[Scott Novak]

I recently purchased a TrustFire EB03 2S-3P, 18650 x 6, Battery Box.
I went on a 14 mile ride with plenty of potholes and the power was NOT interrupted a single time. It has contact springs on both the top and bottom of the Li-Ion cells. It appears that the power interruption issue has been resolved by the new design.

I've disassembled the battery box and I will be making a schematic of the protection circuit. But don't hold your breath. It may still have the problem of circuit damage if the Li-Ion cells are inserted incorrectly.

For now I'm posting photos of the box and it's circuit boards. I'll be adding a few more photos as time goes on. Stay tuned.

Scott Novak

 

[Cat-man-do]

Good show. Looks like this box includes a protection circuit for the output voltage to the lamp. At least that is what the instructions on the box leads you to believe. My 4-cell box ( model EB02 ) has the same instructions on the box so perhaps mine has an output protection circuit as well. Very likely it does. Don't know why I didn't read everything written on the box before. I could of sworn it had indicated that it didn't use protection for the lamp circuit. :idea:

Question; Did yours come with the yellow battery indicators I see on your photos? Wish mine had those.

 

[Velodonata]

Question; Did yours come with the yellow battery indicators I see on your photos? Wish mine had those.

Those yellow stickers come with the case, the 4-cell versions have polarity indicators printed directly on the plastic in silver.

 

[Scott Novak]

Yes, the yellow cell polarity stickers are attached to the inside of the 6-cell case.

I suspect that the same person that designed this previous case design also designed the EB03 box.

The circuit is nearly identical, INCLUDING the incompetent reverse protection circuit! What was changed in the protection circuit is that each series string of cells now has TWO power MOSFETs in parallel and has the addition of a 100KΩ resistor in series between Postive Li-Ion cell terminal and the MOSFET gates. The paralleled MOSFETS will likely slightly lower the voltage loss, but they don't cure the design fault of the protection circuit.

If all cells are installed correctly, the power MOSFETS turn on hard and have very low resistance and the battery pack will operate normally.

If you reverse 3 of the cells so that all of the exposed cell tops are of the same polarity, and the cell voltages are similar, no current will flow, there will be no output, and no damage will occur.

If you reverse the polarity of ALL of the cells, the power MOSFETs will NOT conduct, there will be no current flow through the cells, there will be no output, and no damage to the circuit or to the cells, and the reverse cell protection circuit will work as intended,

The problem occurs if you reverse the polarity of any one pair of series connected cells. It puts the cells in a series connected loop and massive current will flow for a fraction of a second until the tops of the power MOSFETS are blown off! The power MOSFETs blowing up are acting like fuses and are the only thing that will prevent the Li-Ion cells from venting and/or catching on fire.

Even if you reinstall the cells correctly, you have no output from the battery pack and you will be riding in the dark.

The 8.4V circuit has overcurrent protection, overvoltage protection, and undervoltage protection.

HOWEVER, like previous versions, the series connection between the Li-Ion cells is NOT sensed and the protection circuit only sees the average voltage of two Li-Ion cells in series, so voltage imbalances can occur while charging cells inside the case.

I suspect that the USB DC/DC convertor is also current limited. I haven't finished making the schematic diagram, but there also appears to be an additional bipolar transistor and a power MOSFET in the circuit.

The voltage indicator appears to be the same as previous designs.

When you strap the case onto a stem or a bicycle frame member it does slide around. It might be beneficial to glue a grippy pad to the side of the box.

All in all, this battery box appears to have the least amount of problems of any available battery 18650 box, at least for 4 or 6 cell versions. It doesn't appear to have the problem of power disconnection caused by road bumps. Isn't that ironic that you need to be very careful to install the cells with the correct polarity or you will damage the protection circuitry?

Scott Novak

 

[Scott Novak]

Here are the protection circuits for the Trustfire EB03 2S-3P 18650 x 6 Li-Ion Battery Box.

Click on the schematic to view it larger.

The voltage losses consist of the resistance of 4 battery contact springs, and resistance of 3 Power MOSFETs in series with the Li-Ion cells.

