Transformerless Power Supply

[volkov]

Hi all,

I am looking for a reliable Transformerless power supply to run a PIC12F675 and a relay to drive a reticulation solenoid on my farm. The unit is powered from a 30v AC power pack and the circuit will draw something in the order of 10mA when relay is off and less than 25mA when energised.

Have looked at a few circuits but not sure which way to go... any circuits / ideas / calculations would be greatly appreciated. Oh, also the unit can't get too hot as it needs to be sealed.

Thanks

Volkov

[Melanie]

Have tried doing a SEARCH on this forum with the word "Transformerless"?

[psdayama]

Hi all,

I am looking for a reliable Transformerless power supply to run a PIC12F675 and a relay to drive a reticulation solenoid on my farm. The unit is powered from a 30v AC power pack and the circuit will draw something in the order of 10mA when relay is off and less than 25mA when energised.

Have looked at a few circuits but not sure which way to go... any circuits / ideas / calculations would be greatly appreciated. Oh, also the unit can't get too hot as it needs to be sealed.

Thanks

Volkov

Hello Volkov,
It is nice to put the output current requirements but what about the voltage? Is relay used is 24Vdc or 12Vdc? Even smallest "sugarcube" 24V dc
relay will take atleast 25mA !!! The 12V 40mA. U can use buck switching
regulator like TL494 or even old horse UA 723. there are so many books and
articles to check under "switching power supplies" and you will find relevant
info.

[psdayama]

Hi all,

I am looking for a reliable Transformerless power supply to run a PIC12F675 and a relay to drive a reticulation solenoid on my farm. The unit is powered from a 30v AC power pack and the circuit will draw something in the order of 10mA when relay is off and less than 25mA when energised.

Have looked at a few circuits but not sure which way to go... any circuits / ideas / calculations would be greatly appreciated. Oh, also the unit can't get too hot as it needs to be sealed.

Thanks

Volkov

Hello Volkov,
It is nice to put the output current requirements but what about the voltage? Is relay used is 24Vdc or 12Vdc? Even smallest "sugarcube" 24V dc
relay will take atleast 25mA !!! The 12V 40mA. U can use buck switching
regulator like TL494 or even old horse UA 723. there are so many books and
articles to check under "switching power supplies" and you will find relevant
info.

[volkov]

Well looks like I've hit the jackpot with this forum... found stuff that I wasn't even looking for. ;-)

I guess I should have stipulated that the circuit requires 5v DC, basically 30v AC in - 5v DC out and draws 25mA.

Melanie, I was having a look at your circuit in the Feb 2006 thread, and what baffled me is the parallel configuration of the Capacitor - Resistor and the fact that you have the Zenar Diode before D2 - D3.

Basiclly it's exactly what I am after, however can you please explain the workings of the circuit and the formulas to work out the capacitor / Resistor values for 30v.

Many thanks,

Volkov

[mister_e]

you may also do a google search with 'capacitive dropper'

just one part of the theory...
http://www.vintage-radio.com/repair-restore-information/valve_dropper-calcs.html

[Melanie]

The Capacitor value is dependant on the CURRENT you are intending to draw - which will also impact on the current rating of the Zener D2 and VDR1. 470nF is good for about 25/30mA. I find formulas in this case don't hold up too good... simply take your Capacitor and put it across your AC Supply in series with an AC Ammeter. That'll give you a reasonable idea of absolute maxium current. Next step is to build your circuit, and then load the Zener until the input voltage at the Kathode starts to colapse below the holding value. That then tells you how far you can sensibly go with your selected Capacitor.

Without the Zener Diode (D2) and D1 combination, off-Load, the Voltage at the Anode of D3 will quickly reach AC Supply potential (eg 240v) and the poor little 78L05 (which has a 30v max input or thereabouts) and everything else on the board would be history. The Zener clamps the off-load voltage to a safe level. D1 stops the Zener from popping. This design had to be reasonably failsafe, so there is secondary protection in the form of VDR1 should the Zener fail. If Capacitor C2 failed, the combined current through D1, D2 and VDR1 would quickly blow the fuse.

At least that's the theory, nobody's died yet as far as I know (I've been told to refrain from killing off clients as apparantly it impacts on repeat business).

[volkov]

Very logical design approach... I am planning to use a MKT capacitor rated at 63v (for my circuit) is this the same flavour that is used for the "Mains" transformerless Power supplies?

Also, I have seen a few circuits where the input is passed through the capacitor and then a fullvawe rectifier, what would be the advantage of doing it this way.

Well hopfully, will have something built by the weekend. :-)

Volkov

[Melanie]

No problem full or half-wave rectifying after the Capacitor. Full wave recification is better if you are going to have a high current demand as you'll have less ripple on the main smoothing capactor. But for only a few mA you may find it's not worth the extra components and expense.

