i would suggest to put dpdt switch to harness your 6 volt battery when the 36v battery pack dropped to cutoff voltage. when your 36v battery pack is fully charge you will safeguard your controller with max tourque/speed and when it dropped close to cutoff voltage then use the dpdt switch wired in series with your back-up 6volt resulting to the desired 36volts at least. this will give you a turbo boost safely.
If you need more info, just ask. I'll post again in a few days to let you know if it is still working.
If you have 48 volts stamped on the cover with the serial number then the motor is designed to accept the 48 volt controller which you should purchase to avoid problems from circuit differences.
I foresee you having serious problems running 48V through your 36V controller. Both the motor and the controller were warm to the touch after a measly 15kms?
You are going to have a mess of melted wires at some point, pure and simple....
I've worked alot with zener diodes, in defeating the overboost cutoff functions on several turbocharged cars.... It's a nice bandaid...no more.
....you are going to have problems if you continue to abuse your 36V system with 48V.
But to each his own....hammer down :P
Hi pdonahue.
Good news.. Thank you for your report..
Just to confirm. Did you change something in the controller?
(Are you using 4 x 12v batteries, right?)
I will start adding just 6V more... for preliminary tests(42v)(Enhanced version) :)
After that, I will move to 48v(Dragster version..) ;D ;D
Regards
A 48v system should be more efficient,given the same watts rating of the motor.Example;I suspect that is not the way the controller works. Usually the wattage varies with the voltage and not the other way around. You are running a 36V 500W spec'ed motor at 48V you will get the wattage of the beast - not less current.
24v-500w: 500W divided by 24V = 20.83 Amps(Current draw)
36v-500w: 500W divided by 36V = 13.88 Amps(Current draw)
48v-500w: 500W divided by 48V = 10.41 Amps(Current draw)
Also,say we add an extra 12v battery(a 12v-12a SLA for example) to achieve 48v, we also have larger battery storage and capacity for more range,extra weight being the only penalty.
This is one reason 24v systems are dinosaurs,and have very poor range.
This is aslo one reason that electric motors in some other applications run on very high voltage,some submersible oilwell pumps are 600v for example. Or another example is motorised equipment in a workshop such as a lathe or air compressor are cheaper to run on 220v than 110v.
Here are the things to consider.Actually a custom controller is just what you need to do this in a brushless. A reduced output voltage at low speeds prevents saturation, and the extra voltage at "extra speed" ( say 30 instead of 20MPH ) maintains torque at the higher speeds.
1. 48 v vs 36 is 33% increase in voltage and will cause a similar increase in current draw.
2. You will saturate the magnetic structure of the motor and the additional magnetic force will be wasted, with the waste being turned into heat.
3. Generally a motor can take an overvoltage/current situation for a short time. If the motor is already hot then the additional heat from too much power will likely start to burn the insulation on the windings.
4. If it is a brushless design you may exceed the semiconductor ratings in the electronic commutator.
This Tim-the-Toolman approach is not a good idea.
You build it , they might buy it. ;DIf I win the lottery so I do not have to worry about putting food on the table I plan to. ( I am an electronic design engineer )
I hope you have great medical and life insurance! You might need it if a cellphone distracted driver rolls through a stop sign or you eat the asphalt!! :o(which ever comes first). Personally In would opt for a 72 volt , 40 amp comtroller like the other brand "c" has. :)Do you have a link? The lack of specs annoys me.
With what are you measuring your Wh/km?
A 48v system should be more efficient,given the same watts rating of the motor.Example;
24v-500w: 500W divided by 24V = 20.83 Amps(Current draw)
36v-500w: 500W divided by 36V = 13.88 Amps(Current draw)
48v-500w: 500W divided by 48V = 10.41 Amps(Current draw)
Also,say we add an extra 12v battery(a 12v-12a SLA for example) to achieve 48v, we also have larger battery storage and capacity for more range,extra weight being the only penalty.
This is one reason 24v systems are dinosaurs,and have very poor range.
This is aslo one reason that electric motors in some other applications run on very high voltage,some submersible oilwell pumps are 600v for example. Or another example is motorised equipment in a workshop such as a lathe or air compressor are cheaper to run on 220v than 110v.
A lower current only lowers the resistive loss in the wires (Tens of watts at most). The higher speed using 48V means much more wind resistance (hundreds of watts) so the net effect is less efficiency. If we were on the moon though, I would agree that the 48V controller would be efficient. ;)Interesting... this is the first I had heard that they actually have a feedback loop other than the phase sensors.
