B.T.H Mag-Dyno

ClassicBiker

Well Known and Active Forum User
VOC Member
We've started wiring up Doug's Series A Comet. It has a BTH magdyno. When I removed the cover to attach the harness the only place that made sense at the time to attach the harness to the dynamo was to the stud pictured. The lead coming off of the brush of the dynamo, just visible behind the harness, has a red tag. The other brush, barely visible on the left has a lead connected to the case. Have I connected the harness to the correct terminal? I believe I have, but if someone with more experience knows different please let me know. Next question, in another thread I learned that as Series A Comets fitted with BTH magdynos did not have voltage regulators or cut outs to regulate the current the battery receives. It was suggested that a 10W diode would do the job given the lower power of the magdyno. Would the one pictured below be suitable? It is 10W 6.8V cathode cased Zener diode.
Many thanks
Steven

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Simon Dinsdale

VOC Machine Registrar
VOC Member
VOC Forum Moderator
A zener diode will not work and may even damage the Dynamo. Please do not use a zener.
The BTH mag dyno had no automatic regulator but a very crude form of regulation was done via the rider and the headlight switch position. The switch was wired up to this:
Off = no charge at all as the Dynamo is isolated from the rest of the bike.
CH = Dynamo charges the battery through a resistor at half charge rate. No lights on.
H = Dynamo charges the battery flat out and no regulation at all as the headlight is on And the 1/2 charge resistor is bypassed.

The BTH did have a cutout and this was a electro magnet bobbin type which was mounted on top of the toolbox. It looks like it may be a regulator as well but it isn’t. The cutout is purely to stop the battery discharging back through the Dynamo at low RPM and this cutout can be replaced with a basic one direction diode of 10A rating (not a zener). One end of the diode go to the Dynamo output and the other to the harness of the bike but you need to get the polarity of the diode leads the correct way round. Basically cut the wire to the Dynamo and bridge the cut with the diode. I hid the diode behind the toolbox.
if you are not sure then get somebody local to help who knows electronics otherwise you could end up with a burnt out Dynamo and these BTH units are now rare and it’s difficult to find somebody who can repair / rewind them.
 

delboy

Well Known and Active Forum User
VOC Member
A zener diode will not work and may even damage the Dynamo. Please do not use a zener.
The BTH mag dyno had no automatic regulator but a very crude form of regulation was done via the rider and the headlight switch position. The switch was wired up to this:
Off = no charge at all as the Dynamo is isolated from the rest of the bike.
CH = Dynamo charges the battery through a resistor at half charge rate. No lights on.
H = Dynamo charges the battery flat out and no regulation at all as the headlight is on And the 1/2 charge resistor is bypassed.

The BTH did have a cutout and this was a electro magnet bobbin type which was mounted on top of the toolbox. It looks like it may be a regulator as well but it isn’t. The cutout is purely to stop the battery discharging back through the Dynamo at low RPM and this cutout can be replaced with a basic one direction diode of 10A rating (not a zener). One end of the diode go to the Dynamo output and the other to the harness of the bike but you need to get the polarity of the diode leads the correct way round. Basically cut the wire to the Dynamo and bridge the cut with the diode. I hid the diode behind the toolbox.
if you are not sure then get somebody local to help who knows electronics otherwise you could end up with a burnt out Dynamo and these BTH units are now rare and it’s difficult to find somebody who can repair / rewind them.
Nice one Simon.
I would appreciate if anyone out there knows a really good restorer in the UK [not just rebuilder] of these BTH Dynomags.
I could REBUILD it myself, but finding someone who truly understands the wrinkles of these units, as Simon says, is difficult.
Regards,
Delboy.
 

roy the mechanic

Well Known and Active Forum User
VOC Member
There is a guy in Burnham on Crouch who specialises in mags and dynamoes. Cant remember his first name Salmonds surname. I took a mag generator to him a couple of months ago. He knew all about it, thankfully. Phone 01621 784141, tell him I sent you. Roy.
 

ClassicBiker

Well Known and Active Forum User
VOC Member
A zener diode will not work and may even damage the Dynamo. Please do not use a zener.
The BTH mag dyno had no automatic regulator but a very crude form of regulation was done via the rider and the headlight switch position. The switch was wired up to this:
Off = no charge at all as the Dynamo is isolated from the rest of the bike.
CH = Dynamo charges the battery through a resistor at half charge rate. No lights on.
H = Dynamo charges the battery flat out and no regulation at all as the headlight is on And the 1/2 charge resistor is bypassed.

The BTH did have a cutout and this was a electro magnet bobbin type which was mounted on top of the toolbox. It looks like it may be a regulator as well but it isn’t. The cutout is purely to stop the battery discharging back through the Dynamo at low RPM and this cutout can be replaced with a basic one direction diode of 10A rating (not a zener). One end of the diode go to the Dynamo output and the other to the harness of the bike but you need to get the polarity of the diode leads the correct way round. Basically cut the wire to the Dynamo and bridge the cut with the diode. I hid the diode behind the toolbox.
if you are not sure then get somebody local to help who knows electronics otherwise you could end up with a burnt out Dynamo and these BTH units are now rare and it’s difficult to find somebody who can repair / rewind them.
Simon,
Now I understand. The diode is acting as the cutout to prevent discharge of the battery back through the dynamo, diodes being a one way gate for current flow. Do you happen to know what the resistance is of the resistor is?
Thanks
Steven
 
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