Re: What's the transformer voltage at different currents ?
From: John Fields (jfields_at_austininstruments.com)
Date: 02/27/05
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Date: Sun, 27 Feb 2005 15:55:35 -0600
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
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