Is it safe to use the L78M05 to regulate 24 VAC ?

[Voidugu]

I used a bridge rectifier with four 1N4007 diodes to rectify 24VAC. I want to use the L78M05 regulator along with some capacitors before and after the regulator to get a 5VDC out. Is it safe to use the regulator for such an application? (in its datasheet it states clearly that the absolute maximum rating for its DC input voltage is 35 volts)
24 VAC RMS = 34 VAC peak to peak (24/0.707 = 33.9)
34 - 1.4V (forward voltage drop of the diodes) = 32.6V
32.6V is very close to 35V
in addition a non fully loaded transformer rated for 24VAC tends to have a higher voltage.

Thank you very much people
Regards
Void

[Skimask]

It's not the total voltage that'll cause it to fail.
It'll be the heat dissipated that'll put the regulator into thermal overload shutdown once you try to draw any appreciable amount of current.
RTFDS

[Simon]

I may not be a good idea to go over 30V if you need to guarantee reliability. As skimask said you can end up dissipating 50W, it would be best practive to use a SMPS or a SMPS pre regulator. Another way around i use often is another reg, say 18V or a series resistor so that the higher the current taken the more voltage falls across the resistor. If you have a fixed current draw you can throw a zenner in as a pre regulator.

[mariush]

It would work, but probably for a maximum of 20-50mA at best.

32v - 5v = 27v x 0.02a = 0.54w  ...

Most l78m05 have 100C/w thermal resistance junction-ambient and can work up to 150c so with 20-50mA or the chip will work but it will be very hot.

[digsys]

Do you have to run it from that high a DC_in? And waht is your max Current, that's important.
If it's low and you can't reduce the DC_in, you have 2 choices. Wash off several volts with either a power resistor or say 5W Zeners.
Otherwise, use the LM317HV and 2 set resistors. These can go up to 60V In-Out. When they say 30 / 35V Max, it often refers to
V_in - V_out BUT you HAVE to consider a short circuit situation, where V_Out is 0V !!!! In that instance, you'll POP the regulator !!
Irrespective of load at the time !!

[Voidugu]

If i use a big enough heat sink would the regulator be fine handling currents up to a maximum of 100mA?

[mariush]

I wouldn't risk it with a L78m05 ..  i'd use a TO-220 regular 7805 in worst case scenario with a heatsink on it, it's more suitable for heatsinks.

but if board space isn't an issue, i'd probably get a mc34063 or something equally cheaper and get 5v 100mA+  or maybe get 8v out then use a 7805 for very stable output.
Though a mc34063 and a pi filter on the output should be good enough for 5v output. 

[madires]

If i use a big enough heat sink would the regulator be fine handling currents up to a maximum of 100mA?

If you like to use linear regulators (for whatever reason :-) you could cascade two. First one could be 15V and the second one 5V.

[beaker353]

I had pretty much the exact same problem with a open source hardware project I am working on, "HydroGadget".  Basically, it's a home irrigation timer system built around a RaspberryPi.  Same issue, a fixed voltage supply of 24Vac'ish and 5Vdc needed for the Pi.  Ended up using a linear reg to drop just enough voltage for a prepackaged SMPS to drop the rest of the way to 5Vdc.  It's been cooking in my garage for about two months without issues, even when we sustained a week with highs above 115F.  I have full details on my blog http://smokingcircuits.me

- EM

[SeanB]

I have met many 7805 regulators that are very unhappy with a 35V input, and you will either have to use a second regulator ( a 7812 or 7815) as a preregulator, or use an external 1 transistor and zener diode to drop the excess voltage. There are certain versions that are actually specced to 40V input, and they will do around 100mA at 40V differential, but they are not common. The regular run of the mill ones are best if used with an input between 8 to 12V, below that they run out of regulation and above they tend to shut down, especially if the die has for some reason not been fully bonded to the internal thermal pad ( pretty common on cheaper ones) or the tab is making poor contact to the heatsink.  Just because it has thermal shutdown does not mean it will do so gracefully or at all in some cases, this is not tested during production at all, only on a sample basis if they are being lot screened for a high reliability application. Some oscillate at shutdown, others are just very noisy. I had some that blinked the load at 0.2Hz, only after running for an hour or more.

