48V in a 36v system

[myelectricbike]

The resistance that was computed was done with a spreadsheet so I'm not very confident about the number of significant places. However the point is that you can lower the resistance of the load even further so the total load of 1200 watts plus the load of the lead wires is dissipated at 1 volt and 1200 amps without overheating the 16AWG leads and if not then we have found or at least come closer to having an explaination as to how or why 16AWG phase power lead insulation might melt when 48 volts is applied to a motor designed to handle only 36 volts.

[pdonahue]

The resistance that was computed was done with a spreadsheet so I'm not very confident about the number of significant places. However the point is that you can lower the resistance of the load even further so the total load of 1200 watts plus the load of the lead wires is dissipated at 1 volt and 1200 amps without overheating the 16AWG leads and if not then we have found or at least come closer to having an explaination as to how or why 16AWG phase power lead insulation might melt when 48 volts is applied to a motor designed to handle only 36 volts.

at 1200A, the voltage across the wire will be:
V=IR = 1200 * 0.014 = 16.8V so you need at least that much voltage across the wire to get 1200A to flow through it.
As Dan pointed out, at that point, all the power:
 P=I I²R = 1200 * 1200 *0.014 = 20160W
 would be used up in the wire and there would be NOTHING left to power a load. 

In any case, if you would like to continue to believe that you can pass 1200A through the wire, then that is fine...  For me I will continue to obey the laws of physics and this will be my last post on this matter.  I am predicting that this will be the second last post in this thread since you will post a final note (or maybe two depending on how you're feeling!!!). 

Pete "still riding and enjoying the 48volter" D. 

[myelectricbike]

...
at 1200A, the voltage across the wire will be:
V=IR = 1200 * 0.014 = 16.8V so you need at least that much voltage across the wire to get 1200A to flow through it.
As Dan pointed out, at that point, all the power:
 P=I I²R = 1200 * 1200 *0.014 = 20160W
 would be used up in the wire and there would be NOTHING left to power a load. 

Uh pdonahue, we have not quite finished the discussion, so it may be a bit premature to rub your scent glands just yet.   

The purpose of this discussion is to provide a complete answer to the OP's question and for us to reach a complete understanding on his behalf as to why 16AWG phase power leads might glow in the dark if he uses a 48 volt supply to power his HBS-36 hub motor.

According to the performance data the HBS-36 is designed to operate from a 36 volt supply and not one that is 48 volts. We are not here for the purpose of causing anyone to replace their power phase leads or needing to rewind their motor as the result of melted insulation. 

Just because your power phase lead insulation has not melted (yet) and you are not personally affected by the problem does not mean that others will not be affected if they are mislead into thinking that their 16AWG phase power lead insulation will not melt or they will not risk doing other damage if they power their 36 volt hub motor with a 48 volt or even a 42 volt supply. We are not used car salesmen here hoping to make a buck on voided warranty replacement sales resulting from our abundant supply of misinformation.

No problem though, the world is full of outstanding electrical engineers who can pick up where you left off.   

Also from the work we've done here it seems to follow that when we apply 48 volts to a 36 volt motor we are not only faced with the risk of melting our phase power leads but are also at risk (if we take our ebike out on the road) of violating the law, anywhere the law sets the limit on ebike power to 750 watts. (Note: Resistance is for simulated load: R₁=V/(P/V)=1/(1200/1), R₂=V/(P/V)=36/(500/36))

0.00083333 Ohm Resistor/1 Volt Supply
Resistance	Pieces	Resistance per piece	Pieces used	Total resistance	Volts	Amps	Watts per piece	Watts
R1=V2/I1	n	RP=R1/n	m	RT	V	I	WP	W
0.014	2	0.007	2	0.014	1	67.41573034	31.81416488	63.62832976	Leads
0.000833333	1	0.000833333	1	0.000833333	1	67.41573034	3.787400581	3.787400581	Load
0.014833333	1	0.014833333	1	0.014833333	1	67.41573034	67.41573034	67.41573034	Total
.
36 Volt Motor / 36 Volt Supply
Resistance	Pieces	Resistance per piece	Pieces used	Total resistance	Volts	Amps	Watts per piece	Watts
R2=V2/I2	n	RP=R2/n	m	RT	V	I	WP	W
0.014	2	0.007	2	0.014	36	13.81427475	1.335839308	2.671678616	Leads
2.592	1	2.592	1	2.592	36	13.81427475	494.6422124	494.6422124	Load
2.606	1	2.606	1	2.606	36	13.81427475	497.313891	497.313891	Total
.
36 Volt Motor / 48 Volt Supply
Resistance	Pieces	Resistance per piece	Pieces used	Total resistance	Volts	Amps	Watts per piece	Watts
R2=V2/I2	n	RP=R2/n	m	RT	V	I	WP	W
0.014	2	0.007	2	0.014	48	18.419033	2.374825437	4.749650874	Leads
2.592	1	2.592	1	2.592	48	18.419033	879.3639332	879.3639332	Load
2.606	1	2.606	1	2.606	48	18.419033	884.113584	884.113584	Total

...
For me I will continue to obey the laws of physics and this will be my last post on this matter.  I am predicting that this will be the second last post in this thread since you will post a final note (or maybe two depending on how you're feeling!!!). 

Pete "still riding and enjoying the 48volter" D. 

...but hopefully not out on the public roads in violation of the laws of man.

[johnbear]

Pete, could you please photograph where zener diode should go. What are the specs on the zener diode?

Thanks

[pdonahue]

Pete, could you please photograph where zener diode should go. What are the specs on the zener diode?

Thanks

Hi Johnbear...  Sorry for this late reply... I haven't been looking at these forums much in the past while since I put my bike away for the winter.  I used two zeners in series because that was what I had available... They're 4 point something V zeners with a 5W power rating.  You need a total voltage drop across them of about 8-10V if I'm remembering correctly, and they only pass about 100mA so a 1W rating should be OK.  If I were actually going out to buy a zener for the task I would probably get a 10V 1W one. 

The zener is placed in series with the big grey resistor (It doesn't matter which side as long as the band on the zener is pointed toward the positive connection on the board).

Pete

[johnbear]

Thanks, I really appreciate the information!

I will try this out this weekend!