Quick question about virtual heater CT

[mort]

Is 1W plenty strong enough for a pair of 100R resistors as a virtual CT on a 5e3??

On hand all I have are 1W & 10W... don't really want to put in the 10W

[eschertron]

Mort,

P=IE or P=E squared/R
so E (max voltage allowed across a 1W 100R resistor = SQRT(1W x 100R) = 10V
each leg is responsible for 3.15v
so 1W is good. 1/2W would *probably* be OK too (~7.07v max allowed voltage).
at 1/4W you'd cross the 'no-no' design threshold of being less than 200% protected (5v rating/ 3.15v operating)

probably more than you wanted but the answer is yes

[Enzo]

Ever see 10 watters for that in a Fender? I haven't.

The fact they are grounded is irrelevant. What you have is two 100 ohm resistors in series parallel to the 6v winding.

I = 6.3v/200ohm = 32ma

P = 6.3v x 0.032A = 0.2W

That is for both resistors, so each one dissipates 0.1W.

[mort]

Just for the time being I tagged them onto the power tube cathode.. does this change the equation?

[eschertron]

No difference.

they are still 100R and can dissipate the same amount of power. If the voltage from leg to leg changed (on the heater windings), then there'd be an issue.

[mort]

thank you gents!

[The Dude]

That series resistor combination is also in parallel with the heaters, so current through the resistors is negligible.