Can someone suggest a simple power supply schematic to provide power to a 40s type battery set which requires
90 volts and 1 1/2 volt.

I thought the following:|------ R2 ----------- 1 1/2
| | |
| C 3 V
| | -
| G |
| V
| -
| G
AC IN -- R1 --->| -----|----- R3 --- R 4 ---- 90v
| |
C 1 C 2
| |
G G

R1 = 100 ohms C1, C2, C3 = 47 ufd
R2 & R3 = 1000 ohms
R4 = 3900 ohms

The first diode is used to rectify the AC coming in.
Two diodes are used, reversed biased, for .7 volt
drop each to add up to 1.4 volts for the tube
filaments.

Let me know what you think or if you suggest a
better design.

Thanks

Randall

As you can see, the diagram didn't come out as I hoped.

The bottom line: the AC comes in through a small
value resistor ( 100 ohms ) and is rectified by a
diode. One branch of the circuit goes to a 3900 ohm
resistor. On the other side of that resistor is a
filtering cap ( 47 ufd with the negative end going to
ground) and the 1.4 volts needed for the filiments is achieved by reverse biasing two diodes.

On the other circuit branch, there is a 1000 ohm resistor, a filter cap ( 47 ufd - with the negatuve to
ground ) another 1000 ohm resistor. On the other side
of that last resistor another filtering cap and possibly a zenor to regulate the 90 volts.

If you have a better idea, email me back or fax me at
978-264-0494

Thanks Again

: Can someone suggest a simple power supply schematic
to provide power to a 40s type battery set which requires
: 90 volts and 1 1/2 volt.

: I thought the following:|------ R2 ----------- 1 1/2
: | | |
: | C 3 V
: | | -
: | G |
: | V
: | -
: | G
: AC IN -- R1 --->| -----|----- R3 --- R 4 ---- 90v
: | |
: C 1 C 2
: | |
: G G

: R1 = 100 ohms C1, C2, C3 = 47 ufd
: R2 & R3 = 1000 ohms
: R4 = 3900 ohms

: The first diode is used to rectify the AC coming in.
: Two diodes are used, reversed biased, for .7 volt
: drop each to add up to 1.4 volts for the tube
: filaments.

: Let me know what you think or if you suggest a
: better design.

: Thanks

: Randall

Hi Randall

Here are a couple things to think about. You might like to isolate the supply from AC line. How much current do you need from 1 1/2 volts? A portable radio might require 250 ma. If you draw this much current a different value series resistor and larger filter cap will be needed.

Two diodes does make a nice regulator for filaments. A LM317 IC allows you to regulate as well as adjust voltage. You shouldn't need to regulate the 90 volt supply.

Norm

: As you can see, the diagram didn't come out as I hoped.

: The bottom line: the AC comes in through a small
: value resistor ( 100 ohms ) and is rectified by a
: diode. One branch of the circuit goes to a 3900 ohm
: resistor. On the other side of that resistor is a
: filtering cap ( 47 ufd with the negative end going to
: ground) and the 1.4 volts needed for the filiments is achieved by reverse biasing two diodes.

: On the other circuit branch, there is a 1000 ohm resistor, a filter cap ( 47 ufd - with the negatuve to
: ground ) another 1000 ohm resistor. On the other side
: of that last resistor another filtering cap and possibly a zenor to regulate the 90 volts.

: If you have a better idea, email me back or fax me at
: 978-264-0494

: Thanks Again

:
: : Can someone suggest a simple power supply schematic
: to provide power to a 40s type battery set which requires
: : 90 volts and 1 1/2 volt.

: : I thought the following:|------ R2 ----------- 1 1/2
: : | | |
: : | C 3 V
: : | | -
: : | G |
: : | V
: : | -
: : | G
: : AC IN -- R1 --->| -----|----- R3 --- R 4 ---- 90v
: : | |
: : C 1 C 2
: : | |
: : G G

: : R1 = 100 ohms C1, C2, C3 = 47 ufd
: : R2 & R3 = 1000 ohms
: : R4 = 3900 ohms

: : The first diode is used to rectify the AC coming in.
: : Two diodes are used, reversed biased, for .7 volt
: : drop each to add up to 1.4 volts for the tube
: : filaments.

: : Let me know what you think or if you suggest a
: : better design.

: : Thanks

: : Randall