BSA Owners' Club Forum

The BSA Workshop => Twins => Topic started by: Ian haines on 05 December, 2017, 19:21:32

Title: A65 charging
Post by: Ian haines on 05 December, 2017, 19:21:32
Hi all.
         Just a quick question. I took my boys bike for a shakedown run, about 25 miles. But it blood it's fuse on route, I got home and found no diode had been fitted, so got one from brit bits brilliant service and Ray  even found me a second hand heat sink to fit it to. So the question is does the charge fluctuate with the diode, as with it not connected it charges at up to 19volts but with diode in circuit it goes from 14 to 16 but could it be my digital meter or is this normal. Many thanks for your thoughts Ian

Title: Re: A65 charging
Post by: JulianS on 05 December, 2017, 21:00:13
The extract from Lucas Service literature explains the zener.

It is important that the surface of the zener and the heat sink are good to allow the heat to be conducted away, and also the zener has a low torque figure to prevent damage to its brass body - only 2 - 2.3 ft/lbs. See service bulletin.
Title: Re: A65 charging
Post by: Ian haines on 06 December, 2017, 10:01:38
Many thanks again Julian,
            I have tightened it to 2lb ft and checked a good earth, main earth goes direct to frame and heat sink plate. So will give bike another run as soon as Welsh weather brakes, with a bag of tools and spare fuse this time. Thanks again Ian
Title: Re: A65 charging
Post by: Caulky on 06 December, 2017, 11:44:14
You may have cooked your battery, they do not like being over-charged. (and could cause a fire)
The zener is manufactured to limit the voltage to about 15V-ish.
When the rectified Voltage from the alternator reaches the zener cut-off Voltage, It passes current (Amps) to earth.
It gets hot, hence the heat sink.
This limits the Voltage to the battery.
i.e. say the Voltage output (without zener) is 20V, the zener needs to reduce this Voltage by 5V.
In avalanche mode, the resistance of the diode may be very small. Say 1.0 Ohms.
W (Watts) = E squared (Volts) / R (Ohms) so, 5 x 5 = 25/1.0 = 25 Watts.
W = EI so I = W/E, I = 25W / 8V = 3 1/8 Amps.
The diode plus heatsink must dissipate 25 Watts as heat. It needs a good airflow across it to dissipate this heat.
Title: Re: A65 charging
Post by: Ian haines on 08 December, 2017, 10:10:22
Hi caulk,
Thanks for info,checked battery all ok 12.5 v, and no loss of acid, I think the fuse blowing saved it.
We got 3 inches of snow at the moment so might be a while before I get her out to test.
            Happy Xmas everyone Ian