Dead Magic Pie or can I resuscitate it????

[Cornelius]

Ok, you're good the way you tapped the power for the relay.

Are you sure you'll need those diodes for the second battery? They do steal 0.6V each (around 0.4V if they're schottky diodes), and would present an inbalance when power are drained. Besides, I think Bikemad are correct; the diode on the minus side are the wrong way (if drawn correctly, that is. )

Edit:
Also, consider your analyst; it also could be the faulty partner? it does sit in the middle of your main power line to the controller...

[robertozm]

Ok, you're good the way you tapped the power for the relay.

Are you sure you'll need those diodes for the second battery? They do steal 0.6V each (around 0.4V if they're schottky diodes), and would present an inbalance when power are drained. Besides, I think Bikemad are correct; the diode on the minus side are the wrong way (if drawn correctly, that is. )

Edit:
Also, consider your analyst; it also could be the faulty partner? it does sit in the middle of your main power line to the controller...

Everything I have read about a second battery they always recomend to put the Diodes in order not to charge one battery with the other... if you know what I mean.
I was not sure if the diodes were needed in both phases but as I could not find anything about it I put it in both.
About the diode type, herewith I send you a pic, I hope you could tell me if they are the correct ones, and the only thing I can read is 40 HF120 and the simbol of diode.

Long time ago I thought the cycle analyst could be the problem and I email the cycle analyst guys and they told me the have never had a faulty shunt, and that it would be very unlike it...

which of the Solid state relays do you recomend me, the 40, 60 or 80amp?  If I take the 80amp wouldnt be working much more relaxed than the 40amp?

Thanks for your time.
Roberto

[Cornelius]

The diodes... 40A, 1200V max. typical forward voltage drop: 1.3V...

I'd say drop the diodes...
What happens when the diodes are in place are: Power are drawn only from your first battery until the voltage on bat1 drops below 'bat2-1.3V'; only then will you have the power from both batteries...

The only need for a diode between 2 batteries in parallel, would be during charging, and all lithium based batteries with bms should have no equipment connected when charging; it confuses the bms/charger...

Edit:
When 2 equal batteries are connected in parallel, and are of equal age/wear, the load during discharge should be near equal, and should not hurt each other... It is a point to keep the wires from each battery equal in length and size to the point where they're parallelled.

[DirtyGinge]

HI
I  mean remove all obstacles whilst you fault find...then you can add each component back one by one until done or fault found
I use just straight anderson connectors, no switch.... u have built a precharge resistor that I just plug in for 10 seconds, then remove and connect the andersons together.....capacitors are all charged and very minimal spark

I also use 2 battery packs together without any diodes...doesnt block any regen and as long as the batteries are pretty evenly matched, there may be some slight discharge into each other, but fuses are fitted inline to protect from anything drastic like forgetting to charge one battery....1 20 amp inline car fuse per battery , prevents kentucky fried fingers...
At the end of the day, one battery might put out 4AH, 1 put out 3.8 Ah, and there is a small imbalance, but no biggie
You can check the batteries are balanced by doing a set journey with a set speed for each battery, then charge them using an inline turnigy watt meter to measure the AH put back into each battery...they should end up roughly the same
theres a pic of the inline resistor in another post somewhere, less than £10 to built.....2 andersons and a 5 watt resistor 1ohm...stops the lightning bolts, but keeps as much bare metal between the batteries and connectors
As for the cycle analyst, is it standalone ?, yes as long as the shunt has power then the cycle analyst remains on, even if you pedal without a battery the cycle analyst will power due to the power being created by the pie.....if its cutting out, seems like your problems are either  then shunt...or the circuit before the shunt..even if the cycle analyst is faulty, all the shunt does is  feed the bypass current from the shunt to the cycle analyst...

My electronics are not pretty, but build to be problem free ( apart form this weeks short circuit)......one rule we use in I.T. ( and martial arts, well karate for fat people...) when designing large building computer setups..and multilayered networking setups .KISS...keep it simple...i mean why do a massive triple backflip and flying turnaround round kick, when a simple punch in the face will do lol

Hopefully this helps, and doesnt sound too arrogant, don't mean to by any means , I mean im ginger, how bad can life get
regards

[robertozm]

The diodes... 40A, 1200V max. typical forward voltage drop: 1.3V...

I'd say drop the diodes...
What happens when the diodes are in place are: Power are drawn only from your first battery until the voltage on bat1 drops below 'bat2-1.3V'; only then will you have the power from both batteries...

The only need for a diode between 2 batteries in parallel, would be during charging, and all lithium based batteries with bms should have no equipment connected when charging; it confuses the bms/charger...

Edit:
When 2 equal batteries are connected in parallel, and are of equal age/wear, the load during discharge should be near equal, and should not hurt each other... It is a point to keep the wires from each battery equal in length and size to the point where they're parallelled.

