A Battery Eliminator for Battery Portables
Written by Bryce Ringwood   
Note this project is intended to assist with your own design - rather than being a "recipe"to be blindly followed. There are a few simple formulae so that you can adapt the design to your own requirements. Don't get too carried away with the numbers - the supply voltage at my house can be anything between 170 and 320 volts. Capacitors can have a 50% tolerance.  
Battery portable radios and table radios using vacuum tubes have  been around since the beginning of the valve era. Prior to World War 2, Accumulators (Lead-Acid batteries) were used for the filament supply, and the high-tension supply might have been provided by either dry-cells or small accumulators in series. This explains why valve filament voltages are often multiples of 2 volts - approximately.  Note that with battery valves, we talk about the ...
...filament, rather than the "heater" because the valve is directly heated - the cathode is the heater wire, or filament, made of thoriated tungsten wire. 
After World War 2, small B7G based valves with filament voltages of around 1.4 volts became generally available. 
The most notable radio using these was probably the Zenith Transoceanic, first produced in 1942. The Ever Ready and "Vidor" battery companies produced numerous dry battery portables - after all, they helped to sell a lot of batteries.
These battery types used in these sets are long gone. If you try to make-up a battery pack from "D" cells and "PM3"batteries - you will find it is very costly and the batteries will only last for 20 hours at best. 
The answer to this problem is to use a "battery eliminator". 
Normally the radio you are designing for, has the correct set of valves, and you know what the supply voltage and current should be.
The radio we are designing for here has a mixture of valves, some of which have the part number removed, so we can't be certain about the supply current. The supply voltage was marked on the lead wires.
Here is a typical portable radio circuit (This is for a Romac Handbag Radio)
Romac Handbag Radio Circuit
Note that the HT- is connected via a resistor (R14) to ground. In the PAM this had a value of 39 Ohms.
Since I couldn't power the radio without a battery eliminator, the design that follows uses a fair amount of guesswork. The output valve could have been a DL96 with a filament drawing 50mA- or it could have been a DL92 drawing 100mA, so what follows is an example of experimental design, that I hope will assist in your own power supply designs.

Circuit Diagram

This is the final circuit used with the PAM radio

Battery Eliminator Circuit Diagram (Final)

Note that you must to adjust the Rs values in your own application. 


The PAM radio battery eliminator

The PAM radio (see casebook) uses 1.4 volt filament battery valves and  90 Volts between HT+ and Grid Bias -. Unfortunately, at the time of the design of this battery eliminator I was unsure of the filament supply current, since I was unsure about whether the output valve was a DL92 or 96. The HT voltage on one of the leads was 90 volts - but someone had put this piece of information on the GB- lead. The filament voltage (1.6) was also on the negative lead. 
I decided to design the battery eliminator on the assumption that the filament current would be 200mA, and the HT current would be 15mA. The total power consumed by the radio would then be:

\displaystyle {V_{LT} \times  I_{LT} +V_{HT} \times I_{HT} }

where $V$ and $I$ have the ususal meaning for voltage and current,so total power drawn:

$1.5\times 0.2 + .015\times 90 = 1.65$ Watts (possibly!)