[Voidugu]

That's a very nice blog you got there beaker353. I tested the regulator for about half an hour with a heatsink. The whole thing (heatsink + regulator) got warm but not too hot. I mean i was able to maintain skin contact without getting burned.

[Skimask]

How about telling us what you plan on powering with this set up?

Then try it...

Then go out and buy more parts...

[Krzysztof Szymiczek]

Use a pair - first go to lets say 12V, then to 5V, how much current do you need from this 5V supply? If it's not much and for your use, it's probably OK with a pair of regulators, otherwise think about the lost energy transformed into heat - neither elegant not professiolan at all...

[Voidugu]

The regulator has to power a 100 ohm resistor connected in series with a 50 ohms rtd making voltage divider no.1 (the resistance of the RTD varies from 50 to 150 ohms depending one its temperature). It also has to power a 10 k pot making voltage divider no.2. It must also power a comparator (KA393) in which dividers no.1 and no.2 are connected. The output of the comparator is connected via a 360 ohm resistor to a MOC3023 and then from the MOC's internal led to the regulators 5volts. The MOC is used to drive a triac (TIC236) which in turn drives a 24 volt 60 watt soldering iron. The rtd is embedded into the ceramic heater of the iron for closed loop control.

The main power hungry component here is the 100 ohm resistor in series with the RTD. I could raise the voltage of the whole system to say 12 volts but more power will be wasted in the regulator and the rtd resistor combo. 

[megajocke]

in addition a non fully loaded transformer rated for 24VAC tends to have a higher voltage.

You also have mains voltage variation to consider I guess. I don't know where you are, but high mains voltage might be something like +5% or even +10% above the nominal value. Together with your transformer low-load voltage rise you might be out of spec.

[ElektroQuark]

Try a half wave rectifier, it can help too.

[Voidugu]

Yes there is. I am trying to make the unit as compact as possible. Using two transformers takes up more space.

I guess i figured the solution out. After the 24VAC becomes DC by the diodes it has got a peak value of around 32.5 volts. I guess the solution would be to use the appropriate capacitor to smooth it out.

Say with the 5 volts regulator the circuit draws around 60mA. A 9 volt smoothed DC waveform would be pretty much okay to be regulated down to 5 volts. Using the equation C = I/2FV (C = capacitance in Farads, I = current in Amperes, F = AC frequency in Hertz, V = voltage ripple in Volts (ie peak voltage - smoothed DC voltage , 32.5 V - 9V = 23.5V ripple)

C = 0.060A/(2 * 50Hz * 23.5V) = 2.55 x 10^-5 Farads = 25.5uF filter capacitor

This should solve the problem of the regulator's overheating issue.

What do you experts think? (As you must have guessed i am still a beginner/noob in electronics)

[mariush]

You could use a 24v transformer with a center tap.

Then use the full two windings for 24v ac, and take 12v ac between the center tap and one of the other wires, a single shottky diode for half wave rectification and a capacitor would get you a peak of about 8v (0.71 x 12) 

Then you can use a voltage regulator with lower drop like a lm1117/nxp1117/ncp1117/ap1117  (1.0-1.2v drop) or LP2950 with 0.38v drop  or LM2931 with 0.3-0.6v @ 100mA so do the math for at least 6v on the regulator from 8-8.5v ...  basically 0.1 A /(2x50x2v) = 470uF+

A standard 470-820uF 25v capacitor should be enough to keep the voltage above 6v with 100+mA load.

[Voidugu]

The thing is i already bought an 80 watt 24 volt single tap transformer

[TerminalJack505]

Yes there is. I am trying to make the unit as compact as possible. Using two transformers takes up more space.

I guess i figured the solution out. After the 24VAC becomes DC by the diodes it has got a peak value of around 32.5 volts. I guess the solution would be to use the appropriate capacitor to smooth it out.

Say with the 5 volts regulator the circuit draws around 60mA. A 9 volt smoothed DC waveform would be pretty much okay to be regulated down to 5 volts. Using the equation C = I/2FV (C = capacitance in Farads, I = current in Amperes, F = AC frequency in Hertz, V = voltage ripple in Volts (ie peak voltage - smoothed DC voltage , 32.5 V - 9V = 23.5V ripple)

C = 0.060A/(2 * 50Hz * 23.5V) = 2.55 x 10^-5 Farads = 25.5uF filter capacitor

This should solve the problem of the regulator's overheating issue.

