Errr, make that that "silicon diodes", I keep gettin' 'em confused with the
girls! grin

Warmest regards,
John

"John Smith" wrote in message
news | 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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