Note that the power MOSFETs intended for reverse polarity protection are paralleled to reduce voltage loss.

Also, U2 and U3 are dual power MOSFET packages and they are also paralleled for lower voltage losses.

I haven't measured the voltage losses yet.

The reverse polarity circuit may require a second MOSFET in series to work correctly.

Note: Schematic edited because of an error.

Scott Novak

 

[-Archie-]

Many thanks for schematic! :thumbsup:

I've ordered 4-cell version of that box; probably the circuit is identical except one 2s part.

 

[Scott Novak]

Trustfire EB03 2S-3P 18650 x 6 Battery Box
LED Battery Voltage/Charge Indicator Circuit.

Click Schematic for Larger View

Note the circuit board layout error. They connected C4 to the wrong part of the circuit. C4 should be connected from the enable input to ground (across the switch contacts) to reduce general noise problems and eliminate switch contact bounce noise problems specifically.

Scott Novak

 

[ledoman]

Scott, great job! Looks like some student has designed those things for his homework. Took things from somewhere and put (copy/paste) together. Then tried: does it works, yes, then let's do it on massive scale. No checking what is actually happening.
Now only USB charging part of circuit is missing.

 

[Scott Novak]

EB03 2S-3P 18650 x 6 USB Converter Circuit

Click Schematic for Larger View.

Note the addition of a MOSFET switch to power up and down the converter circuit. It should reduce the idle current drain on the Li-Ion cells.

Also note that both the USB converter circuit and the LED battery charge/voltage indicator circuit draw their power from the Li-Ion cells BEFORE the battery protection circuits. It may be possible to drain the Li-Ion cells to the point of permanent damage.

As I really don't have much use for USB power I'll probably disable the USB converter.

Note: The schematic layout has been edited for greater clarity. No actual circuit changes have been made.

Scott Novak

 

[Scott Novak]

Looks like some student has designed those things for his homework. Took things from somewhere and put (copy/paste) together. Then tried: does it works, yes, then let's do it on massive scale. No checking what is actually happening.

That's it in a nutshell. The reverse voltage protection circuit (Right out of an application note) will only work with ONE series string of cells, not two stings of cells or 3 strings of cells.

Nobody bothered to test the operation, or even checked to see that the PCB layout was correct.

The circuit board layout is also poor. The freewheeling diode currents are flowing through the ground of the convertor IC introducing noise and otherwise reducing the stability of the converter. I can't say that the decoupling of the power supply is much better.

It's a far cry from the testing that a reputable US manufacturer does. I'd spend 3 weeks just performing the safety testing on a product, and that's not counting the hundreds of hours spent testing the actual performance of the product to make sure that it's meeting all specifications.

What is so frustrating is that there aren't even any better battery box options available at any price.

Scott Novak

 

[ledoman]

Can you show on the PCB picture where to cut traces to do that. It would be good for someone who would like to do the same. How much will reduce parasitic drain when cutting off USB output?

 

[Scott Novak]

Unless the USB converter is turned on, there will be virtually no parasitic drain caused by the converter, as it is switched out of the circuit by the power MOSFET Q1.

At full charge there is 42µA of current flowing through the voltage divider R2 & R3 of the battery protection circuit. I don't have any specs on the operating current of the 7022 IC or the voltage indicator IC.

I don't want to have an extra USB jack dangling on my bike as I'm not using it anyway. I will probably use my previous battery box, without the battery spring contacts on the bottom, for USB power if I ever need, it, but not for use while mounted on my bike.

The easiest way to disable the USB converter is to unsolder the Source of the MOSFET Q1 (which supplies power to the converter.) and lift it up off of the circuit board.

Scott Novak

 

[Scott Novak]

BTW, does anyone have any idea which IC is being used for the LED voltage/charge indicator? I can't see any markings on it.

Scott Novak

 

[garrybunk]

Nice work Scott!

