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<font face="David">I seconds that.<br>
without this resistor the voltage at low currents will go sky high
(~20V)<br>
the 12V is usually close to the maximum current.<br>
<br>
Alon.<br>
</font><br>
On 7/14/2010 11:46 AM, Erez D wrote:
<blockquote
cite="mid:AANLkTim93PXfbPMcBEbJlvt3iR35csylOkciHp2qGM-C@mail.gmail.com"
type="cite">
<div dir="ltr">The transformer gives 12V DC RMS, The peak voltage
however is higher (but with lower current)<br>
<br>
I guess the reason for the resistor is to reduce the peak voltage and
keep it closer to 12V when there is little or no load, or when
connecting the load.<br>
<br>
<br>
cheers,<br>
erez.<br>
<br>
<div class="gmail_quote">2010/7/14 Shachar Shemesh <span dir="ltr"><<a
moz-do-not-send="true" href="mailto:shachar@shemesh.biz">shachar@shemesh.biz</a>></span><br>
<blockquote class="gmail_quote"
style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
<div dir="ltr" bgcolor="#ffffff" text="#000000">
<div class="im">geoffrey mendelson wrote:
<blockquote type="cite"><br>
<br>
I just took a DC adaptor and removed the components that coverted the
AC to DC. I did that for some modems that needed AC but came without
adaptors. <br>
<br>
<br>
</blockquote>
</div>
That's what I ended up doing. Now I'm having second thoughts whether I
did that correctly.<br>
<br>
The DC adapter had the voltage selector, connected to a diode bridge,
connected to a capacitor and a resistor in parallel to the connection
to the polarity selector. I understand what the diode bridge and
capacitor were doing. The diode bridge was keeping the positive end of
the AC sine wave, and the capacitor was leveling off the signal. Making
it more like a DC line. What I'm not sure is what the resistor was
doing.<br>
<br>
This was a 460Ohm resistor, which at 12V would mean about 0.4 Watt.
Seems like a lot of wasted energy for a 12Watt power supply.<br>
<br>
I checked the output signal after removing all of the above (resistor
included), and a volt meter and an oscilloscope both shows exactly
what I was hoping to see - an AC line giving the correct voltage. The
only problem is that I'm not sure whether it was correct to remove the
resistor.
<div class="im"><br>
<br>
Thanks,<br>
Shachar<br>
<br>
<pre cols="72">--
Shachar Shemesh
Lingnu Open Source Consulting Ltd.
<a moz-do-not-send="true" href="http://www.lingnu.com" target="_blank">http://www.lingnu.com</a>
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