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 Post subject: Battery Radio Power Supply
PostPosted: Nov Sat 09, 2019 11:01 pm 
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Joined: Dec Wed 24, 2014 7:34 pm
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Location: Weimar, Texas
I moved this to this Category and deleted the other one

The other posts on this topic (that I found with search) are old. I made a prototype based on this schematic and it's a work in progress:

Image

The schematic was redrawn in Autodesk Eagle but I found it on here.

It's not my design (so it has a much better chance of working). I think I need to make the following point. The schematic defines the board in Eagle. You literally cannot wire a board different than the schematic. So if something looks odd on the board, it's odd in the schematic and has to be fixed there. My schematic says I used a 1N4004 diode where the original used 1N4002. I used a 1N4002 when I built the prototype.

This is the current board layout. The first version of this board was 3.0" x 3.5" and it had some component footprint issues (luckily they were just too big) that I have since fixed. This board is 3.0" x 3.0"

Image

Questions:

(1) The 4.7k dropping resistor (thankfully I used a 2W resistor!) was a 220°F when I tried a no load power up. B+ was only 40V. Filament voltage was at 3.5V with no load. The voltage regulator wasn't hot (shouldn't have been). AC current was nil (as expected).

Transformer Vin: 120V
DC out to the resistor: 130V
DC out: 44V
Voltage drop: 86V
Resistance: 4.7k but I didn't pull it to get an accurate measurement
18mA / 1.5W if I used Mr. Ohm's law correctly

I need a 40V drop to get to 90V BUT how much sag am I going to see and how much power does the resistor need to dissipate. I get about .7W and 2.2k resistor.

(2) Is there enough RF filtering to allow this to be used near an AM radio? The board will be in a metal box

(3) I didn't make the DC grounds common.

(3) I have not as yet made what is called a copper pour and assigned it to GND. GND in this case is AC ground for the power cord. I normally use a ground plane on both sides of a board but I don't normally have 120VAC on the board. Should I ground the board with the AC power cord ground and then use metal standoffs to tie it to the metal case or just bolt the ground to the metal case? I could also just use a two wire polarized cord (like my test cord) Currently the ACG (AC GND) pin doesn't go anywhere. If I add a top and bottom copper pour the ACG will connect to the pour or I could make one side the %V negative and the other side the 90V negative.

Once this thing is working right:

I can supply the Eagle SCH and BRD files as well as a parts list with MOUSER part numbers on it. I can also provide a link to order the board from Osh Park (I don't get a dime from that and don't care to). If there's interest let me know. A lot of suppliers will take Eagle board files so there are much cheaper board options out there but the minimum is usually 10 boards.

Here's a shot of the first (larger) version of the board. The GND connection doesn't go anywhere. I have two more of these boards. If you want one of them let me know where to send it (deal of the century. They cost me $20 each and I'm giving them away, if they work. The board vendor I use, Osh Park, isn't cheap but they are fast and high quality. I only have two.

Image

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It's hard to solve an equation if every term is an unknown.

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Mike


Last edited by Mike6158 on Nov Sun 10, 2019 2:13 am, edited 1 time in total.

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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Sat 09, 2019 11:02 pm 
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Joined: Dec Wed 24, 2014 7:34 pm
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Location: Weimar, Texas
First problem solved. I was never able to get more than 40V output with 130V DC to the dropping resistor. I finally jumpered it out and it was still only 40V. So what was dragging the supply down? C4. It's supposed to be 100µF 200V and somehow I managed to find a 100µF 25V electrolytic. It was 180°F when I checked it. ESR is a thing :D

It's working like expected EXCEPT the 4.7k resistor has 350-900 mV across it. I suppose that because there's no load it's not stable?

With 125VAC (simulating a higher than normal AC supply voltage) there is 170VDC at the B terminals. This is a head scratcher. How do I determine how much the load will pull the voltage down and therefore how much resistor I need (or if the 4.7kΩ is correct, I doubt it)?

I've got a nice electronic load that I can put on it but it's 975 miles from me...

Could I put a 90V zener in place of the resistor?

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It's hard to solve an equation if every term is an unknown.

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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Sun 10, 2019 1:16 am 
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Location: Austin, Texas
I see an immediate problem with the transformer rating. The 3FD-410 is rated at 6VA IF you use the two primary windings in parallel as it was designed to operate. Using one of the primary windings as an output will lower the transformer output capability. It would be marginal at 6VA so it will probably be operating too hot once you get the loads attached.

