What's the transformer voltage at different currents ?

Signal's transformers tend to be "soft", high copper resistance, which they make up for by using high-temperature insulation; that tradeoff saves copper and iron. So at light loads you'll see way more DC than

1.414 * rated RMS voltage. Unloaded output has to be measured; there no way to predict regulation of a purchased transformer (except, maybe, calculating watts per pound as a crude guideline.)

You could run the relay coils off rectified, unfiltered DC. The relay will then see an effective voltage of 0.63 times peak, close to 24.

And yes, 1.3 watts will get a TO-220 pretty warm.

John

A sine wave peaks at 1.414 times the RMS (what you read on an AC volt meter) voltage. Small transformers don't have very good regulation, so that may produce 10% or more excess voltage unloaded, compared to running at rated current.

You are correct that 1.3 watts is too much for a TO220 without a heat sink. Though you could move the heat elsewhere by adding a resistor between the rectifiers and the filter capacitor (which you haven't mentioned).

However, most DC relays will run pretty well on rectified but unfiltered AC, so you might try this with yours and see if it gets too hot. you can lower the voltage to the relay by adding a series resistor, instead of a regulator, since the load is constant.

--
John Popelish

--- To answer your last question first, you have to know how well the transformer can regulate. For small transformers of the type you're using it usually runs about 30%, which means that when the transformer isn't loaded its output voltage will rise to about 1.3 times what it is when it's fully loaded. It varies from manufactrurer to manyfacturer though, so you'd need to check with whoever your vendor is. Transformer output voltages are usually specified as "RMS" volts, which is the voltage out of the transformer which would be required to heat a resistor to the same temperature as DC would heat the resistor.

That is, if 24VDC heated a resistor to 100°C, then 24VRMS would also heat that resistor to 100°C. However, since the voltage on the output of the transformer _isn't_ DC and varies by going from some positive peak voltage through zero volts and then to a negative peak voltage equal in amplitude (but opposite in polarity) to the positive peak and then through zero volts and then... forever, the peaks have to be higher than the RMS value in order to compensate for the times the voltage is below 25V.

The relationship between peak and RMS is:

Vpk = VRMS * SQRT2

Now, what happens in a power supply like you have:

+-----+ +-------+ MAINS>-----|~ +|----+---|IN OUT|--->VREG | | | +---+---+ | | [C] | | | | | MAINS>-----|~ -|----+-------+------->GND +-----+

Is that the peak AC coming out of the transformer is:

Vpk = 24VRMS * 1.414 ~ 34V

That voltage goes through the rectifier where it encounters two diode drops (~1.4V) if it's a full-wave bridge, so the capacitor sees peaks of:

Vin = 34V - 1.4V = 32.6V

every 8.3ms, (for 60Hz mains) and that's what it tries to charge up to.

Worse yet, since the transformer is rated for 800mA, with a light load on it, like 100mA, its output voltage will rise to significantly more than 24VRMS, and that's why you're sitting with 37V on the input of the regulator.

Getting back to the regulation thing, if your transformer is rated for

24V @ 800mA, and you're getting 37V across the cap with a 100mA load, then adding in the diode drops and working backwards gives us:

Vout + 1.4V 37V + 1.4V VRMS = ------------- = ------------ ~ 27VRMS 1.414 1.414

So if the transformer is rated at 24VRMS out, loaded, and you're getting 27VRMS out of it with a light load on it, that looks like about a 3 volt increase over 24V, which is about 12.5% regulation. Not bad.

Since there's really no need to electronically regulate the relay voltage, my suggestion, at this point, would be for you to get a transformer which would supply the right voltage to start off with and dump the regulator.

Working backwards, you know that the relay is going to need 24V at

100mA, and that if you use a full-wave bridge you're going to need another ~1.4V on top of that, so the transformer needs to put out

Epk = 24V + 1.4V = 25.4V

at 100mA. Except that since, for a capacitive input filter, the transformer has to both charge the cap and drive the relay at the same time during part of the AC cycle, it would be a good idea to get a transformer rated for 200mA.

Still, what you need is 25.4VPK, which is

VPK 25.4 VRMS = -------- = ------- ~ 18VRMS SQRT(2) 1.414

For a first cut. Now, keeping in mind that US mains voltages can vary between 108V and

132V and that the relay's must-operate voltage is probably 75% of 24V, or 18V, and it becomes clear that the transformer you need needs to be able to put 18VDC into the relay with low mains at whatever current the relay needs with 18V across it, plus whatever current the cap needs to be charged during the voltage peaks. The relay looks like

E 24V R = --- = ------ = 240 ohms I 0.1A

So at 18VDC, (mains at 108V) it'll draw

E 18V I = --- = ------ = 0.075A R 240R

and at 22VDC (mains at 132V) it'll draw:

22.0V I = ------- ~ 0.092A 240R

Since, with low mains in, the transformer needs to put 18VDC into the relay, it needs to be rated at 10% higher than that to realize 19.8V with nominal mains in.