Also, moving to 48V does not mean keeping the same output level. More realistic would be:
24v-500w: 500W divided by 24V = 20.83 Amps(Current draw)
36v-750w: 750W divided by 36V = 20.83 Amps(Current draw)
48v-1000w: 1000W divided by 48V = 20.83 Amps(Current draw)
since the current draw is generally limited by the current feedback loop of the controller and so does not decrease as the voltage is increased.
Pete
A lower current only lowers the resistive loss in the wires (Tens of watts at most). The higher speed using 48V means much more wind resistance (hundreds of watts) so the net effect is less efficiency. If we were on the moon though, I would agree that the 48V controller would be efficient. ;)Interesting... this is the first I had heard that they actually have a feedback loop other than the phase sensors.
Also, moving to 48V does not mean keeping the same output level. More realistic would be:
24v-500w: 500W divided by 24V = 20.83 Amps(Current draw)
36v-750w: 750W divided by 36V = 20.83 Amps(Current draw)
48v-1000w: 1000W divided by 48V = 20.83 Amps(Current draw)
since the current draw is generally limited by the current feedback loop of the controller and so does not decrease as the voltage is increased.
Pete
The Golden graphs are implying almost totally resistive losses of 220W on the 500W model at full load. I therefore would not expect much more than 220W lost on a 48V controller assuming it is actually current limited.
Dan
Thinking about it, I guess all the losses are resistive (some in the coil, some in the core) except for the loss due to friction which I would assume would be quite small. So that would blow my "tens of watts at most" out of the water :-[ But I have noticed that the actual current draw at both 36V and 48V is about the same (Peaks at around 30A on startup and then drops to around 20A during operation.)LM358 would be an opamp. I would have to see it to know the the thing is. As to losses there are copper resistance losses, and magnetics losses in the form of eddy currents (normally not an issue) and hysteresis losses (which would show up at higher frequencies so at 60Hz should also not be an issue).
There is a metal bridge on the controller board leading to the neg. rail. I assume that it is being used as a current sense resistor. I haven't traced out that part of the circuit, but I am guessing that it leads back to an op-amp (I thought there was an LM356 on the board, but it may have been a LM358... It's hard to see the markings.) to cut back the duty cycle when the current gets over 20A.
Pete
At 320 RPM and 46 poles Golden sensors switch at 122.66 times per second per phase, however, their are 3 phases to consider and 2 directions of current flow per phase. Golden yokes are laminated of course to minimize Eddy currents. Surprised no one has yet mentioned back EMF.Nice, a three phase adjustable HV drive would work then. I have mentioned it indirectly previously: motor voltage is directly proportional to motor speed minus the IR drop ( current times resistance = voltage )
However, again in relation to the topic copper conductors with smaller crossections represent greater electrical resistance and it is greater electrical resistance which results in the constraint imposed upon the HBS-36 not being suitable for use with 48 volt controller without upgrading either the power phase lead insulation or the power phase lead gage, still with the risk of thermal damage to the windings without upgrading them as well.
Unfortunately Federal law prohibits ebike top speed of over 20 MPH, and modification is required to HBS-36 wiring to gain low end torque and acceleration to forestall thermal damage when 48 volts is applied.Indeed unfortunate, unless you care to register as a motorcycle.
The consumer market, unfortunately dictates wheel size and even though a smaller rim would allow higher RPM to be achieved faster during acceleration and reduce the duration of the load and wattage demand, the HBS-36 has 60c to 90c insulation which prohibits handling heavy or sustained loads with 48 volts, much less 72 volts, without wiring modification. Furthermore, if such modification makes an ebike motor capable of handling more than 750 watts then it can not longer be used.Why do you insist on missing the point? CONSUMER controllers would indeed cause a melt down, I am not denying this. Wire temperature is related to CURRENT not POWER, were it not so they could not use ALUMINUM in high tension wires.
Maximum current capability ratings are normally base on line voltage with maximum current ratings based on insulation temperature and ultimately wire size.Ummm...NO, you can NOT use higher currents. The rotor resistance is constant, if we ignore the copper tempco, and the power dissipated in the winding resistance is the current squared times the resistance.
The reason you can use higher current at lower voltages is because wattage is a function of current times voltage. Thus wattage or power in the presence of changing voltage or current is used to rate wire capacity, still based on insulation temperature and ultimately wires size, i..e., wire diameter or crossection.
In Oregon electric bikes are limited to 20 MPH and motors to 1,000 watts. It is interesting in hearing that there is a federal law limiting electric bikes (what happened to state rights?). I have yet to see a federal TRAFFIC cop patrolling the highways around these parts to enforce any such laws.
