Lead/acid battery charger
YES!!!! Found the perfect solution for hum caused by filaments.... But what happens after a few hours when the batteries start getting drained?!
The logical follow-up of battery heated filaments is of course a battery charger. The circuit below I got from a friend who used them in his own 26 preamp (nice tube, eh!?). It is very simple to build requiring only a handful of cheap parts and an extra winding on your power tranny (or a separate tranny). Since the charger described is not active during sound reproduction, I have made no criteria on parts quality, so el-cheapo was the way to go.
There are many circuits available for charging batteries, but many are for NiCd or NiMH batteries. They will not work for lead/acide batteries, so don't use these!!! The best way to charge a battery is with a current limited voltage regulator. This sets a maximum current (say 1A) that will flow to charge the battery. If the current rises over this set current, the regulator will be forced to put out a lower voltage. Since voltage drops, so will the current; hence current limited. While the battery is charging, the current should decrease slowly while voltage starts to increase. In the end the current will be next to zero and the voltage will be equal to the set voltage.
Use the following to design a charger:
1) The charging current should be kept to around 0.1 times the capacity of the battery. So a 10Ah battery should be charged with 1A of current (10 x 0.1 = 1). The battery will not be forced (quick charged) this way and assures a longer lifetime.
2) The charging voltage should be set to 2.3 - 2.4V per 2V cell. So a 12V battery (6 cells of 2V) is charged at 6 x 2.3 = 13.8V.
3) The transformer winding for the charger supply is chosen at the charging voltage plus 3V regulator drop plus 1.4V rectifier drop (two diodes) plus 10% safety. So for a 2V battery, the winding (AC) is 2V + 3V + 0.7V + 0.7V = 6.4V + 10% = 7V.
4) The charging current is limited by the small resistor in the common leg of the charger. The value for this resistor can be calculated with: R = 0.6V / max current. So if I want a maximum current of 0.5A, I will need 0.6V / 0.5A = 1.2 ohms. The 0.6V is the voltage required for the transistor to go fully into conduction. Between 0V and 0.6V the transistor will adjust the regulator to increase or decrease voltage depending on the current it is passing.
5) The charging voltage can be set using the potentiometer. Just hook up a volt-meter to the charger (without the battery attached) and adjust the output voltage until it macthes the charging voltage. Use a 1K pot for 2V batteries, use 5K for 6V and 12V batteries.
6) Use an extra fuse on the charger, about two times the maximum current you are charging with. Fusing is critical for safety reasons. A battery can deliver 100+ amperes during a short, so it can cause serious damage or even fires when something goes wrong with the charger. Put the fuse behind the regulator in the lead going to the battery. This fuse is not to protect the supply from a short in the charger, but to protect everything else from the battery in case of a short anywhere. Don't forget to fuse the primary winding of the charger as well...
7) You're set up to charge...
The diode over the regulator in my schematic prevents damage to the regulator when the supply is turned off. In this situation the battery that is still attached offers a negative voltage over the regulator, as the voltage on the output is greater than the voltage on the input. The diode shorts the regulator in this situation and avoids any problems. It is just a safety.
Tip: If you hook the battery up to a dual pole, dual position (DPDT) switch you can change between charger and filament. This way the batteries will be charged when the amp is not playing.
DIY seems to be returning to my life, what better way to refresh some old ideas...
The battery charger circuits above have been in use for quite some time during my 26/battery fetish period. Making chargers for my B+ battery stack left me melancholic with the thought of repeating the soldering task up to 35 times. Looking through some old datasheet books I discovered a little critter that combines the circuits above into a single chip. Granted, it only removes the BC547 from the picture, but it makes soldering so much easier as there are less routes to be made. Just solder on the desired resistor values and you're off.
The chip in question is the L200. It can supply up to 2A of output current (plenty for any battery) and is setup almost the same as a regular LM317, it just has two extra pins (3 and 5).
Pin 1 : Input supply
The output voltage is calculated as follows:
2.77V x (1 + (Pot / 1K))
The current limit is calculated as follows:
Imax = 0.45V / R1
Say you want 1A of maximum
output, you would need a 0.45 ohm resistor (0.45V / 1A). The 0.45V is a
given taken from the datasheets (V5 - 2).