What do you experts think? (As you must have guessed i am still a beginner/noob in electronics)

I like some of the other solutions better.  A dropper resistor is probably the simplest.  You could use a 220R/3W+ resistor and have it drop about 22V @ 100mA.

Page 7 of this datasheet explains.

[SeanB]

Run the comparator from the 12V rail from the first regulator. That will ease the power requirements and as well the comparator will be better able to operate at the ends of the input range. Some even just run the comparator off the unregulated rail ( provided it is below 36V) and leave it at that.

[Paul Price]

You can use the simple circuit to preregulate the up to 36P-P AC voltage into the diode on the left to a rectified <12V on the output of the TIP31(or any other NPN medium power xistor) feeding the 78M05.  You can adj R1 1K resistor for any voltage desired >7.5 V out to feed the 78M05 reg input. The NPN at the bottom of the circuit could be a 2N3904 or any approx. equiv.

Since the TIP31 conducts only when the AC voltage in is < approx 10V, or approx 30% of an half wave duty cycle, then the total on-time duty cycle is approx 15% so the power dissipated by this pre-reg circuit is quite low.

[slburris]

Why not just use a 78XX replacement module like the
OKI-78SR-5/1.5-W36-C?  $4.34 qty 1 at Mouser:

http://www.mouser.com/ProductDetail/Murata-Power-Solutions/OKI-78SR-5-15-W36-C/?qs=sGAEpiMZZMtwaiKVUtQsNemMZL4TplJBqOl7845nWHA%3d

Up to 36V in, supplies up to 1.5amps.  It's the same form factor as a 78XX regulator.
And it's a switching module, so no heatsink and it won't matter as long as the input
voltage is between 7 and 36v.

Except for stuff which requires really quiet power, I've stopped fussing with
power supply specifics and use switchers now.


Scott

[C]

The regulator has to power a 100 ohm resistor connected in series with a 50 ohms rtd making voltage divider no.1 (the resistance of the RTD varies from 50 to 150 ohms depending one its temperature). It also has to power a 10 k pot making voltage divider no.2. It must also power a comparator (KA393) in which dividers no.1 and no.2 are connected. The output of the comparator is connected via a 360 ohm resistor to a MOC3023 and then from the MOC's internal led to the regulators 5volts. The MOC is used to drive a triac (TIC236) which in turn drives a 24 volt 60 watt soldering iron. The rtd is embedded into the ceramic heater of the iron for closed loop control.

The main power hungry component here is the 100 ohm resistor in series with the RTD. I could raise the voltage of the whole system to say 12 volts but more power will be wasted in the regulator and the rtd resistor combo.

You might want to rethink your circuit
The MOC3023's Led needs a constant current. When using a voltage regulator to get this, a change in the voltage regulator voltage is just a resistor value change.
The KA393 has a Vcc limit of 30V Vcc
If you put the RTD and the pot in a Wheatstone bridge http://en.wikipedia.org/wiki/Wheatstone_bridge
you can have this part of circuit not voltage source sensitive and also lower the voltage across the RTD to save power.


In image RX & R3 are the RTD & 100 ohm resistor
The pot is R2
Note that you state that the RTD = 50 to 150
If you put a 5K resistor in series with the pot and tie the wiper to one end of pot, you have a scaled difference in resistance the same as the RTD
Using the same scaling you select R1 to be the scaled version of the 100 ohm
When the RTD's resistance (which changes with temp) = the pot's  side of bridge, Points B & C will be equal( the inputs to the  KA393)
Adding a series resistor to point A would allow you to lower the voltage to the bridge.
Your voltage regulator would only have to handle the current changes not the total current with the remaining current supplied with a resistor.
You can now use any voltage regulator less than 30V that can hold regulation over the min current to max current change the circuit creates( RTD change + Pot change + led change + KA383 change.

C   

[Zero999]

I like some of the other solutions better.  A dropper resistor is probably the simplest.  You could use a 220R/3W+ resistor and have it drop about 22V @ 100mA.

Page 7 of this datasheet explains.

You mean like this?

220R will probably drop too much voltage when you take the ripple into account but it's a good idea and will work, try 100R