Maximum current capability ratings are normally base on line voltage with maximum current ratings based on insulation temperature and ultimately wire size.Ummm...NO, you can NOT use higher currents. The rotor resistance is constant, if we ignore the copper tempco, and the power dissipated in the winding resistance is the current squared times the resistance.
The reason you can use higher current at lower voltages is because wattage is a function of current times voltage. Thus wattage or power in the presence of changing voltage or current is used to rate wire capacity, still based on insulation temperature and ultimately wires size, i..e., wire diameter or crossection.
Dan
Maximum current capability ratings are normally base on line voltage with maximum current ratings based on insulation temperature and ultimately wire size.Ummm...NO, you can NOT use higher currents. The rotor resistance is constant, if we ignore the copper tempco, and the power dissipated in the winding resistance is the current squared times the resistance.
The reason you can use higher current at lower voltages is because wattage is a function of current times voltage. Thus wattage or power in the presence of changing voltage or current is used to rate wire capacity, still based on insulation temperature and ultimately wires size, i..e., wire diameter or crossection.
Dan
I am refering to published wire current ratings for line voltage. Wire rated at 10 amps at 120 volts can carry 20 amps at 60 volts, in terms of insulation temperature and ultimately wire size because the wattage is the same.
16 AWG wire is rated at 10 amps at 120 volts and can carry 1,200 watts. Reduce the voltage to 1 volt and 16 AWG can carry 1,200 amps because the derived wattage rating is not exceeded.Post a reference... If you have one other than your own faulty logic then you should share this gem with all the wire manufacturers, none of whom appear to know this.
See: Reply #73 on: Today at 11:10:37 AMOk so you have a reference to ohms law and watts law, that still has nothing to do with your erroneous claims.
Except that I2R = V2/R and R=V/I.Not true. The back EMF of the motor shows up as a battery between one terminal and the resistance of the winding. If you care to on a BDC, sit it on a rack, spin it up to 10MPH and take a voltmeter to it. Subtract that number from the rotor voltage to get the "winding" voltage.
P then equals (V^2)/(V/I) and I=P/V.
16 AWG wire is rated at 10 amps at 120 volts and can carry 1,200 watts. Reduce the voltage to 1 volt and 16 AWG can carry 1,200 amps because the derived wattage rating is not exceeded.
Show me the calculation then when voltage is zero.
I've cut a 3 ft length of 16AWG and its sitting right here on my desk. Its not red hot or melting, in fact it even feels cold.
Wonder how can that be when your calculations clearly show it should be burning a hole through my desk.
Show me the calculation then when voltage is zero.
I've cut a 3 ft length of 16AWG and its sitting right here on my desk. Its not red hot or melting, in fact it even feels cold.
Wonder how can that be when your calculations clearly show it should be burning a hole through my desk.
When the voltage is zero, the current through the wire is:
I = V/R = 0/0.0014 = 0, so using the same formula as in my post, the heat dissapated in the wire is:
POWER = I^2 * R = 0 * 0 * 0.0014 = 0 Watts
0 Watts is enough power to do absolutely nothing to your wire. I assume that is why it is not red hot or melting.
Pete
So what voltage then would have to be applied to each end of my wire get 20,520 watts?
What equation did you use to obtain a resistance of .0014 per 3 ft of 16AWG and what were the values for voltage, current and power?
R=P/I2
R=V/I
R=V2/P
| P=watts | I=amps | V=volts | R=ohms | |
| P=I2R | 1200 | 1200 | 1 | 0.000833333 |
| P=IV | 1200 | 1200 | 1 | 0.000833333 |
| P=V2/R | 1200 | 1200 | 1 | 0.000833333 |
| V=P/I | 1200 | 1200 | 1 | 0.000833333 |
| V=IR | 1200 | 1200 | 1 | 0.000833333 |
| V=(PR).5 | 1200 | 1200 | 1 | 0.000833333 |
| I=(P/R).5 | 1200 | 1200 | 1 | 0.000833333 |
| I=V/R | 1200 | 1200 | 1 | 0.000833333 |
| I=P/V | 1200 | 1200 | 1 | 0.000833333 |
| R=P/I2 | 1200 | 1200 | 1 | 0.000833333 |
| R=V/I | 1200 | 1200 | 1 | 0.000833333 |
| R=V2/P | 1200 | 1200 | 1 | 0.000833333 |
| P=watts | I=amps | V=volts | R=ohms | |
| P=I2R | 1200 | 10 | 120 | 12 |
| P=IV | 1200 | 10 | 120 | 12 |
| P=V2/R | 1200 | 10 | 120 | 12 |
| V=P/I | 1200 | 10 | 120 | 12 |
| V=IR | 1200 | 10 | 120 | 12 |
| V=(PR).5 | 1200 | 10 | 120 | 12 |
| I=(P/R).5 | 1200 | 10 | 120 | 12 |
| I=V/R | 1200 | 10 | 120 | 12 |
| I=P/V | 1200 | 10 | 120 | 12 |
| R=P/I2 | 1200 | 10 | 120 | 12 |
| R=V/I | 1200 | 10 | 120 | 12 |
| R=V2/P | 1200 | 10 | 120 | 12 |
PeteBut that is NOT how it works! The resistance of the motor is more like 0.5 ohm. You MUST take out the motor voltage BEFORE you can calculate the resistance. Energy can neither be created nor destroyed. If the resistance were 12 ohms you would NOT have and energy left to propel the bike as it would all be lost in the winding resistance.