The transformer has no built in thermal protection and no specification on the insulation between the primary windings. You will be risking primary winding over heating and the possibility of shorts between the AC power and the 90VDC output.

I think you should look for a better design.

Jay


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Sun 10, 2019 3:18 am 
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Joined: Dec Wed 24, 2014 7:34 pm
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Location: Weimar, Texas
It's not my design. I linked to the original builders website above. His article says it'll supply the 2W required. I don't know. There are older posts on this forum that indicate others have been successful with this design. Again, I don't know. I'm was just trying to figure out the ground plane for the board.

25mA per primary winding. The 90V side power requirement will be 90V at or near 12mA. Filaments need 1.4V at 250mA-ish. I just powered the filaments on the Farm radio that I just restored and it consumed about 240mA at 1.4V.

I'll put it on my Dynaload and load load it to 2W when I get home. I have a Fluke IR camera and I'll shoot the transformer (and board) while it's under load. If I remember to click the trigger I'll upload a photo.

Here's another design, also not mine, even more loaded, with a Hammond transformer, also 6VA

Image

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Mike


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Sun 10, 2019 3:27 am 
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My supply uses a Tamura 3FD-410 power transformer, and puts out correct voltages. I've built 3 of them, in fact.


Attachments:
farm radio PS.jpg
farm radio PS.jpg [ 94.65 KiB | Viewed 4812 times ]
three.jpg
three.jpg [ 188.95 KiB | Viewed 4812 times ]

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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Sun 10, 2019 3:43 am 
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Location: Austin, Texas
I calculate the transformer load at about 3.8 watts using your measured loads. You have to multiply the load currents by the actual transformer output voltages to get the load on the transformer and those voltages should be about 7 volts and 170 volts.

The VA load will normally be about twice the DC wattage so that would put the VA at about 7.6 VA.
Using one of the transformer primary windings as a load will probably reduce the transformer capability to about 4 VA.
You may get away with it, but I think you will be running the transformer at about twice it's design load.

You will need to take temperature measurements about every 30 minutes until you get a stable reading.
If you measure more than 60C on the outside of the primary windings, I would have concerns about the internal temperatures.

Edit: Since the transformer has no specification on the insulation between the primary windings, I think it would be a good idea to connect the safety ground directly to the B- output. That would give a high probability of blowing the fuse if there is a winding-to-winding short. This assumes the fuse is on the hot wire.

Jay


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Sun 10, 2019 4:32 am 
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Location: Weimar, Texas
fifties wrote:
My supply uses a Tamura 3FD-410 power transformer, and puts out correct voltages. I've built 3 of them, in fact.


Nice!

JnTX wrote:
I calculate the transformer load at about 3.8 watts using your measured loads. You have to multiply the load currents by the actual transformer output voltages to get the load on the transformer and those voltages should be about 7 volts and 170 volts.

The VA load will normally be about twice the DC wattage so that would put the VA at about 7.6 VA.
Using one of the transformer primary windings as a load will probably reduce the transformer capability to about 4 VA.
You may get away with it, but I think you will be running the transformer at about twice it's design load.

You will need to take temperature measurements about every 30 minutes until you get a stable reading.
If you measure more than 60C on the outside of the primary windings, I would have concerns about the internal temperatures.

Edit: Since the transformer has no specification on the insulation between the primary windings, I think it would be a good idea to connect the safety ground directly to the B- output. That would give a high probability of blowing the fuse if there is a winding-to-winding short. This assumes the fuse is on the hot wire.

Jay

I'll add the connection to the B- as suggested. Yep, I will make sure that the fuse and switch is in the hot.

Thanks for the feedback. I will not let the power supply out of my site until it's run in. I'll refer to this post (memory aid) and log the temps in case someone runs across this thread. The Transistor Devices load is pretty handy for tests like this. I'm going to take another look at voltages in the morning.

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It's hard to solve an equation if every term is an unknown.