-Garry

 

[Cat-man-do]

First, just want to say...Great work Scott! Nice to have someone on here that knows something about electronics. Now for the big question; Any idea yet on how much output current the protection circuit is going to allow to the lamp output? If you're using a six cell box I'd like to think these might become useful once they're hooked up to a lamp that requires a good bit of juice.

I've yet to test my 4 cell box on a ride. Was going to do it today but accidentally stepped on my glasses when I woke up today. That meant I had to rush on down to the local Lens Crafters to have them fixed and then try to do another errand that had to be done. By the time I was done running around I was so tired I fell asleep on the couch ( again ) and didn't get up till about 9:30pm. Really wishing now that I hadn't agreed to work for one of my co-workers on Saturday...:nonod:

( edit...Man...I am so tempted to play hooky from work tomorrow....:devil: )

 

[Scott Novak]

Any idea yet on how much output current the protection circuit is going to allow to the lamp output?

Unfortunately, only 2.0Amp output current maximum. The reason for this is that there is only one protection circuit. It must protect against over current under the worst case conditions.

Many 18650 Li-Ion cells have a maximum 2.0A discharge current specification. The worst case condition is when only two Li-Ion cells are installed in the box. The protection circuit MUST limit at 2.0A maximum.

Installing additional cells only allows longer for runtime, but NO increase in output current limit. Additional cells also reduce the average internal resistance which allows longer runtime in subfreezing conditions. But you are using the same current limit circuit for all cells in parallel, so you are stuck with a maximum of 2.0A current draw.

For higher than 2.0A output current, each series string of Li-Ions cells should have it's own protection circuit limiting the output to 2.0 A maximum. The total current from each parallel string of Li-Ion cells would add up. 4-cells in the box could safely output 4.0A and 6-cells could output 6.0A maximum and still be safe.

If I remember correctly, the output power connector is only rated for a maximum of 6A. Someone please correct me if I'm wrong.

There isn't enough room for 3 protection circuits on the PWB. There would be if the USB Converter circuit was eliminated.

If you really needed a USB output, it would make more sense to have a separate USB converter that would operate from 8.4VDC and plug it into the battery pack when you need it.

Scott Novak

 

[Scott Novak]

It should also be noted that, like most previous Chinese battery boxes, the cells are NOT monitored individually. Charging inside the battery box might not result in evenly charged cells. As such, each series pair of cells should be closely matched so that they charge evenly.

If you look at the protection circuit schematic, R2 and R3 are connected in series across the series Li-Ion cells to simulate the junction of the series cells that connects to the sense terminal of the 7022 protection IC. At least they used 1% tolerance resistors.

Some of you might be tempted to wire the junctions of all three series Li-Ion cell strings together and connect the junctions directly to the protection IC, so that all cells will be charged to equal voltages. But unless you plan to incorporate reverse polarity protection for EACH individual cell, don't do it! Reversing the polarity of just one cell will result in huge current flowing though the cell which can cause it to vent, or explode, or catch on fire.

This modification is also of very little benefit when you are charging the cells inside the battery box. Because there is only ONE protection circuit, you are limited to a charging current of only 2.0 Amp. With only 2.0 A of charging current available, if you are using high capacity 3,500 mAH Li-Ion cells, you can only charge those cells at a maximum rate of .2C (C = the cell capacity rating.) and it will take forever and a day to charge them.

For longest cell life the general wisdom is to use a charge rate of a maximum of .8C. This means that in an external battery charger you could be charging the cells with four times the current and four times faster than you could inside the battery box!

So unless you are willing to make a new circuit board with 3 individual protection circuits, add 3 separate contacts to connect the junction of each series pair of cells individually to it's respective protection circuit, just use the battery box as is with matched Li-Ion cells.

This battery box not perfect but it's now usable, provided that you are careful to install the cells with the correct polarity. It would be best to remove the cells and charge them inside an external battery charger. If you have an 2.0 Amp 8.4 VDC charger that you can connect to the battery box, expect it to take overnight and then some to charge.

Scott Novak

 

[Cat-man-do]

Unfortunately, only 2.0Amp output current maximum. The reason for this is that there is only one protection circuit. It must protect against over current under the worst case conditions.