Since that 19.8V is after two diode drops, and is DC, that comes out to be:

Vout + 1.4V 19.8 + 1.4V VRMS = ------------- = ------------ ~ 15VRMS 1.414 1.414

With high mains on the transformer the relay will draw 92mA, so if we double that to 182mA in order to keep the filter cap charged up and look for a transformer that puts out 15VRMS at 182mA, Signal has two pretty good choices, the ST-4-16 with secondaries in series and the ST-4-36 with secondaries in parallel. Of the two, I think I'd go with the ST-4-16, but I'd have to run the numbers again to be sure...

Then there's the question of selecting the proper filter capacitor. Do you need help with that?

-- John Fields

--
LOL! Should, of course, be:


         +-------+   +-----+        +-------+
MAINS>---|PRI SEC|---|~   +|----+---|IN  OUT|--->VREG
         |       |   |     |    |   +---+---+
         |       |   |     |   [C]      |
         |       |   |     |    |       |
MAINS>---|PRI SEC|---|~   -|----+-------+------->GND
         +-------+   +-----+

Hi all,

I bought a transformer (Signal Transformer model st-6-48, 24VAC@0.8A, parallel connection). I needed 24vdc to drive the coil of a relay. I thought

24vac after rectification and filter it should go to 30vdc so a 24vdc (7824) regulator should be right. Well... the voltage at the input of the regulator is 37vdc. It gets so hot you could fry an egg on it. I'm only drawing like 100mA. So ... I think (37-24)/0.100=1.3w is too much for a to-220 with no heat sink ? and the more important question I have is how do you find the DC voltage after rectification and filter when you are not drawing the full 0.8 amps ?

Thanks

Multiplying by the square root of 2 will convert RMS to peak.

You don't need a 78xx regulator just for a relay - the coil voltage isn't that critical.

You might be able to salvage the situation by forgetting about the 7824 and connecting a dropper resistor in series with the relay across the 37V supply.

That would, of course, have to be a high wattage resistor.

I read in sci.electronics.design that Larry Brasfield wrote (in ) about 'What's the transformer voltage at different currents ?', on Sun,

27 Feb 2005:

The 24 V is almost certainly at full load, so under light load the r.m.s voltage will be up to 10% high. So multiply by sqrt(2) and then by 1.05 to 1.1.

After all, 24 x sqrt(2) = 34 V, not 37 V. and that 37 V should be increased to 38.4 V (not 39 V, I suggest) to allow for diode voltage drops at low load In this case, the 24 V has risen by 13% on low load, which is a bit unusual. Maybe the mains voltage was high when 37 V was measured.

2/3 of a half cycle = 60 degrees conduction angle. You get this with quite a lot of resistance in the transformer. I suggest a quarter cycle, especially at low load.

--
Regards, John Woodgate, OOO - Own Opinions Only. 
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

You take good measurements because ass-uming 10% transformer regulation,

120VAC line, and full-wave bridge w/less than inordinately large C, puts you at 24 x 1.1 x 120/115 x 1.414 -2=37VDC. At low line you can expect (37+2) x 105/120-2= 32VDC, and at high line 40VDC. If you went with the 18VAC transformer at same VA, you would have 30VDC at high line and 24VDC at low line and median of 26VDC for respective dissipations of 0.6W, .15W, and 0.2W. You let the regulator dropout at low line ( an infrequent brownout condition), by 1- 1.5V, ending up with 22.5-23V across the relay which is 94% rating and of no effect- unless possibly some other electronics requires regulated 24VDC. This is still an enormous waste of transformer.

I read in sci.electronics.design that John Larkin wrote (in ) about 'What's the transformer voltage at different currents ?', on Sun, 27 Feb 2005:

You can measure the DC resistances and calculate (approximately) the currents. You can't easily calculate the magnetizing current, but for a

20 W transformer it may be negligible. However, if Signal's designs are pushing the core towards saturation, the mag current will perhaps not be negligible. Since it's almost in quadrature with the load current, however, it doesn't make much difference with a significant load on the transformer.

RMS secondary current is approximately 1.8 times the DC load current for a bridge rectifier.

Primary current is smaller by a factor equal to the turns ratio.

So, 0.8 A DC implies 1.44 A r.m.s. secondary current. If the secondary resistance is 1 ohm, the voltage drop due to it is 1.44 V. The primary

*ought* to be designed to give the same voltage drop measured at the secondary, i.e. another 1.44 V.

At the primary side (assuming 120 V mains), the current is 1.44 x 24/120 = 0.29 A. If the primary resistance is 25 ohms, the voltage drop at the primary side is 25 x 0.29 = 7.2 V and thus referred to the secondary side it is indeed 1.44 V.

So the *off-load* r.m.s. secondary voltage, with 120 V input, should be

24 + 2.9 = 26.9 V, and the DC output on no load is about 26.9 x sqrt(2)

- 1.0 (2 diode drops at very low current) = 37 V!

