DC, to maintain the lowest point of the ripple voltage before the regulators high enough.