Power supply confusion

[stingray_65]

OK i got this old chassis and pulled the iron and tubes.

The power transformer reads 353-0-353 volts 5.2 volts and 6.6volts when connected to 120VAC (no load)

So out of boredom I hooked up an RCA 5Y3GT and my voltage dropped to 300VDC.

on the bench was an 8uf 350V cap so I put that across the leads like in the attachment

When i put my meter on it it raised to 407VDC. i shut it down and discharged the cap and removed it. back to 300 VDC.

I was afraid the cap was under rated and scrounged up an old 40-40uf 450v can capacitor. When i installed it the output is up to 470VDC.

Whats going on here?

[Old Tele man]

...without a "load" the cap was charging to the voltage 'peaks' instead of the commonly "loaded" levels of 0.638 (Vavg) or 0.707 (Vrms).

...hang a "reasonable" load across that cap (50-100mA) and you'll see the DC-voltage reading go down.

[loudthud]

Without any capacitor, your meter just reads the average of the pulsing DC coming out of the 5Y3 which is .636 times the peak voltage. When you install a cap, it charges up to the peak with no load and your meter reads that. This is normal.

[stingray_65]

OKEY DOKEY

so being braver than I'm smart and not knowing what would suffice for a 1 or 2 milliamp load, i started off with a 1 meg 1/2w resistor and a small fuse and saw a 2v drop. took that out and with the same fuse clipped in i put in a 470K 1/2 watt and saw a bigger drop.

ok so now theres this big honkin 220ohm 5W wire wound resistor and clip that in with the fuse.

it got hot REAL HOT like it was gonna fail soon and the voltage dropped to zero.

so being brave at this point cause the only thing getting hot was the resistor and not the wires i put my amp clamp on the HVtap and got 500 milliamps
and 3.9A on the mains.

I guess that answered my question of how many milliamps the tranny could deliver.

well i dismantled everything before i got too stupid with the stuff, after all i want to use the tranny for my budget build.

thanks for the input guys! i had fun messin round.

[Enzo]

Please learn Ohm's Law. It is simple, or just write it on a card and tape it over your work area.

Volts = amps x ohms

You can move that around to its equivalents

Amps = volts / ohms


Ohms = volts / amps


When I learned this we used E for voltage, but I think they have moved over to V these days, but amps is still I and ohms is R. SO where I learned E = IR, today you learn V = IR, and I = V/R, and R = V/I


Whenever you have two of those amounts, you can figure the third one. This is fundamental electronics, nothing is more basic and critical than Ohm's Law. If you have 400v and put 1000 ohms across it, then

I = V/R = 400/1000 = .4Amp = 400ma.

Of you have 400v and want a 100ma current draw, then you need a resistor load.

R = V/I = 400V/.1A = 4000 ohms

REmember, Ohm's Law units are always full amp, full volts, and full ohms, so keep in mind that 100ma is really .1amp. You COULD calculate in millivolts or milliamps as long as you keep track of that 1000 factor.


And before you scream I HATE MATH at me, take a look. This ain't math, this is just arithmetic, and dollar-store pocket calculator will do it if your head can;t.

[stingray_65]

Enzo!
I'm always glad to hear from you!

I was hoping some one with patience would bring this up!

now I'm NOT being a smart a$$ when i throw some questions to you but i've understood the ohms law on paper but have a real hard time with the practical aplication.

ok that said, lets look at this schematic again.

so I would take a reading from TP1 to ground to get my voltage

then TP1 to TP2 to get my resistance

i plug that into the formula V/R gives me the amperage of what?

OH DAMNIT! i get it! i was looking to put a specific draw on the circuit,
so i could have went:

470V (measured at TP1 to ground) and then guestimate a resistance of say 4.7K

plug that in 470V/4700ohms=.1 A=100mA

Then knowing a SANE load put the 4K7 resistor in and see my voltage drop that Tele Man said would happen.

I think i might put it back together again and play with it some more tommorow

[Albert Kreuzer]

And remember that Watts = Volts x Amperes, so with 470V and .1A your resistor has to dissipate 47 Watts. A 5W resistor will get really, really hot

Cheers,
Albert

[tubeswell]

If you don't want to fry the resistor too fast you could try using 5 x 10W 27k resistors in parallel. (That will give you 5k4 50W)