Guys:

I want to replace the power transformer in my Stewart Warner Series 900 receiver. The original is shorted ( primary is shorted).

The original has two 2.5 V windings, a center tapped plate winding for the type 80 rectifier and a filament winding for the type 80 as well.

Can I estimate the voltage and current requirements for this transformer so I can replace it properly? Obviously the filament draws can be estimated but how do I estimate what the needed plate to plate voltage should be?

Also can I use a transformer with a 5 volt, center tapped filament winding to feed the 2.5 volt tubes (seems obvious)?

Thanks
Lou

PS: Just need a little guidance here since I am going back to my college textbooks to get the answers.Just thought one of you may have recently done this.

You should be able to get the "plate to plate voltage" from the Rider data.

Lots of early AC-powered sets still used 2-volt tubes, so you might be able to salvage a power transformer from a junk chassis. Of course, it probably won't have the perfect form factor but, if you can make it fit mechanically, it should work OK electrically. Plus, you can forget about needing a ballast.

I will have to see if I have the appropriate Rider manual for this. If not are they available in the local library?

Lou

:You should be able to get the "plate to plate voltage" from the Rider data.
:
:Lots of early AC-powered sets still used 2-volt tubes, so you might be able to salvage a power transformer from a junk chassis. Of course, it probably won't have the perfect form factor but, if you can make it fit mechanically, it should work OK electrically. Plus, you can forget about needing a ballast.
:

This info is in the Resources section. Plate voltage on the #45 audio-output tubes appears to be 226 volts per the Rider info, with a 115 VAC line input to the set. Normal plate current for the #80 rectifier tube is 90 mA, from the same Rider data. That info should allow you to "size" the replacement transformer.

Hi Lou,
Oddly enough you cannot power the other tube filaments with the same supply as the 80. The filament of the 80 will have the B+ on it. Most of the other tubes have cathodes surrounding their filaments at near ground potential. This exceeds the specs for all the tubes for the heater to cathode insulation.
What you will get is a high rate of tubes with H-K short failures and open filaments.
When you get a transformer from a scrap chassis, it should be equipped with a separate filament supply for the other tubes.

Best regards,

Bill Grimm

Thanks for that insight Bill

What I was thinking was that if I get a transformer with multiple 5 volt, center tapped windings, can I use one side of each winding to power the separate RF string for example.

Thanks

Lou

:Hi Lou,
: Oddly enough you cannot power the other tube filaments with the same supply as the 80. The filament of the 80 will have the B+ on it. Most of the other tubes have cathodes surrounding their filaments at near ground potential. This exceeds the specs for all the tubes for the heater to cathode insulation.
: What you will get is a high rate of tubes with H-K short failures and open filaments.
: When you get a transformer from a scrap chassis, it should be equipped with a separate filament supply for the other tubes.
:
:Best regards,
:
:Bill Grimm
:

Yes, assuming that the nominal center-tap-to-end voltage is 2.5V and that the winding has sufficient current capacity to handle the tubes assigned to it.

Thanks again

So rather than rebuild the one I have I think I will go on a search for one with 2 2.5 V windings , a 5v windings and the center tapped HV winding or something equivalent with 5v center tapped windings ( after some calculations of course!!)

Thanks
Lou

:Yes, assuming that the nominal center-tap-to-end voltage is 2.5V and that the winding has sufficient current capacity to handle the tubes assigned to it.
:

One other thing - in order to estimate the transformer voltage I need to determine whether I have a choke input filter or a capacitive input filter. According to the diagram, there are some leading filter caps in front of the two series chokes - does this make it capacitive input? Thanks Again, Lou

:Yes, assuming that the nominal center-tap-to-end voltage is 2.5V and that the winding has sufficient current capacity to handle the tubes assigned to it.
:

Appears to me that the 80 rectifier tube will "see" a standard pi filter (C-L-C). The 80 tube is good for 350 VAC plate-to-center-tap. Seems to me that transformer voltages in the not-quite-300 VAC neighborhood are pretty common for this type of set, giving a loaded B+ voltage of about 275 VDC.

Thanks CV

I got about 250 V CT with my calculations so far. Think I still need the ballast on the primary? Reading some literature that seems to point in that direction.

I also noticed that when I tried to sanity check the detector plate voltage and RF plate voltage across one of the PS series resistances the math did not add up. The plate volts on all tubes are measured with the volume control on full.Does this have any impact? ( Sorry this one is keeping me up tonite ) Ex: the 75000 ohm det resistor has almost 40 V on one end and 135 V on another. Assuming only detector current flows in it ( 2.6 ma) the voltage drop does not come out right. Am I missing something?

