What's the transformer voltage at different currents ?

[Rodo]

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

 

[Andrew Holme]

Rodo wrote:

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.

 

[Andrew Holme]

Andrew Holme wrote:

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.

 

[HKJ]

"Rodo" <dsp1024@yahoo.com> wrote in message
news:NJoUd.46918$uc.14480@trnddc08...

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 ?

You can download this program to design a simple power supply:
http://hjem.get2net.dk/hkj/miscelPowerSupply.html

 

[John Woodgate]

I read in sci.electronics.design that John Larkin <jjSNIPlarkin@highTHIS
landPLEASEtechnology.XXX> wrote (in <d4b42119kq9fj2f8c71nvl3otoqfgk67au@
4ax.com&gt about 'What's the transformer voltage at different currents
?', on Sun, 27 Feb 2005:

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 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

 

[John Fields]

On Sun, 27 Feb 2005 18:56:45 GMT, "Rodo" <dsp1024@yahoo.com> wrote:

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 ?

---
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

 

[John Fields]

On Sun, 27 Feb 2005 15:55:35 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Sun, 27 Feb 2005 18:56:45 GMT, "Rodo" <dsp1024@yahoo.com> wrote:

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 ?

---
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
+-----+

---
LOL! Should, of course, be:


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

 

[Rodo]

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]

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

--
John Fields

 

[Fred Bloggs]

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.

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 | ->

-----|~ +|----+-----|IN OUT|--/\/\----+--+--------+
| | | + | | R2 | | | +

24VAC | | [C] | GND | | | -
| | | Vc +-------+ | | |/|

-----|~ -|----+ | | | 240R|/| 24V
+-----+ | - +--------------+ | coil|/|

| - |/|
| ^ -
| / |1n4002 | -
+----------------o o--------+--------+
|
---
///



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

Vc-1v
At Vc<=~37V, IREG= ------- + I = 90mA at low line
130+240 bias

 

[Larry Brasfield]

"John Fields" <jfields@austininstruments.com> wrote in message
news:lhp6211qgcdiq18a4dhob5n7hnj1r1eqlo@4ax.com...

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.

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.

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.

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.

 

[Larry Brasfield]

"John Fields" <jfields@austininstruments.com> wrote in message
news:i16721pf365othskn0fjqe980nabj5ae00@4ax.com...

On Mon, 28 Feb 2005 13:43:13 -0800, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:
....
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.

A pretty safe couple of bets then, I think, would be that that relay
was spec'ed to a fare-thee-well, and that it wasn't all that cheap...

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.