It all ties up quite nicely if you know how to fiddle the figures. (;-)

--
Regards, John Woodgate, OOO - Own Opinions Only. 
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

thought

(7824)

regulator

DC

0.8

You can download this program to design a simple power supply:

Sure, why stop now :-) ?

BTW thanks for the previous post. It was very informative. I guess I missed the RMS rating on the transformer output.

Again, thanks.

[snip]

If the relay is the only load on this supply then you can reconfigure to a low dissipation ckt w/7812 and drive it like so- multiple relays would use multiple 7812's, 1N4002's, 130R, and 22uF boost caps, and make C=470uF/100mA loading:

View in a fixed-width font such as Courier.

22u 25WVDC +--------||--------------+ | + | 1.5A | 12V | 250PIV | 7812 + - | IREG +-----+ | +-------+ 130 2W | ->

MAX dissipation 7812 = (40-36)V*0.1A=0.4W

Vc-1v At Vc

The spec relevant for determining a minimum voltage for the cap is the maximum "may release" voltage. The relay could release at a somewhat higher voltage than is "must release" spec.

True enough, but there are issues with movement that does not constitute release. Contact wear and audible buzzing are some side effects to beware of if this shortcut is taken.

I have no nits to pick from the rest of your post.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

I suppose this can be extended to a variety of relay types: View in a fixed-width font such as Courier.

22u 25WVDC +--------||--------------+ | + | 1.5A | 12V | 100mA 250PIV | 7812 + - | IREG +-----+ | +-------+ 130 2W | -> +-|IN OUT|--/\/\----+-/\/\-+--------+ | | | | | | + | | GND | | | - | +-------+ | | |/| | | | | 240R|/| 12V | +--------------+ | coil|/| | - |/| | ^ - | / |1n4002 | - | +-o o--------+--------+ | | | --- | /// | | | 10u 25WVDC +--------||----------+ | + | | 12V | 25mA | 7812 + - | IREG | +-------+ 510 0.5W| 680 1W -> +-|IN OUT|--/\/\----+-/\/\-+--------+ | | | | | + | GND | | | - +-------+ | | |/| | | | 200R|/| 5V +--------------+ | coil|/| - |/| ^ - / |1n4002 | - +-o o--------+--------+ | --- ///

--- You're welcome.

With no regulator your circuit would look like this, assuming you're using some kind of switch to turn the relay on and off:

+-------+ +-----+ ---- MAINS>---|PRI SEC|---|~ +|----+--O O---+-------+ | | | | | |K | | | | | [C] [1N4001] [COIL] | | | | | | | MAINS>---|PRI SEC|---|~ -|----+---------+-------+ +-------+ +-----+

Also, assuming your relay has a "must release" spec of 25% of its nominal 24V rating means that it _must_ drop out at 6V. Therefore, the cap's job is to make sure that the voltage across the relay coil never falls below 6V.

In truth, however, you may not even need the cap since the inertia of the relay armature and the time it takes for the magnetic field around the coil to decay to the point where the armature is allowed to drop out may be greater than the time the rectifier's output dips below 6V. You can get an idea of how long it takes the relay to de-energize by looking at the 'release time' spec on the relay data sheet.

But, assuming that you _will_ need a cap, we can determine the minimum capacitance you'll need from:

I dt C = ------ dV

Where C is the required capacitance I is the relay coil current dt is the period of the rectified waveform, and dV is the allowable ripple

Looking at low line, the output of the rectifier will be

Vin(lo) * VPkout(nom) Vout = ---------------------- - (1.4V) Vin(nom)

108 * 24 = ---------- - (1.4V) ~21V 115

So if the relay's dropout voltage is 6V, then the acceptable ripple will be

dV = Vout(min) - Vdroput = 21V - 6V = 15V

and we can write:

I dt 88mA * 8.3ms C = ------ = -------------- ~ 4.9E-5F ~ 50µF dV 15V

Electrolytics generally come with a negative tolerance of 20%, so to make sure you get that 50µF you'd need to get a 60µF cap.

Now you need to make sure that the capacitor can handle the ripple current.

To find out how to do that go to:

and find "Calculation Examples"

-- John Fields

....

No. However, with consideration of the magnetic circuit gap(s) before and after the relay operates, together with the specified minimum required operating voltage and the spring geometry, it would be reasonable to estimate a conservative number for "may release" voltage. (I say "gap(s)" because the construction may contain gaps, intended or not, other than the one which moves the armature.)

I have seen a relay application where human safety/health was at stake and the relay was held operated with half the operating voltage in order to save battery power.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

....

The solenoid that determined drop-out was relatively cheap. The relay itself was custom designed, built, and tested in-house. I can attest that the drop-out was tested for each unit. I vaguely remember a 6 V spec for a nominally 24 V part.

My point was only that relying on drop-out behavior is reasonable and that the voltage at which it happens is comfortably below the minimum operating voltage, as we all expect who have played with magnets.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

--
Heh... Good point, but I've never seen that number specified
_anywhere_ .  Have you?