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I'm trying to convert mains AC to DC while dissipating as little heat as possible. Some ripple is tolerable if it cuts down on heat so I'm not sure if I need the regulator. I also don't know how well the values I've chosen for the various components will work together. Perhaps there's even a much easier way to go about it.

Finding parts is another concern. For instance, can a 22 volt regulator can even be bought?

Any helpful advice is greatly appreciated.

Here's the circuit diagram:

enter image description here

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    \$\begingroup\$ what are the specs on the bridge rectifier? also, directly feeding a large-ish cap like 1500uF, realize that on power up, the load seen by that bridge is essentially a short circuit, and will pull a huge transient current through the bridge. \$\endgroup\$
    – JustJeff
    Jun 28 '11 at 9:32
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    \$\begingroup\$ @JustJeff - Correct, but this is how it is done in 99% of linear power supplies, literally billions of product-hours, and they do fine. In practice it's not a problem because the transformer will resist the peak. \$\endgroup\$
    – stevenvh
    Jun 28 '11 at 12:48
  • \$\begingroup\$ electronics.stackexchange.com/questions/515 \$\endgroup\$
    – endolith
    Jun 28 '11 at 13:15
  • \$\begingroup\$ @stevenvh - yeah, i know. the point, though, is "don't be stingy with your margins re current when selecting a bridge" \$\endgroup\$
    – JustJeff
    Jun 28 '11 at 20:18
  • \$\begingroup\$ @JustJeff: Again true, but rectifier diodes and bridges have short term peak current specs quite a few times higher than the max average operating current. Yes, there will be a large inrush for up to a few ms, but the diodes are generally specified to handle that. And as @stevenvh said, other impedances in the system will limit the inrush. Of course you need to check the specs, but this sort of circuit is very very common. \$\endgroup\$ Jun 29 '11 at 12:18
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I think it will barely work as shown.

22V is an oddball voltage. Use an adjustable regulator where the output voltage is set by the ratio of two resistors.

Peak input voltage is \$\sqrt2 * 18\$ minus diode losses (~1.5 volts), so about 24 volts. So if your regulator works with 2V or less of dropout, you will be OK. (This is kind of a tight dropout spec, not impossible, but not every regulator you find will be OK. Micrel MIC29303 appears to be OK.)

Now, how much current can you draw? Assuming 60 cycle power, let's say we can allow the supply to droop 1V in the 1/120 seconds between voltage peaks. Then the voltage regulator will have 1 volt headroom. \$I = C * {dv\over dt}\$ so if \${dv\over dt}\$ is \$1\over1/120\$, then \$I = 120 * C\$. For a current of 0.5A, then, choose C of about 4000 μF.

All of this design is cutting pretty close: If the line voltage is just a little high, you will exceed the regulator's input voltage limit. If the line voltage is just a little low, you will drop out of regulation. I suggest you find a transformer with a little higher output and use an LM317 regulator instead of a low-dropout type. Additional headroom will allow a smaller filter capacitor as well.

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  • \$\begingroup\$ Also need to ensure a minimum load is maintained on the secondary. \$\endgroup\$
    – Mark
    Jun 28 '11 at 2:53
  • \$\begingroup\$ @Mark please explain why \$\endgroup\$
    – markrages
    Jun 28 '11 at 4:04
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    \$\begingroup\$ @markrages - good analysis. But my numbers are even worse. You rightly mention tolerances on line voltage. In Europe the 230V can be up to 6% lower, I assume the same value for the US. My voltage drop over the diodes is also higher; I usually work with 1V per diode. Then the rectified transformer voltage is 22V, and even the MIC29303 won't cut it. \$\endgroup\$
    – stevenvh
    Jun 28 '11 at 6:26
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    \$\begingroup\$ Going by the '317 data sheet you should have a second cap at the far right end of the circuit (I think to respond quickly to transient current changes) \$\endgroup\$
    – Majenko
    Jun 28 '11 at 8:53
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    \$\begingroup\$ @Matt Jenkins: that's one reason, but most regulators will need it for stability; without they may oscillate. \$\endgroup\$ Jun 28 '11 at 9:32
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I think @BarsMonster sortof has the right idea. Instead of a 12V power supply, I'd get a 24V one. That's also a common value. Then you can linearly drop the extra 2V if you really need to. A adjustable regulator tweaked to 22V should do it. Depending on how accurate the 22V needs to be, it could be as simple as the 24V supply with a few diodes in series.

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  • \$\begingroup\$ Tweaking the switcher's voltage should be possible. In my experience it's only low-voltage devices which have a fixed output voltage. For higher voltages the output voltage is set by a resistor divider. \$\endgroup\$
    – stevenvh
    Jun 28 '11 at 14:37
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Short answer:

Use of a SMPS (switched mode power supply) is likely to be preferable if heat generation is a genuine concern.

However, actual heat dissipation in the power supply is liable to be under 2 Watts in most cases. Whether this is significant depends on your application.


Long answer:

When attempting to provide any solution, knowing as much as possible about the actual requirement will greatly assist people who attempt to help. A poorly understood problem is liable to lead to less useful advice than would otherwise be possible.

  1. Why do you want 22 Volts. How precise is this requirement? How much variation is acceptable either in the initial voltage or between loaded and unloaded values.

