On Tue, 2 Mar 2010 12:34:55 -0800 (PST), George Herold
wrote:

On Mar 2, 12:23*pm, John Fields wrote:
On Mon, 1 Mar 2010 13:53:13 -0800 (PST), George Herold

wrote:
I've got a tankless electric water heater from eemax. that stopped
working the other day. *eemax won't provide any component level
support, but I've ordered a new board for $45.00.

When I opened up the unit, after switching off the circuit breaker on
the 240V AC line, I observed that a big (maybe 3-5 watt) (metal film?)
resistor was discolored and was an open circuit. *The markings look
like 100 ohms, but because of the discoloration it's hard to be sure.
I say metal film because the resistor is pale blue in color. *The
resistor feds a neon indicator bulb... (And probablly more of the
circuitry.)

My question. *Is 100 ohms a good value as a current limiting resistor
for a small neon lamp running off 240 V AC? *

---
No.

For a small neon lamp like an NE-2, (GE 3AD) with a recommended current
of 0.3mA and a nominal holding voltage of about 60V, peak, that's about
42VRMS, so the resistor would have to drop about 200V at 300 A:

* * * * *Vs - Vne * * *200V
* * R = ---------- ~ ------- ~ 670k oms
* * * * * *Ine * * * *3e-4A

and it would dissipate:

* * P = IE = 3e-4A * 200V ~ 0.06 watts

If that resistor were *100 ohms, then with 200V dropped across it
there'd be 200 amperes through it _and_ the neon lamp, which they'd both
hate for a little while.

OTOH, if the resistor is 100k it'll drop about 200V with a neon lamp in
series, which will let about 2mA through when the lamp fires. *

That'll cause the resistor to dissipate about 0.4 watts, which doesn't
explain why it got toasted, since even into a short it would only
dissipate about:

* * * * *E * *240 V
* * P = --- = ------- = 0.576 watts *
* * * * *R * * 1e5R

Just for grins, let's say it's a 3 watt resistor running full bore in
series with a load (like a high current neon lamp) drawing 5mA.

Then the value of the resistor would be:

* * * * *E * *200 V
* * R = --- = ------- = 13,333 ohms ~ 13k
* * * * *P * * *3W

and if there was some circuitry downstream from the lamp, which shorted,
then the resistor would dissipate:

* * * * *E * *240 V
* * P = --- = ------- = 4.43 watts
* * * * *R * * 13e3R

Not huge but, anyway, without a schematic it's all just conjecture.
---

(60 Hz if that matters.)

I'm not sure what the I-V curve for the lamp will look like. *(Is the
one shown here OK?)

http://en.wikipedia.org/wiki/Neon_lamp

---
Only in a very general way; here's a much better one for your
application:



JF

Thanks John, I must admit it's hard to understand how a 3 watt 100k
ohm resistor across the 240 V lines would blow even if everything down
stream of it was shorted. (As your calculation shows.) A new circuit
board is in the mail, but this won't solve the mystery. The board has
been redesigned and is no longer analog but digital. Once I put the
new board in I'll look at the busted one and see if I can figure
anything out.

Say could you resend that link to the neon bulb I-V curve? What ever
you posted got turned into gobble-de-gook by google.

"


George H.


http://www.infinitefactors.org/docs/...resistance.pdf

Jon