I can´t have "wires from each battery equal in length and to the point where they're parallelled", since one of the batteries is inside the box and the other is in the pannier.

They are not the same age and wear, there is a difference of nearly a year (the one in the box is a year older) and I don't keep a record of the times I have charge it, but yesterday was the first time I charge completely the battery 2.

During charging they are not disconected of the circuit but there is no load since everything is swicht off.  The only thing I can do when charging is to unplugg the circular conector that conects the battery 2 to the circuit.

What about the regen function, will recharge both batteries?

Given this data, do you still think it is better to drop the diodes?

Roberto

[DirtyGinge]

diodes will block regen 100%...

The purpose of the diode is to only allow power to flow one way, so if you have them going from battery to controller, then the power cannot flow from wheel back to battery......

Are the batteries both GM batteries ?

Even if just for testing, drop the diodes, drop the relay, have a straight connection between battery and wheel, test, then add back cycle analyst, test, then add relay, test, then add diodes, test etc etc

good luck

[Bikemad]

I'll get back to this later as I'm just off to work now, but the diodes in your diagram do not allow the second battery to give any assistance at all.

I'm back again.

Roberto, as your batteries are connected in parallel, there should not be any need for the diodes to be fitted in the circuit at all.
Diodes are sometimes fitted to BMS protected batteries that are wired in series (as shown in the attachment below) to prevent problems if one battery reaches it Low Voltage Cutout limit before the other.
Diodes are also used to ensure that any regen current is directed through the charging side of the BMS instead of entering the battery through the output wires of the BMS.

I've tried to indicate on your diagram below, how the diodes could be connected, but your connections to the BMS do not appear to be correct, so I could not show it properly, but it will hopefully give you a rough idea.

I would try connecting the batteries without the diodes and see if the cutting out still occurs, as it could be a problem with "battery 1" BMS or even faulty cells.

Alan
 

[robertozm]

I'm back again.

Roberto, as your batteries are connected in parallel, there should not be any need for the diodes to be fitted in the circuit at all.
Diodes are sometimes fitted to BMS protected batteries that are wired in series (as shown in the attachment below) to prevent problems if one battery reaches it Low Voltage Cutout limit before the other.
Diodes are also used to ensure that any regen current is directed through the charging side of the BMS instead of entering the battery through the output wires of the BMS.

I've tried to indicate on your diagram below, how the diodes could be connected, but your connections to the BMS do not appear to be correct, so I could not show it properly, but it will hopefully give you a rough idea.

I would try connecting the batteries without the diodes and see if the cutting out still occurs, as it could be a problem with "battery 1" BMS or even faulty cells.

Alan
 

OK I am going to follow Yours and Cornelius advice and forget about the Diodes.

You said, and in your drawing I can see, that both phases +/- goes into the BMS, but I can´t do that with mines, because the have only - leads.  You can see the pic attached and this is the same type of BMS I have in Battery 1, it has one - lead going in, 1 - for the recharging post, and the - lead for the load, because the BMS in Battery 2 only have 2 wires one - going in, and - going out, so the recharging and the load goes through the same wire.

I am going to follow the advice of DirtyGinge as well and I am going to start with the battery/bms directly conected to the hub first bat1, then bat2, then both and if it works I will be adding the cycle analyts, the swith relay, the lights etc...   So I hope I will find what is driving me crazy.

Nevertheless, I am comitted to change the electromagnetic relay for the solid state one, as I understand it is more reliable, less bulky and also that I can have a better heat transfer through the aluminium box to the outside than with the other which is encloused inside a plastic box, but I would like to have your advice about which amperage is better, the 40, 60 or 80amp, because I don't know if it is better to oversize it or doing it so I will have some disadvantage.  So please could you tell me, which one of these three will suit me better given that the maximum amperage record by the cycle analyst has been nearly 30 amps?

http://cgi.ebay.com/Solid-State-Relay-SSR-DC-DC-40A-3-32VDC-5-220VDC-/250793127324?pt=LH_DefaultDomain_0&hash=item3a646f6d9c
http://cgi.ebay.com/Solid-State-Relay-SSR-DC-DC-60A-3-32VDC-5-220VDC-/390300012140?pt=LH_DefaultDomain_0&hash=item5adfb18e6c
http://cgi.ebay.com/Solid-State-Relay-SSR-DC-DC-80A-3-32VDC-5-220VDC-/370495498474?pt=LH_DefaultDomain_0&hash=item564340a8ea

Again thanks for your deatailed answers and the time you are taking.