On the "mains" supplies, I recommend the use of X2 rated Capacitors only. If you already have a low-voltage supply, that's not so much of a concern. Put in a Capacitor whose voltage rating is at the minimum twice that of the off-load volatge of your AC Supply.

[psdayama]

Very logical design approach... I am planning to use a MKT capacitor rated at 63v (for my circuit) is this the same flavour that is used for the "Mains" transformerless Power supplies?

Also, I have seen a few circuits where the input is passed through the capacitor and then a fullvawe rectifier, what would be the advantage of doing it this way.

Well hopfully, will have something built by the weekend. :-)

Volkov
It is always better to use fullwave rectifier as capacitor may buid up charge
in half wave supplies. 63V MKT capacitors can work with 30V ac source.No need of expensive
X2 capacitors. Only need is to check ampere rating of capacitors.
U have to consider capacitor as lossless impedance and calculate required value as per current requirement. Please note this arrangement works as current source so U have to provide requisite load all time. I am attaching
diagram for same: The unmarked resistor has same value as relay coil.

[volkov]

Hi everyone,

Just a quick note to thank everyone who helped out and an update... well after several weeks I finally found the time to do some experimentation.

I found that the capacitance required was in the order of 4.7mF which meant big MKT caps... as space was a premmium I opted for a Bipolar Electrolytic with a 50Vw rating. (Even though they are not recomended for AC circuits) Anyway, todate the PS has been running without a beat. If in time I have a disaster on my hands I'll let you know.

Thanks again

Volkov

[Luciano]

The unit is powered from a 30v AC power pack and the circuit will draw something in the order of 10mA when relay is off and less than 25mA when energised.

Hi,

Also possible when the rectified voltage is above 35V which is the maximum input voltage for a 7805:

3-Terminal Adjustable Voltage Regulator
http://www.national.com/pf/LM/LM317HV.html

The maximum input voltage of a LM317HVT is 60 volt.

Best regards,

Luciano

[psdayama]

OK, "nothing is impossible!!!" is Adidas ad.
But have U considered the power dissipation in regulator?
Capacitor voltage divider is 'lossless' so there is hardly any dissipation
in dropper.
Anyway we expect Volkov to announce his result.Although his use
of Bipolar instead of MKT is little deviation I think his ckt will work OK.
Best Luck to all who want to use same method. I have constructed
10s of units with such supply and are working OK last 2-3years.

[Luciano]

Hi,

With a current of 25mA the power dissipation will be less than 1 watt.

30V AC - 1.2V (rectifier loss) = 28.8V
28.8V x 1.41 = 40.608V DC
40.608V - 5V = 35.608V
35.608V x 0.025A = 0.8902 W

* * *

Not in this project but maybe one day you will need a voltage
regulator with an input voltage up to 60V.

http://www.national.com/pf/LM/LM317HV.html

Best regards,

Luciano

[psdayama]

Well estimated 1watt loss can heat LM317HV quite a bit.
I think that in the price of that regulator U can buy 2MKT capacitors
easily (Atleast in this part of world )
Please make calculations as follows:
30V ac = 30 x 1.41 = 42.42 volts DC
- 1.2volts drop = 41.22 volts - 5volts output = 36.22V
x .025A = 0.9055 Watts.

One needs big heatsink for this.

The purpose of circuit is to get lower voltage without transformer
and not much regulated voltage is required.

When required I will certainly use HV regulator if called for.
otherwise I will use switching one to make it "cool" !

Thanks a lot for suggestion.

[Luciano]

One needs big heatsink for this.


Your statement is not accurate.

1 Watt of power dissipation without heat sink is possible with a LM317HV/TO220.
Below you can see how to calculate the maximum allowable power dissipation without heat sink.

Best regards,

Luciano



How to calculate the maximum allowable power
dissipation without heat sink:


PD MAX = TjMAX − TaMAX
_____________
Rja

PD MAX = Maximum Power Dissipation (Watt)


TjMAX = maximum junction temperature (°C)
For the LM317HV the operating junction temperature range is 0°C to +125°C.


TaMAX = maximum ambient air temperature (°C)
Room temperature or temperature inside enclosure.


Rja= thermal resistance from junction to ambient without heat sink (°C/W)
For the LM317HV/TO220 is 50 °C/W.


Example ambient air temperature 45°C:

125°C − 45°C
__________ = 1.6 Watt
50°C/W


With an ambient temperature of 45°C, the maximum power dissipation
of the LM317HV / TO-220 without heat sink is 1.6 Watt.

With an ambient temperature of 75°C, the maximum power dissipation
of the LM317HV / TO-220 without heat sink is 1 Watt.