Thanks for all the inputs - trying to get "back into the books"

Lou

Lou

:Appears to me that the 80 rectifier tube will "see" a standard pi filter (C-L-C). The 80 tube is good for 350 VAC plate-to-center-tap. Seems to me that transformer voltages in the not-quite-300 VAC neighborhood are pretty common for this type of set, giving a loaded B+ voltage of about 275 VDC.
:

If you choose a replacement transformer that has a primary designed to operate on 117VAC, there is no reason why an input line ballast would be needed. My guess is that it was probably there originally so that the manufacturer could use up a stock of low-input-voltage transformers, or maybe to provide a wider potential range of operation for greater compatibility with regional line power. (AC distribution standards were not as monolithic in 1930 as they are now, with the national power grid.)

When the set is asked to operate at high volume levels, the output stages will draw more current which will tend to lower the B+ voltage; but if you have a robust power supply (able to provide ample current) then the effect will be very small.

I think that the "32" volts in the table has a dropped digit and should be "132" volts as it is with several other entries. If that's the case, and B+ is 250 VDC, the 75K resistor can source up to 3.3 mA for the detector tube. Given that there is an additional resistance (coupling transformer) in series with the detector tube plate, 2.6 mA tube current seems to be in the ballpark if the above missing-digit explanation is accepted.

Forget the preceding explanation- I just woke up when I typed it. 32 volts plate voltage is probably correct, per the table. If B+ is 250 VDC and the detector tube plate voltage is 32 volts, that means that a current of 2.9 mA is flowing through the 75K ohm resistor (250-32)/75000= 0.0029. Which is in the ballpark for detector tube current of 2.6 mA, since the detector tube and the coupling transformer that feeds its plate are the only two items drawing current through the 75K ohm resistor.

Hey CV

Let me try again..

If you look at the schematic, there is ONE resistor, 75000 ohms, that carries ONLY the detector plate current. One end of that resistor is tied to a voltage = 135 V - the same voltage that is on the RF tubes. The other end of that resistor is tied ONLY to the detector plate, which the schematic says is at appx. 40 v. Well 135 V - 40 V = 95 V, and 95 V dropped across 75000 ohms is 1.2 ma. This MUST be the detector plate current since that is the ONLY thing this resistor is tied to. BUT the table in the schematic says the detector plate current is 2.6 ma, as measured. WHY?

Thanks Once Again
Lou

:Forget the preceding explanation- I just woke up when I typed it. 32 volts plate voltage is probably correct, per the table. If B+ is 250 VDC and the detector tube plate voltage is 32 volts, that means that a current of 2.9 mA is flowing through the 75K ohm resistor (250-32)/75000= 0.0029. Which is in the ballpark for detector tube current of 2.6 mA, since the detector tube and the coupling transformer that feeds its plate are the only two items drawing current through the 75K ohm resistor.
:

As far as I know, no one has repealed Ohm's Law. My guess is that there is an error in the documentation, possibly in the detector plate current figure shown on the chart. Equally possible is that the value of the detector plate resistor was changed by the factory to something other (less) than the 75K shown on the schematic, and some draftsperson dropped the update ball 84 years ago. I'd file this away under the "unsolved mysteries" folder and revisit it when I have the set working and can take some real measurements.

Thanks

I thought I was going to throw my degree away there!!

Lou

:As far as I know, no one has repealed Ohm's Law. My guess is that there is an error in the documentation, possibly in the detector plate current figure shown on the chart. Equally possible is that the value of the detector plate resistor was changed by the factory to something other (less) than the 75K shown on the schematic, and some draftsperson dropped the update ball 84 years ago. I'd file this away under the "unsolved mysteries" folder and revisit it when I have the set working and can take some real measurements.
:

:Thanks
:
:I thought I was going to throw my degree away there!!
:
:Lou
:
::As far as I know, no one has repealed Ohm's Law. My guess is that there is an error in the documentation, possibly in the detector plate current figure shown on the chart. Equally possible is that the value of the detector plate resistor was changed by the factory to something other (less) than the 75K shown on the schematic, and some draftsperson dropped the update ball 84 years ago. I'd file this away under the "unsolved mysteries" folder and revisit it when I have the set working and can take some real measurements.
::
:
:
Looks like I dropped the ball, too. I am glad CV got you straitened out.

Best regards,

Bill Grimm