[Bikemad]

Roberto, your BMS appears to use a common positive connection for the battery, power output and charging. It controls the charging and power output by switching the negative side of both circuits.
Although there isn't a heavy duty + cable going to the BMS, it shares the battery + connection through the balance lead connection.
With this particular BMS you would need to fit the diodes into the Negative battery leads and also on the negative charging leads to enable the BMS to control the regenerative charging.

Regarding the solid state relays, I have not had any experience of them, so I'm not sure how efficient they are, but I'm guessing they will generate more heat and be less efficient than a standard relay, but not having any physical contact damage caused by arcing should be a major benefit.

All of those relays are the same physical size and each would be carrying the same current in your application, but the more expensive 80A version might be better at handling the initial surge of current that charges the capacitors in the controller when you first switch on.

I am however concerned about over-volting the control voltage, as 40V is 25% higher than the maximum recommended!
I would suggest using a suitable resistor (560? 1/2 watt might be somewhere near) to drop the voltage, or alternatively an LM713T regulator chip could be used to reduce the voltage to around 30V.

Check out this link with regards to Using The LM317T To Regulate Voltage.

Alan
 

[robertozm]

Roberto, your BMS appears to use a common positive connection for the battery, power output and charging. It controls the charging and power output by switching the negative side of both circuits.
Although there isn't a heavy duty + cable going to the BMS, it shares the battery + connection through the balance lead connection.
With this particular BMS you would need to fit the diodes into the Negative battery leads and also on the negative charging leads to enable the BMS to control the regenerative charging.

Regarding the solid state relays, I have not had any experience of them, so I'm not sure how efficient they are, but I'm guessing they will generate more heat and be less efficient than a standard relay, but not having any physical contact damage caused by arcing should be a major benefit.

All of those relays are the same physical size and each would be carrying the same current in your application, but the more expensive 80A version might be better at handling the initial surge of current that charges the capacitors in the controller when you first switch on.

I am however concerned about over-volting the control voltage, as 40V is 25% higher than the maximum recommended!
I would suggest using a suitable resistor (560? 1/2 watt might be somewhere near) to drop the voltage, or alternatively an LM713T regulator chip could be used to reduce the voltage to around 30V.

Check out this link with regards to Using The LM317T To Regulate Voltage.

Alan
 

I hope not to abuse of your kindness but given the circuit diagram I post were exactly would you place the diodes.  Do I need the 3 diode you draw in order to "allow the regen voltage to go through the bms even if the realy is swithched off".  Because if the relay is swtiched off I wont have any regen going on since I will be motionless
Another question.  If I install the diode directly to the relay and from the diode I get two wires out one going to the fuse and another to the circular conector in the box I would save 1 diode, isn´t it?  And the diode wont suffer because the current is 36volt and the max amperage would be 30amp, isnt it?
I am going to order the 80amp solid state relay, right now and I will have a look at the post you suggest to solve the problem with the over-volting...

Also I am going to wait to install the diodes until you tell me  something.


Thanks for your advice.

Roberto

Edit:  I have change my mind about putting another electronic device (that will die someday) and  needs of another device to correct the input voltage (pair of resistance) that could and will also die one day and also both devices (relay+resistance) will generate heat inside a closed box with a battery and a bms that are not keen on heat...  therefore even if it means that I will have to drill new holes and cover old ones I will go for the more "basic" solution of a battery isolator switch.
I have found one at a marine spare part shop, for 17€ but it is quite big and the guy at the shop has told me that on Monday he will speak with his supplier to look for a smaller one.  
I am going to try to keep it as simply as possible as I have done reducing the connectors.  Less electronics/connections = less posibilities of something going wrong.
 

[Bikemad]

Roberto, the two diodes shown in the attachment below are there to ensure that regen current is controlled by the BMS.
One diode blocks the regen from going back through the Power Output side of the BMS and the other directs the regen through the charging side of the BMS.

Unless you are going to set off downhill with a fully charged battery, the diodes are probably not needed, but if you live at the top of a steep hill, it might be a good idea to fit them.

Alan
 

[robertozm]

Roberto, the two diodes shown in the attachment below are there to ensure that regen current is controlled by the BMS.
One diode blocks the regen from going back through the Power Output side of the BMS and the other directs the regen through the charging side of the BMS.

Unless you are going to set off downhill with a fully charged battery, the diodes are probably not needed, but if you live at the top of a steep hill, it might be a good idea to fit them.

Alan
 

Hi Alan:
I would like to know if there is any problem if I forget about battery 2 for the regen function.

I ask you this because the BMS in battery 2 hasn´t got any C- Imput.  The recharge goes through P- and the +lead, therefore I can not wire up the circuit as you propouse, because I have get another wire out of the aluminium box, and because the overall regen current I get is not really important.

