Yes. I frequently find I overtax a components max. ratings (the smell of
burnt electronics is an acquired taste, like tobacco smoke and beer
grin)... I tend to pick up large quantities at auctions and surplus
outlets... replacing "sane design methods" with brute force and the
wholesale slaughter of un-counted components...



And, one should use caution with my labeling of components--I have boxes of
silicone diodes which are "similar" to 1n914's--and pounds of mis-valued
resistors/caps (the manufacturers just don't realize the value of a resistor
marked as one-ohm--when the actual value is one meg-ohm!!!)



"Whatever works!" sometimes becomes, "A list of what doesn't work!"



Quite often, the answer lies in "what I haven't tried..."



I don't expect all to be able to appreciate my "methods"... but, it breaks
the monotony of having to be so precise--which my field places upon me...



Warmest regards,

John



"Tam/WB2TT" wrote in message
...
|
| "Cecil Moore" wrote in message
| ...
| John Smith wrote:
| Well, If there is a voltage, there is a current (albeit, at times very
| small)--the opposite is also true, ohms law is standing proof...
|
| Yes, but the amplitude and phase relationship of current to
| voltage can have any possible value and there are an infinite
| number of possibilities. In the equation, Z = V/I, you cannot
| determine Z unless you know BOTH V and I.
|
| Why can't I measure the standing wave ratio as a ratio of power, ...
|
| You need both voltage and current to determine power. Your design
| senses only current.
| --
| 73, Cecil http://www.qsl.net/w5dxp
|
|
| Cecil is right. Also, the 100K implies 200V to get 2ma of LED current. The
| 1N914 won't hack that.
|
| Tam/WB2TT
|
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