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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Sun 10, 2019 9:35 pm 
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Posts: 119
Location: Texas, U.S.A.
1.5 V x .25 A = 375 milliwatts (mW) filament power
90 V x .012 A = 1.1 watts plate power
1.1 W + .375 W = 1.5 watts radio power

Using zener string across C4 to provide shunt regulation for 90 volts:
Zener string = 22 + 24 + 22 + 22 = 90 V (nominal, +/- ~ 5% with zeners 2EZ22D5, 2EZ24D5 from Mouser or Digi-Key).
Giving 12 mA to zener string: 90 V x .012 mA = 1.1 watts.
With no load connected to 90 V output, zener power will be 90 V x 24 mA = 2.2 W. Zeners should have at least 3/8 inch lead length and good ventilation should be provided.

120 V ac secondary voltage x 1.4 = 170 V ac peak volts = dc voltage across C2 with no load.
For loaded condition, guessing C2 voltage will be 80% of peak = 153 V across C2.
153 V - 90 V = 63 V across R1.
63 V / 24 mA = 2.625 k ohms for R1 (R1 = 2.7 k 10% okay). Should be okay even if C2 voltage is 140 V.
63 V x 24 mA = 1.5 watts. At least 2 watt rating for R1 and good ventilation should be provided.

153 V x 24 mA = 3.7 watts.
3.7 W / 0.81 transformer utilization factor = 4.6 VA at 120 V winding, unless I've made a blunder.
4.6 VA = 153% of the winding VA rating if the winding is rated 3 VA (transformer is rated 6 VA, the two 120 V windings in parallel are usually used as primary, so 3 VA each, possibly this is faulty).

Reference designators refer to those in schematic shown in initial post of this topic.

-----------
WB5HDF


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Mon 11, 2019 2:18 am 
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Posts: 2047
Location: Weimar, Texas
infzqi wrote:
1.5 V x .25 A = 375 milliwatts (mW) filament power
90 V x .012 A = 1.1 watts plate power
1.1 W + .375 W = 1.5 watts radio power

Using zener string across C4 to provide shunt regulation for 90 volts:
Zener string = 22 + 24 + 22 + 22 = 90 V (nominal, +/- ~ 5% with zeners 2EZ22D5, 2EZ24D5 from Mouser or Digi-Key).
Giving 12 mA to zener string: 90 V x .012 mA = 1.1 watts.
With no load connected to 90 V output, zener power will be 90 V x 24 mA = 2.2 W. Zeners should have at least 3/8 inch lead length and good ventilation should be provided.

120 V ac secondary voltage x 1.4 = 170 V ac peak volts = dc voltage across C2 with no load.
For loaded condition, guessing C2 voltage will be 80% of peak = 153 V across C2.
153 V - 90 V = 63 V across R1.
63 V / 24 mA = 2.625 k ohms for R1 (R1 = 2.7 k 10% okay). Should be okay even if C2 voltage is 140 V.
63 V x 24 mA = 1.5 watts. At least 2 watt rating for R1 and good ventilation should be provided.

153 V x 24 mA = 3.7 watts.
3.7 W / 0.81 transformer utilization factor = 4.6 VA at 120 V winding, unless I've made a blunder.
4.6 VA = 153% of the winding VA rating if the winding is rated 3 VA (transformer is rated 6 VA, the two 120 V windings in parallel are usually used as primary, so 3 VA each, possibly this is faulty).

Reference designators refer to those in schematic shown in initial post of this topic.

-----------
WB5HDF


Thank you. That's what I was looking for.

_________________
It's hard to solve an equation if every term is an unknown.

73
NE5U

Mike


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Mon 11, 2019 4:26 am 
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Joined: Nov Tue 14, 2017 5:09 am
Posts: 2025
Location: Austin, Texas
Mike,

I did a rough simulation of the circuit and I think you may need to reduce the value of R1. If you can measure the resistance of the two primary transformer windings, I can do a better simulation and give you a new value for R1.

Jay


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Mon 11, 2019 4:51 am 
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Joined: Dec Wed 24, 2014 7:34 pm
Posts: 2047
Location: Weimar, Texas
JnTX wrote:
Mike,

I did a rough simulation of the circuit and I think you may need to reduce the value of R1. If you can measure the resistance of the two primary transformer windings, I can do a better simulation and give you a new value for R1.

Jay


That's pretty cool. I checked the transformer soldered to the board but without a fuse or power cable installed and one of the uninstalled transformers that I have. As expected they were almost the same

On board
Primary 1: 307Ω
Primary 2: 365Ω

Off board
Primary 1: 303Ω
Primary 2: 360Ω

_________________
It's hard to solve an equation if every term is an unknown.

73
NE5U

Mike


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Mon 11, 2019 6:58 am 
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Joined: Nov Tue 14, 2017 5:09 am
Posts: 2025
Location: Austin, Texas
Mike,

These are the results of my simulation. I assumed a temperature rise of 25C in the transformer so the windings would be at 50C. That much temperature rise would make the winding resistances increase by about 10%.