I had transformer laminations and a bobbin for a transformer with a core area 0.8 sq ins. (Transformer cores and laminations have imperial measurement sizes, so I will stick with the imperial units formula.)
The formula for core area required in sq. inches =
\displaystyle {Ac = {{1\over 0.72}\times \sqrt {{W_{out}}\over f}   }}
$A_c$ is the core area in sq. ins.
$W_out$ is the power the core can handle for a 50 deg  C rise in temperature
The transformer made on this core could provide up to $(0.72 \times 0.8)^2 \times 50$ Watts, for a 50 deg C rise in temperature $= 16.6$ Watts. (It was purchased as a 14 Watt transformer core.) - plenty big enough.
The next step is to calculate the primary turns required:
\displaystyle {N_p = {{3.49 E_p\times 10^6 }\over {f A_c}B_m}}
$N_p$ = Number of turns required on the primary
$E_p$ = primary voltage
$f$ = Mains frequency (50Hz in SA)
$A_c$= Area of transformer core in sq.ins
$B_m$= Permeability in Gauss (I use 12000).
The number of turns on the primary required =
\displaystyle {{(3.49 \times 10^6 \times 230)\over (50 \times .8 \times 12000)}=1676}
This assumes the flux density to be 12000 Gauss and a 230 volt 50Hz input supply.
I decided that the design output voltages should be 80 volts and 2.0 volts for the HT and LT windings. This gives $(80/230)\times 1676 = 582$ for the 80V section and $(2.0/230)\times 1676=14.5$ turns for the filament supply. I made it 15 turns of 0.5mm enamelled copper wire. (Which I had in my workshop).
There was a reel of 0.14 mm (measured) dia wire in the workshop - I decided to use that for the primary and the 80 V secondary. The area of the primary and secondary windings = ${(1676+582)\times .14\times .14}$ mm = 44 sq.mm = 0.07 sq. in , ignoring the 2V secondary, for the moment. and a further .1 sq in would be needed for the LT winding, leads etc. In practice, just about the entire winding area of 0.25 was used up with layers of tape, leadout wires etc. The primary was wound first of all, then a layer of tape, then the secondary. The flament winding was given a half  turn extra and was wound as the final layer.
The transformer was wound on a Micafil coil winder. For those wishing to wind their own - you can cobble something together using a veeder-root or electronic turns counter and a hand or electric drill, or you could get a local transformer company to do it for you  (Better idea). It isn't practical to hand wind it, because you will lose count and/or break the wire. N.B. I am not a transformer company - please don't ask.
After assembly, the transformer was tested and the correct voltages of 80 and 2.0 volts(yes, really!) were measured on a true RMS voltmeter(see section on multimeters). The laminations were hammered tight and varnished to stop it humming. The transformer did not get hot under load (unsurprisingly, given the over size core).
In South Africa, transformer laminations are available from RSE Electronics and transformer winders tape is available from Mantech. Mantech also sell reels of wire. You probably should use .25 mm dia for the primary and maybe .2mm for the secondary. The wire I used was extremely fine.
The measured primary resistance $R_p$ was $300$ Ohms and the secondary $R_s$ measured $ 170$ Ohms. We will need these figures later. 

The High Tension 90 Volt Supply

Portable radios were designed to operate with a HT voltage between 90 and 70 or so volts - so there is quite a margin for error. I used a 1000V 4A bridge rectifier for both HT and LT initially (because they were in my junk box). As we will see, it was a really bad choice for the LT circuit. 
Calculation of Loss of Voltage due to the Resistance of the Transformer Windings
The transformer primary is effectively a $300$ Ohms resistor in series with the primary. The power drain is $I^2 * R = 1.65$ Watts, so the current will be $\sqrt(1.65/300) = 0.074$ Amps. The effective primary voltage will be $230Volt - 300*.074=207.8$ Volts RMS.
There will be a secondary voltage drop of $0.015$ Amps$*170$ Ohms $= 2.6 $ Volts RMS.
The total RMS voltage from the secondary will be $207.8 Volts* 582/1676= 72.16$ Volts$ - 2.6$ (voltage drop above) $= 69.7$ Volts RMS under the assumed load. N.B. This is sort of OK for this design -but really the transformer should have a much heavier gauge primary winding. 
Loss of  Voltage Due to Diode Drops
The voltage on the capacitor will be $(69.7-1.2)*1.414= 96.9$ Volts because the silicon diodes will remove roughly $1.2$ volts. I used a 150 volt woking capacitor.
Loss of Voltage Due to Ripple
The ripple voltage will amount to $I/(f*C) = 0.015*(10^6)/(100*100)=1.5$ Volts
This will reduce the measured voltage to $96.9 - 1.5/2 =96.15$ Volts
Loss of Voltage Due to Series Resistor
Placing a 680R series resistor will further drop the output voltage to $96.15 - .015*680 = 86$ Volts DC. 

Take the calculations with a large grain of salt. At this point, we don't  even know the load resistances accurately. We are, at least in the correct ball-park for the HT voltage.  It also gives us some understanding of the losses in a power supply.