Okay then at line voltage of 120 volts AC (but lets say DC) we are told 16AWG has a current rating of 10 amps.10 amps at 120 volts gives us a wattage rating of 1200.
My heat gun is 1200 watts and if I use a 16AWG extension cord you can feel it getting warm with prolonged use and the heat gun seeming to not have as much heat or power.
If we use the formula R=V2/P or the R=P/I2 formula to calculate resistance then R= 12 ohms, so I think the 10 amp rating refers to the wattage of the appliance rather than to the wattage a shorted wire can handle.
I think this is where you misunderstand. I am not talking about a shorted wire load but rather the load of the appliance, which a 16AWG wire with only a .014 resistance can carry.
Click here (http://goldenmotor.com/SMF/index.php?topic=137.msg1599#msg1599) Dan.Actually you have regularly implied that the entire voltage drop and all the energy delivered to the motor is dissipated by motor resistance.
Aside from this I have not yet addressed winding resistance other than to mention briefly the dumping of stored inductive charge when PWM power to a BLDC motor is turned off under no load and load conditions.
Pete
Okay then at line voltage of 120 volts AC (but lets say DC) we are told 16AWG has a current rating of 10 amps.10 amps at 120 volts gives us a wattage rating of 1200.
My heat gun is 1200 watts and if I use a 16AWG extension cord you can feel it getting warm with prolonged use and the heat gun seeming to not have as much heat or power.
So how does this apply to a 1 volt, 1200 amp load?
Again we are not talking about a shorted 3 ft. length of 16AWG but rather a 1200 watt load.
Imagine we have a 1200 watt resistance load designed not to operate at line voltage but rather at exactly one volt and 1200 amps. To simulate this load let us use our equations to see just what the results will be.
Starting with P=IV we get R=V/I or 1/1200=0.000833333
0.000833333 ohms then is the resistance of our appliance load.
Lets select a 1200 watt, 0.000833333 ohm wire wound ceramic resistor to simulate this load.
Now we can cut our 3 ft piece of 16AWG in half, attach it to each end of out appliance load and operate our 1200 watt appliance all day long at 1 volt and 1200 amps using our 3 ft length of 16AWG, 10 amp, 120 volt, .014 ohm piece of wire.
Here is the full set of equations for 10 amps at 120 volts and 1200 amps at 1 volt.
P=watts I=amps V=volts R=ohms P=I2R 1200 1200 1 0.000833333 P=IV 1200 1200 1 0.000833333 P=V2/R 1200 1200 1 0.000833333 V=P/I 1200 1200 1 0.000833333 V=IR 1200 1200 1 0.000833333 V=(PR).5 1200 1200 1 0.000833333 I=(P/R).5 1200 1200 1 0.000833333 I=V/R 1200 1200 1 0.000833333 I=P/V 1200 1200 1 0.000833333 R=P/I2 1200 1200 1 0.000833333 R=V/I 1200 1200 1 0.000833333 R=V2/P 1200 1200 1 0.000833333 P=watts I=amps V=volts R=ohms P=I2R 1200 10 120 12 P=IV 1200 10 120 12 P=V2/R 1200 10 120 12 V=P/I 1200 10 120 12 V=IR 1200 10 120 12 V=(PR).5 1200 10 120 12 I=(P/R).5 1200 10 120 12 I=V/R 1200 10 120 12 I=P/V 1200 10 120 12 R=P/I2 1200 10 120 12 R=V/I 1200 10 120 12 R=V2/P 1200 10 120 12
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.
| 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. ;D ;D ;D
Pete, could you please photograph where zener diode should go. What are the specs on the zener diode?
Thanks