  2. You say "... while dissipating as little heat as possible." Can you please explain why you have this requirement and how much heat you can tolerate. eg:

    • You may care about the energy efficiency, or

    • You may want to enclose this in a space where there is very little cooling, or

    • you may wish to run it in a refrigerated space where heat produced is a load on the refigeration system, or
    • ... ?

Knowing the reason for the requirement is important in understanding it. Also, stating how much heat you can tolerate may assist us providing a good answer. If you can ask a perfect question then we can provide a perfect answer ;-).

Most of the solutions so far either are based on versions of what you showed in your circuit or do not make it clear what technology is used. If minimising heat generation is genuinely important then use of a SMPS ("switch mode power supply") is probably a good idea. These can operate at in excess of 90% efficient and will automatically adjust for variations in mains voltage. At say 90% efficient and a 22V, 0.5A output, the energy lost as heat will be Energy lost = 22V x 0.5A x (10%/90%) = 11 x 1/9 =~ 1.2 Watts. That is a small amount of thermal energy in most contexts. Whether it is small enough in yours depends on your requirement.

As a guide to what you can expect from a "linear" supply of the type you have shown: Lossesof a componentare roughly proportional to the voltage drop they cause. For example, if voltage drop across diodes was 1 Volt average and Vout = 22VDC then diode losses would be about 1V/22V ~= 4.5% of ouput power. There are reasons why this simple analysis is not completely accurate, but it is good enough to get a feel for losses.

Losses are (mainly) caused by transformer, rectifier, regulator, wiring and "headroom".

  • Losses in transformer and wiring are generally small enough to be ignored in this context.
  • Rectifier losses are caused vy voltage rop across the diodes. Diode drop is a minimum of 0.6 V per diode and can be more like 0.8V typically and in higher current circuits than this one can be over 1 V. Use of "Schottky" technology diodes can reduce this to 0.3V to 0.5V per diode. In the circuit shown there are 2 diodes in series conducting at all times 20 you get 2 x diode drops or about 10% losses. By chaning to a centre tapped transformer you can halve this to about 5%. By changing to Schottky diodes and a centre tapped transformer you can reduce diode losses to about 2.5%. Whether this is worthwhile depends on your requirement.
  • "Headroom" losses are voltage drops which have to be built into the system to make it work correctly in all cases.

    eg if your mains voltage variation is +/- 5% then you must allow 5% more voltage than is necessary at the transformer output when the system is running at nominal input voltage, so that it will still function correctly when voltage input falls by 5%. This extra 5% designed voltage increase is a direct loss during normal operation. - Also, if heat dissipation is absolutely crucial you may wish to allow for the +5% mains situation when calculating heat dissipation. Under 230V +5% conditions the system will dissipate +10% more energy (approximately) - 5% due to the needed 5% overdesign to allow for low voltage !, and 5% because the mains is 5% higher than usual.

    The regulator will require some input to output voltage drop to function. This is part of overall "headroom". The LM317 regulator mentioned is an extremely useful IC but requires about 2V "headroom" to operate. Data sheet: http://bit.ly/DS_LM317 This is about an additional 9% loss in this circuit (!) A genuine LDO (low dropout) regulator can have almost zero dropout voltage if absolutely necessary.

    Finally, as noted by others, the input capacitor will have some "ripple voltage" as its voltage falls when it supplied current when the transformer output drops towrds zero on each half cycle. The amount of ripple can be designed - the larger the capacitor the lower the ripple. Headroom losses due to ripple will be about 2.5% per 1 Volt peak to valley ripple. (This is half loss due to a 1V DC drop as the average value of the ripple is about half its peak to valley magnitude. Assume for now we design for 1V p-v capacitor ripple or 0.5 V average.

Adding these up we get Rectifier + mains voltage variation allowance + regulator dropout + headroom MJixing % losses and voltage drops as explained above.

Best case 0.3V + 5% + 0 + 0.5V = 0.8V + 5% =~ 9% =~ 1 Watt Here, using Schottky rectifier, centre tapped transformer, 1V ripple and perfect LDO the allowance for mains variation dominates.

Typical case using circuit as shown, LM317 and 1V ripple. Losses = 1.2V + 5% + 2V + 0.5V = 3.7V + 5% =~17% losses.

Thermally, if output = 22V x 0.5A = 11 Watt, losses would be 1.9 Watt.

So we get:

1.2W SMPS at 90% 1 W linear best case. 1.9W linear typical.

For most purposes the differences would be of low importance or even insignificant. In some cases they may be crucial.

In many cases a single winding transformer, intelligently sized winding voltage, 4 diode bridge and as large a capacitor as sensibly possible (allowing a lower transformer voltage) would be adqeaute. HOWEVER it is easy to get far greater dissipation if eg the transformer woltage is not well designed. eg a 18V winding would have an RMS voltage and so a nominal rectified DC voltage of about 25.5V. The additional (25.5-18) = 7.5V ~= 4 Watts has to be dissipated in some manner.

So, use of a smps is likely to be preferable if heat generation is a genuine convcern

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I would recommend to get any off-the-shelf 12V power supply, and connect it to DCDC 12->22V (also, you may find adjustable off-the-shelf modules). This will produce less heat, no ripple, and you won't need to hassle with mains voltage.

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