Subject: Re: 24 VAC to 9VDC supply
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NNTP-Posting-Date: Mon, 09 Dec 2002 21:24:05 GMT
Organization: Insight Broadband
Date: Mon, 09 Dec 2002 21:24:05 GMT
On 4 Dec 2002 15:29:08 -0600, Mark Lloyd wrote in
>On 4 Dec 2002 02:30:38 -0000, Steve W.EE
>>What is the quickest, dirtiest, cheapest way to get
>>9 volts DC from a 24VAC source.
>You won't need a transformer, but can connect the 24VAC directly to a
>> Need about 150ma
>>of current but the DC does not need to be regulated.
>>Half-wave rectifying will give too much DC voltage,
>>it would burn up a regulator since I don't have a
>>place to heatsink one.
>I doubt it. The power dissipated in the regulator is:
>Where Vi is the input voltage, Vo is the output voltage, and I is
>current. In your case.
>(Vi-Vo)*I = (24V-5V)*.15A = 19V*.15A = 2.85W
For a full wave rectifier + capacitor, the voltage drop for the 5 vdc case
Mark Lloyd analyses is:
(24VAC x 1.4VDC/VAC)-(2 x 0.65VDC) - 5VDC = 27.3VDC not "19V" as Mark Lloyd
Furthermore, 27.3 VDC * 0.15A = 4.1 watts which is not "2.85W" as Mark Lloyd
>Less than 3 watts isn't that much. You could even use one of the
>low-power regulators (78L05). These handle up to 150mA, are in TO-92
>cases, and don't use heatsinks.
The National 78L05 is rated for 100ma, not 150max.
Furthermore, the power dissipation is limited to between 1 and 0.5 watt
depending on the length of the leads and ambient temperature. See
http://www.national.com/ds/LM/LM78L05.pdf so "3 watts" really is *way* too
Furthermore, why use a TO-92 package when T0-220 and TO3 devices are readily
What Mark Lloyd recommends is _guaranteed_ to shut-down or self destruct and
is incorrectly analyzed, poorly informed, bad advice.