Voltage on C1: 4.5VDC
Voltage on C2: 119VDC
Voltage on C4: 90VDC (Assumed to be clamped by the string of zener diodes)
R1: 1.8K, 1 watt
Zener current: 4mA to 16mA depending on the radio load

Primary 1 current: 44mA RMS
Transformer load: 5.3VA

You can check your results against my simulation when you get your supply going.

The main concern with the above is the primary 1 current. The transformer is designed for a current of 25mA for each primary winding. The temperature rise in primary 1 will be more than 3 times the design point.

Jay


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Mon 11, 2019 10:58 am 
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If you are designing for a specific radio, using salvaged transformers. One PSU I made for three UX201A's used fixed regulators 7805 for filament and 7905 for bias on separate windings. HV was a reversed 240V primary transformer and used two LR8's. One for HV and one for 45V 7805 cooled using a small CPU heatsink fan blowing onto its heatsink but also venting the entire housing. Totally uncomplicated.

Marc


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Mon 11, 2019 4:12 pm 
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This is the PSU I usually build for battery radio sets. The PSU shown in the image below was built for an OSRAM 4 TRF receiver.


Attachments:
PSU for battery radio set.JPG
PSU for battery radio set.JPG [ 55.38 KiB | Viewed 4627 times ]
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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Mon 11, 2019 6:17 pm 
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JnTX wrote:
Mike,
These are the results of my simulation.
Jay


Jay,

Very illuminating to see the magnitude of the winding resistance and effect on secondary voltage.

I realize that you were not addressing other forum participants, but I'm curious as to the "simulation" technique to which you refer.
I would be grateful if you could show the calculations or cite a source.

Thank you,
WB5HDF


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Mon 11, 2019 8:47 pm 
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infzqi wrote:
I realize that you were not addressing other forum participants, but I'm curious as to the "simulation" technique to which you refer.

Thank you,
WB5HDF


I use a free program called Microcap.
http://www.spectrum-soft.com/download/download.shtm

This is the way the input looks for something like the battery radio supply. This circuit is a variation I tried with the caps stacked and a common A- and B- point.
The change did not make a significant reduction in primary current.
Attachment:
Bat_PS_Ckt.jpg
Bat_PS_Ckt.jpg [ 87.02 KiB | Viewed 4596 times ]

The second primary transformer winding isn't needed since it is 1:1 with the first primary.
The voltages under the caps eliminate the time needed to get them up to normal operating level.

You can ask for different types of analysis like the actual voltages or a frequency response curve of an amplifier.
These are plots of the ripple voltage on the capacitors and the average value that a voltmeter would measure.
Attachment:
Bst_PS_caps.jpg
Bst_PS_caps.jpg [ 193.1 KiB | Viewed 4596 times ]


Lots of functions are available like the RMS value of a waveform or the power dissipation in a component.

Jay


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Mon 11, 2019 11:17 pm 
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One of the things that I have issues with, is the failure to provide for protection from transformers (back EMF) & particularly ecaps on the output side of the HV regulators, on some of these units. On loss of source power the caps will, given the opportunity, discharge via the regulator possibly destroying it.

A 7805 can be used for 4V Philips tubes by simply putting diodes in series with the OP. A trap with both 7805 /7905 (and LR8) is that they must draw current on the OP side to regulate. In the case of 7905 on a grid, it needs a 1K minimum to compensate for no grid current. 7805 same deal for filaments that can be shut off. (LR8 0.5mA)

Marc


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Tue 12, 2019 1:24 am 
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Thank you Jay for the information about the Micro-Cap program and showing how the winding resistances etc. are added to the diagram for simulation. I will try it out.
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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Tue 12, 2019 1:57 am 
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Location: Austin, Texas
infzqi wrote:
Thank you Jay for the information about the Micro-Cap program and showing how the winding resistances etc. are added to the diagram for simulation. I will try it out.
-----------
WB5HDF

For the typical 60Hz power transformer, an ideal transformer model works well with only the need to add the DC resistances of the windings. It isn't necessary to worry about the coupling coefficient between the windings because the transformer regulation is dominated by the winding resistances. An inductance can be added across the primary of the ideal transformer to improve the model but it only has a small effect for a well designed transformer.

