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Hoffman Amps Forum image Author Topic: Squeezing the most out of two PTs  (Read 2180 times)

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Offline TIMBO

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Squeezing the most out of two PTs
« on: January 07, 2013, 04:20:23 am »
Hi guy's, I've cut away the crap from my two Aussie scores and i am unable to find any info on the PTs (current capacity), so i've added up the current draw of the tubes used.

Score #2
Plate current is approx 71mA
Heater current 1350mA

The first problem is that marked on the tag board attached to the PT is saying that the HT leads are 385v each which i would think is way too high for a single 6AQ5 that it was running (this voltage will be checked before i commence), so............

The new lineup may be 2x6V6, PI 12a_7 and 2x 12a_7
Plate current is approx 80mA
Heater current 1800mA

Heaters marked on PT as 3A so the heaters are covered
A bit SHORT on the plate current
Q. Does the rect tube added to to plate current load of the other tubes? 5Y3 with separate 5v 2A tap
 This amp came from a Radio/record player
Q. Would the radio part suck more than 10mA out of the PT as it was combined with the record player amp??

Score #1 Same problem
Plate current approx 92mA
Heater current 2000mA

New lineup may be 2x6aq5, PI 12a_7 and 2x 12a_7
Plate current approx 104mA
Heater current 1800mA

Heater current seems to be covered,if reading the data on the 6X4 rectifier said that the heater current is 800mA,i added this in cause the heater is 6.3v connected to the other heaters

Again the current draw is SHORT, so.............
Q. Does the 6X4 current draw (not heaters)add to the tubes load on the PT (6x4 plate current was NOT added to the plate current draw
Q. If the above does not apply , what would the chance be that the PT might have an extra 15mA or better kick'n around

Any simple test to find the current capacity of the PTs. Thanks
« Last Edit: January 07, 2013, 11:58:21 pm by TIMBO »

Offline jazbo8

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Re: Squeezing the most out of two PTs
« Reply #1 on: January 07, 2013, 06:25:02 am »
Don't know about simple, but if you follow the instruction included in Rod Elliot's transformer application, you can determine all the parameters of an unknown transformer. You can get the application here.
« Last Edit: January 07, 2013, 06:27:07 am by jazbo8 »

 


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