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 16:42:02 -0600
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
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