When modeling transformers operating at 20KHz or more, coupling coefficients and leakage inductances become significant along with skin effects that increase the effective winding resistances. Capacitance across and between windings can also be significant. You can get good results with a crude transformer model at 60Hz but things get really messy at higher frequencies.

Jay


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 Post subject: Re: Battery Radio Power Supply
PostPosted: Nov Tue 12, 2019 7:05 pm 
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...
New calculations, accounting for winding resistance, using Schade's full wave center tap graph that I believe to be appropriate also for the full wave bridge [1].
Designing for 16 mA shunt zener regulator current at no load on 90 volt output. 12 mA maximum load on the 90 V output.
The results are very similar to JnTX's Microcap simulation result. I'm sure that either method is close enough.
It would be interesting to learn the actual measured voltage across C2 with the radio load on the supply.

Results:
Dc voltage across C2: 131 volts (JnTX gives 119 V)
R1: 2.2 k 10% 2 watt. (JnTX gives 1.8 k 1 W)
Primary VA: 5.75 (192% of VA rating of 120 V winding of Tamura 3FD-410) (JnTX gives 5.3 VA).
120 V secondary VA: 3.4 (113 % of winding rating).
LM317 dissipation: 1.35 watts.

(For a new design a transformer such as a Triad FP10-1200 or a Signal DP-241-5-10 would be more suitable than the Tamura 3FD-410).

Method:
Dc voltage at C2:
120 vac secondary voltage x 1.4 = 170 V ac peak volts.
Rs = secondary resistance + reflected primary resistance. Using hot resistances given by JnTX, = 333 + 396 = 729 ohms Rs.
(reflected primary resistance = 396 ohms because primary to secondary ratio is assumed to be 1.0).
Using JnTX's figure of 16 mA as nominal current to be supplied.
Estimate of Rl at C2: 140 V / 16 mA = 8700 ohms.
Ripple angular frequency x C2 x Rl = 308
Estimate Rs / Rl = 729 / 8700 = 0.084
Estimate Edc / Eac peak = 0.78 (prone to variation in reading of graph)
Estimate C2 voltage = 0.78 x 170 V = 132.6 V across C2.
Revised Rl = 132.6 V / 16 mA = 8287 ohms.
Revised Rs / Rl = 729 / 8287 = 0.088
Revised Edc / Eac peak = 0.77
Revised C2 voltage = 170 x 0.77 = 131 V (12 volts higher than 119 V Microcap simulation figure)

R1 resistance:
131 V - 90 V = 41 V across R1.
41 V / 16 mA = 2562 ohms = R1.
Subtracting 10% from the 2.562 k R1 resistance in order to permit 16 mA at maximum R1 resistance gives 2.306 k. 2.2 k 10% is commonly available. 2.2 k - 10% = 1980 ohms. 41 V / 1980 = 21 mA. 21 mA squared x 1980 = 0.87 watt dissipated by R1.

120 volt secondary VA:
131 V x 21 mA = 2.75 watts dc power.
2.75 W / 0.81 transformer utilization factor for full wave bridge rectification [2] = 3.4 VA
3.4 VA / 3 VA = 113 % of winding rating.

5 volt secondary VA:
5 x 1.4 = 7 V dc
As wired, the 5 V secondaries are parallel, rated at 1.2 amps total. The assumption is made here that the resistance is small enough to be disregarded in this application.
250 mA filament current + 20 mA LED current = 270 mA
7 V x 270 mA = 1.9 watts.
1.9 W / 0.81 TUF = 2.35 VA

120 volt primary VA:
3.4 VA + 2.35 VA = 5.75 VA
5.75 VA / 3 VA = 192 % of winding rating.
5.75 VA / 120 V ac = 48 mA (very close to JnTX's figure)

LM317 dissipation:
7 V dc - 1.5 V dc = 5.5 V drop across LM317
5.5 V x 0.25 A = 1.35 watts.

[1] Landee, Davis, Albrecht, Electronic Designers' Handbook, 1957, pp. 15-10 to 15-18
[2] R. Visintini, "Rectifiers", Power Converters for Particle Accelerators, CERN course book, 2004, p. 141

Hopefully no significant errors.
-------------
WB5HDF


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