Voltage Loss/Gain Formula Result Comment
Loss in Primary Winding $I_p \times R_p\times N_s/N_p$ 7.7 Take into account filament current
Loss in Secondary Winding $I_s \times R_s$ 2.6  
Loss due to diode drops $0.6 \times 2$ 1.2  
Peak Voltage at Capacitor $1.414 \times V_s - losses$ 96.9 The capacitor charges to peak voltage
Loss due to Ripple $0.5 \times I_s /{f*C}$ 0.75 Use 2 times mains frequency for f
IR Loss from Series Resistor  $I_s \times R_s$ 10.2 Giving 86 volts

The Low Tension Filament Supply

I was not sure of the valves used, or the ones that should have been used. When I looked at the valve data, I was surprised to find that the filament voltage  had a polarity. I guess that changing the polarity must alter the grid bias of the valve. I had to correct the wiring of the battery eliminator to allow for this. 
The other nasty surprise was that I was only getting a filament voltage of about 0.8V under a load provided by a test resistor. Were my transformer calculations all wrong? 
The culprit was the junk-box silicon rectifier. Under even a modest forward current, the diode drops managed to be around 1.7 volts altogether. Fortunately, I did not have to rewind my transformer for a 4.0 volt secondary. I substituted the bridge rectifier for 4 individual 1N5819 Schottky barrier rectifiers, which gave 1.7 and a bit volts under load. At that stage, I just had a single 10 000uf and 1.8 Ohm resistor to drop the voltage to somewhere between 1.2 and 1.5 volts. In the days when this radio was made, battery eliminators used copper-oxide rectifiers. 
Applying the eliminator to the PAM radio worked somewhat unsatisfactorily because of  hum. Although it wasn't bad, it wasn't good enough. Substituting a battery for the filament supply showed that the LT supply had to be much smoother than I thought originally. I put another 10 000uF capacitor after the 1.8 Ohm series resistor as shown in thecircuit diagram - and the problem went away. The capacitors were made up from 3 3300uF 16 Volts working caps. I was very tempted to try out a super cap with a 5 Farad capacitance. You will need to check and adjust the value of Rs by starting with a high value - say 5 Ohms, and reducing the value until you get a filament voltage of between 1.2 and 1.4 volts. Note that the set will work with a filament voltage of 0.8 - but it is better to stick to the recommended value for the valve.


The power supply was constructed on stripboard (Veroboard). This was cut to slide in to slots in a small plastic case. Four binding posts in colours corresponding to the PAM radio power lead were used. 
The Battery Eliminator
The design has also been used with a National Panasonic radio. In this case there was loads of room to put the entire supply inside the receiver cabinet.  I used a slightly higher LT secondary winding so that the ubiquitous 1N4007 rectifiers could be used. Alas, someone had conected the valves to a high voltage supply and the all had to be replaced. These battery portable sets sound great. The National also worked well on shortwave.


In practice with the radio on half volume, the filament voltage was 1.55 volts (higher than I would like, but less than the 1.6 volts specified on the input lead itself.) A new zinc-carbon or alkaline battery has an open-circuit voltage of 1.6 volts. Presumably the resistance of the supply leads and the internal battery resistance would be sufficient to drop the supply to between 1.2 and 1.4 volts.
The HT voltage weighed in at almost exactly 90 volts under load at medium volume. The measured HT current was somewhat less at 11mA than the 15mA allowed for in the design.
The entire project took about 4 hours (including winding the transformer).

In use

The battery eliminator worked without significant levels of hum and distortion. The PAM delivered surprisingly high levels of volume. See the video.

Appendix - Using Voltage Regulators

I suppose many people will be puzzled by the fact I have designed this as an unregulated supply.  There has been a suggestion that the LM317 could be used. The LM 317 will supply a minimum voltage of 1.25 volt, but would work with (at least) a 5 volt input - which would have to be ripple free, so you would still need those large capacitors. The minimum input voltage for the LM317 IS 3.7 Volts - so the "trough" of your ripple voltage must be greater than this, and you will have to design accordingly.

This is illustrative - there may be regulators better suited to the task (beware of the switchmode types, as they will cause sharsh in the receiver).

For a "once off" battery eliminator for a specific radio, the use of a simple series resistor and smoothing capacitor is a much cleaner solution. If you are making a "Universal Battery Eliminator" for a collection of battery portables- then you probably want to have a variable HT and LT output where using IC regulators makes more sense., but that's a different project for another day.

I will put an LM317 calculator on this web site as soon as I've cleared the backlog.


"Radio Data Reference Book" - Howard W Sams publishers

"The Art of Electronics"-Horowitz and Hill - Cambridge



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