Here is my layout you can see the 270k vishay bleeder resistor after the first cap (between the first two caps at top right). The "standby" is after the first cap per the VVR instructions from Dana.
Q1:
I added the resistor as a bleeder. What happens when the stdby is off for a while? I am wondering if the first cap gets charged then everything goes through the resistor? Can the resistor get too hot in a situation like this if the stby is left off for long periods?
On the valve wizard's site he has an article about putting the standby on the AC side of the circuit so the DC side doesn't cause arcing on the contacts and make the standby switch fail as quickly: http://www.freewebs.com/valvewizard1/standby.html
I realize that this doesn't answer your question, and I don't know why Dana instructs the standy to be on the opposite side of the filter caps. Maybe to keep the filter caps charged so the inrush current doesn't degrade or fry his circuit as quickly I.E. "Long Live the VVR" . You would have to ask Dana. But the article is an interesting read and has some good ideas. It may even bring up more questions for you.
If the bleeder is around 200k or so, then it won't be enough to load down the B+ anyhow.
Here's an exercise - Check the voltage drop across the resistor with the standby off and with it on. (V/R = I) Then you will see that there isn't much difference in current through the resistor worth worrying about.
Its 270k but I'm not sure how to determine the current without measuring which would be after it is built and turned on...which is what I was worried about. Since you suggest it I assume its okay. I'll measure it once its going anyway. It might be good to know one day.
Under no load conditions (this is your worst case i.e. voltage is maxed), with a full wave rectifier, your voltage is approx SQRT(2) times the rms voltage of your PT. If you have a 260-0-260 PT, you should get ~368VDC. So from there you can calculate current and power dissipated by your resistor.
surfsup wrote:Its 270k but I'm not sure how to determine the current without measuring which would be after it is built and turned on...which is what I was worried about. Since you suggest it I assume its okay. I'll measure it once its going anyway. It might be good to know one day.
I'll also ask dana. Thx.
You can easily guestimate the effect with a simple armchair exercise. e.g.: dropping 500V across 270k will be 500V/270,000R = .00185A or say 1.9mA. 1.9mA x 500V = 0.95W
whereas dropping 450V across 270k will be 450V/270,000R = .00166A or say say 1.7mA. 1.7mA x 450V = 0.77W
Not a lot of difference even if there was a 50V jump, which there probably won't be. Either way you'd keep the smoke inside by using a 2W resistor.
738]http://chicagocadcam.com/ChrisHahn/schem/x10_option.jpg[/img]