Yesterday, I got the Battery Isolator Swicth (14€), and I am waiting for the anderson powerpole unipolar conectors (pp45, and pp30) to arrive to start making the new wiring, but I would like to know if I can do it as shown in diagram attached,  and I would like to know what AGW it is recomended, because I have been looking and searching, but I haven´t find anything "really clear".  Iam thinking to use a 10 AwGfor power, and 15 Awg for recharge, lights and horn.   As you know I don't have much electrical knowledge so I don't know if it is bad to oversize the thikness of the wires...  I am thinking to buy Teflon insulated wire as I understand is the thinnest insulation jacket and can stand a lot of heat, to rewire 2 Hbs36 hub motors I have against the wall and if succesful maybe I will venture to rewire the Magic Pie...  but I am not sure about this 

Thanks,
Roberto

[Bikemad]

Hi Alan:
I would like to know if there is any problem if I forget about battery 2 for the regen function.

Roberto, if the battery is normally charged through the P- connection, then any regen current would simply follow the same path as the charger, so this should not be a problem.

I would have connected the horn and lighting supply on the output side of the cycle analyst's shunt so that it would be able to monitor all of the power being used from the battery. If you were using 3 x 50Watt halogen lights in series, you would have more than 4Amps being consumed, which would not be measured by the cycle analyst, but I'm assuming you're using the GM LED lighting which only use around 1/20 Amp (50mA) for both front and rear lamps.

I would have thought that 12AWG wiring would be sufficient for your power wires, just so long as they are not too long, but 10AWG will have less voltage drop over longer distances. If the batteries and motor are both on the rear, I would use 12AWG, as it will be cheaper and lighter.

15AWG will be fine for the charging sockets, but smaller 20 or even 22AWG wire might be easier to route for the horn and LED lighting.

I would try to use as few Anderson connectors as possible, as each connection is another possible point of failure and could add more resistance to the circuit, especially when the connectors start to deteriorate with age. A decent soldered joint would be better if the connections do not need to be separated very often.

You should also make sure your Anderson connectors are properly coloured and labled to avoid any possibility of accidentally reversing the polarity of the battery or controller cables etc.

Alan
 

[robertozm]

Hi Alan:
I would like to know if there is any problem if I forget about battery 2 for the regen function.

Roberto, if the battery is normally charged through the P- connection, then any regen current would simply follow the same path as the charger, so this should not be a problem.

I would have connected the horn and lighting supply on the output side of the cycle analyst's shunt so that it would be able to monitor all of the power being used from the battery. If you were using 3 x 50Watt halogen lights in series, you would have more than 4Amps being consumed, which would not be measured by the cycle analyst, but I'm assuming you're using the GM LED lighting which only use around 1/20 Amp (50mA) for both front and rear lamps.

I would have thought that 12AWG wiring would be sufficient for your power wires, just so long as they are not too long, but 10AWG will have less voltage drop over longer distances. If the batteries and motor are both on the rear, I would use 12AWG, as it will be cheaper and lighter.

15AWG will be fine for the charging sockets, but smaller 20 or even 22AWG wire might be easier to route for the horn and LED lighting.

I would try to use as few Anderson connectors as possible, as each connection is another possible point of failure and could add more resistance to the circuit, especially when the connectors start to deteriorate with age. A decent soldered joint would be better if the connections do not need to be separated very often.

You should also make sure your Anderson connectors are properly coloured and labled to avoid any possibility of accidentally reversing the polarity of the battery or controller cables etc.

Alan
 

Hi Alan:
When I recharge battery 2, I already do it through P- connection (since I don't have other option) but I am going to forget about getting regen current going to batery 2, since in my usual use of the regen I don't get much, so I will just leave this regen going to battery 1.

I am going to follow your advice about the conecction of the horn and light on the output side of cycle analyst, it is something I didn´t think about it.

The distance between my battery and the hub power phase connection is about 50cm so it is not much, in the future when I change the inside phase hub motor power wires I will do it with 12 AWG.  As for the wiring from the battery to the hub I will use the 10AWG Hi-fi wire I already have.

About the anderson connectors, my intention is to make the minimum possible in order to if something is broken to be able to change it quickly without cutting or soldering again, because that is what I have been doing and I think it has not be a very goog idea, and it has ended a little bit messy.
And of course I will buy and install them in the correct color, black or red.  I will use the 45amp for power, and the 30amp (because they are cheaper than the 15amp) for the light, horn and recharge, but I will have to wait to do it, next month 

I hope I will have the time to make the trial with the isolator battery swicht this weekend.  And crossing fingers wishing the fairlure was on the relay...

Again, thanks for your time.

I will keep this updated with the progress of the repair and rebuilding. 

[Bikemad]

I hope I will have the time to make the trial with the isolator battery switch this weekend.  And crossing fingers wishing the failure was on the relay...

It sounds like you will soon be finding out for sure.



Alan