Many 18650 Li-Ion cells have a maximum 2.0A discharge current specification. The worst case condition is when only two Li-Ion cells are installed in the box. The protection circuit MUST limit at 2.0A maximum.

Installing additional cells only allows longer for runtime, but NO increase in output current limit. Additional cells also reduce the average internal resistance which allows longer runtime in subfreezing conditions. But you are using the same current limit circuit for all cells in parallel, so you are stuck with a maximum of 2.0A current draw.

For higher than 2.0A output current, each series string of Li-Ions cells should have it's own protection circuit limiting the output to 2.0 A maximum. The total current from each parallel string of Li-Ion cells would add up. 4-cells in the box could safely output 4.0A and 6-cells could output 6.0A maximum and still be safe.

If I remember correctly, the output power connector is only rated for a maximum of 6A. Someone please correct me if I'm wrong.

There isn't enough room for 3 protection circuits on the PWB. There would be if the USB Converter circuit was eliminated.

If you really needed a USB output, it would make more sense to have a separate USB converter that would operate from 8.4VDC and plug it into the battery pack when you need it.

Scott Novak

Okay, now I'm scratching my head... I did test the box's main lamp output when I first got it using my ITUO XP3 which I'm pretty sure ( with three XML2's ) must be drawing more than 2A ( *although I could be wrong about that if the LED's are connected in series ). Didn't notice any dimming or lack of output when I first hooked it up. Now I'm going to have to confirm that by measurement. That means I have to find the wire's and clips I use for amp battery testing which I haven't used in quite some time.

The box that the battery holder came in does state that the output is suppose to be 2100ma which ( in how it was worded ) I took to mean that the USB circuit was limited to 2.1A and that it was on a separate loop from the main lamp output. All I can say at this point is that I hope you are wrong.

 

[Scott Novak]

8.4VDC @ 2,000 mA is the specification. The accuracy of that specification is up for speculation and ought to be measured.

The current limit is determined by the voltage drop across the power MOSFETs inside U2 & U3. That voltage drop also changes with the MOSFET temperature because the Rds On changes resistance with temperature. So the designer needs to be careful with power MOSFET selection to achieve the necessary voltage drop to obtain the desired current limit.

It's possible that there may be different versions of the protection circuit available that use a different voltage to trip the current limit that will allow the use of power MOSFETs with a lower Rds On.

It's also possible that when using both the 8.4V output AND the USB output you may exceed the recommended maximum discharge current of some Li-Ion cells. This is just more reason to disable the USB DC/DC converter circuit.

After I test to see whether or not the USB DC/DC converter circuit can draw the Li-Ion cell voltage to an unsafe level, I am going to disable the USB circuit in my battery box. I may keep the voltage indicator intact.

Scott Novak

 

[Scott Novak]

I finally got around to reassembling the battery box after someone stole the Li-Ion battery pack off of my bicycle. Not great loss as it was made from Li-Ion cells from a dead computer battery pack. But it still left me without light and that pissed me off.

Anyway, I put two 18650 Li-Ion cells into the battery box, hooked it up to my bike light and let it discharge until the Li-Ion protection circuit disconnected power to the light. However, as the schematics show, the USB supply will still continue to draw power from the Li-Ion cells after power to the lights is shut off.

I connected my cell phone's USB port to the battery pack and it charged my cell phone for a short time. Then it stopped charging my phone. An LED on the battery box was blinking and I just let it continue. I finally pulled the Li-Ion cells and measured one at 3.21 VDC and the other at 2.06 VDC. That's an average of 2.635VDC per cell. The LED was still flashing when I pulled the cells and would have continued to discharge the cells.

Anyone feel safe with a battery box drawing voltage down to 2.635VDC per cell?

I will probably buy a second battery box for a backup. But I'll need to be extra cautious while inserting the Li-Ion cells as a polarity reversal will cause the protection FETs to explode. Eventually I'll get around to disconnecting the voltage indicator circuit as well as the USB supply.

Scott A. Novak