Junk Driver?

Nev · 2019-04-20T18:44:48.000Z

I wouldn’t drive while junk , it’s against the law & very dangerous.

Sirstinky · 2019-04-21T06:03:49.000Z

Sirstinky · 2019-04-21T06:04:43.000Z

Lexel:

if the inductor gets too hot to touch within 7 seconds something is wrong on this driver

What do you think is the issue?

Barkuti · 2019-04-21T10:11:10.000Z

I always pay attention to the current sense resistor stack value on buck or boost drivers, by Ohm's law the sense voltage is easily obtained. I prefer to stay away from drivers with high sense voltage values, makes them less efficient and less overcurrent capable.

By the way, generally a buck driver designed for 2S emitters should also work good for 1S ones. ?

:-)

Lexel · 2019-04-21T15:05:25.000Z

Barkuti:

I always pay attention to the current sense resistor stack value on buck or boost drivers, by Ohm’s law the sense voltage is easily obtained. I prefer to stay away from drivers with high sense voltage values, makes them less efficient and less overcurrent capable.

By the way, generally a buck driver designed for 2S emitters should also work good for 1S ones. ?

this is not true 2S Buck driver with variable switching frequency for low dropout do switch faster when In- and Output voltage difference gets higher
too high swichting frequency adds a lot losses on the MOSFET

Barkuti · 2019-04-21T20:41:58.000Z

Maybe you overshot my previous question a bit Lexel. I think one of those drivers meant for XHP70s with up to 3S cells input voltage should work nicely with 2S cells driving an 1S emitter.

Sirstinky · 2019-04-23T19:18:17.000Z

So I tried doing the current measuring again. I soldered the driver output to the short leads from the LED and ran it off a 3S li-ion battery. I saw 3.5A for about 4 seconds on 100% until the driver dimmed the light and it started fashing. I’m not sure if it’s overheating or the battery is too low? The battery is 10.95 volts so it doesn’t make sense. It can’t be drawing the battery down that much.

Barkuti · 2019-04-23T19:43:57.000Z

What actually matters to the driver is the amount of voltage which reachs its input, Sirstinky. The driver will try to draw well above 3A or even more from the cells in order to serve 33+W at the emitter. How are the cells connected to the driver? If you are using alligator clips that for sure causes a lot of contact resistance at the input. Photos?

Sirstinky · 2019-04-23T21:08:42.000Z

Yes you’re right on the money. I’m using alligator clips to connect the battery to the driver. I figured that wasn’t the best way from what everyone’s saying. I will try another way. I will send photos when I very home from work. I’m learning more about these drivers all the time! It seems they are quite different than the CC/CV drivers I’m used to!

You mentioned the current sense resistor? If I put up a picture of the driver, could you point it out?

Thanks again.

Barkuti · 2019-04-24T01:02:47.000Z

Sirstinky sense resistors usually start with an R letter (is a decimal separator) or its figures are underlined (acts as a milli prefix). They are in series with the load, usually at the low side.

Sirstinky · 2019-04-24T03:53:49.000Z

He are pictures of the driver. I tried to hey pictures of the resistors on the PCB. Which one is the sense resistor? How do you use it to measure the current? When I test the driver again with a full battery I'll take a picture of it.

Thanks

Barkuti · 2019-04-24T06:09:17.000Z

Here is the first post photo you posted above. I can see an R120 leaning out under the inductor, next to the red wire which connects to the emitter. The driver may be using high side sensing, check this by testing continuity between the positive lead and one of the R120's leads.

If the driver is what it claims to be current wise I am pretty sure there are more sense resistors in parallel with that one, at least 3x more. This is because they're likely to be ¼W rated resistors and at 5A the dissipated power at each one would be P = I²R = (5 / 4)² × 0.12 = 0.1875W.

In order to determine how much current does the driver handles you need to find a way to check the voltage drop across that leaning out R120 (this will be a very low voltage value, less than 150mV as I see), and also check what is the sense resistor stack combined value. You may need to remove the inductor to see what is going on under it. Use Ohm's Law plus common sense.

Sirstinky · 2019-04-24T22:06:00.000Z

Excellent thank you for the detailed response! Sounds like you know these driverseeally well. I did probe one of the resistor contacts and the positive lead for continuity and got a short. I might take off the inductor as it looks to only be connected on one side. If I do, I’ll take a picture so you can see.

How do I check the voltage drop?

Lexel · 2019-04-24T22:32:39.000Z

Barkuti:

Here is the first post photo you posted above. I can see an R120 leaning out under the inductor, next to the red wire which connects to the emitter. The driver may be using high side sensing, check this by testing continuity between the positive lead and one of the R120’s leads.

If the driver is what it claims to be current wise I am pretty sure there are more sense resistors in parallel with that one, at least 3x more. This is because they’re likely to be ¼W rated resistors and at 5A the dissipated power at each one would be P = I²R = (5 / 4)² × 0.12 = 0.1875W.

In order to determine how much current does the driver handles you need to find a way to check the voltage drop across that leaning out R120 (this will be a very low voltage value, less than 150mV as I see), and also check what is the sense resistor stack combined value. You may need to remove the inductor to see what is going on under it. Use Ohm’s Law plus common sense.

1206 package is easy to have 0.5 or even 0.75W rating on good power sense resistors
but on those cheap chineese drivers you never know
for usual 5A driver likely 3-4 are in parallel so 30-40mOhm with the low power rated resistors

Sirstinky · 2019-04-25T00:07:00.000Z

Thanks. I bet they are under the inductor then because I didn’t see any other resistors in the surface.

Sirstinky · 2019-04-28T03:12:37.000Z

Barkuti:

Here is the first post photo you posted above. I can see an R120 leaning out under the inductor, next to the red wire which connects to the emitter. The driver may be using high side sensing, check this by testing continuity between the positive lead and one of the R120's leads.

If the driver is what it claims to be current wise I am pretty sure there are more sense resistors in parallel with that one, at least 3x more. This is because they're likely to be ¼W rated resistors and at 5A the dissipated power at each one would be P = I²R = (5 / 4)² × 0.12 = 0.1875W.

In order to determine how much current does the driver handles you need to find a way to check the voltage drop across that leaning out R120 (this will be a very low voltage value, less than 150mV as I see), and also check what is the sense resistor stack combined value. You may need to remove the inductor to see what is going on under it. Use Ohm's Law plus common sense.

I finally got around to looking under the inductor! There are 3 of the resistors underneath. It looks like they are the same value in parallel? I measured one with my multimeter and got 0.1 ohm. I’m not sure how accurate that is. How would I figure the current value?

Barkuti · 2019-04-28T07:36:35.000Z

Sirstinky:

… I finally got around to looking under the inductor! There are 3 of the resistors underneath. It looks like they are the same value in parallel? I measured one with my multimeter and got 0.1 ohm. I’m not sure how accurate that is. How would I figure the current value?

…

3 resistors in parallel, that is. Combined sense resistor value is 40mΩ.

What do you mean when you say I measured one with my multimeter and got 0.1 ohm? In order to properly measure an onboard resistor you need to desolder at least one of its terminals, if you don't you will also be measuring any other component resistances in parallel simultaneously. A standard multimeter cannot measure such low value resistors accurately.

To determine the output current value use your multimeter to measure the voltage drop at the sense resistor stack terminals (any of them will do) while the driver is operating, milivolt scale of course. Then use Ohm's law: I = V / R.

^:)

Lexel · 2019-04-28T12:48:26.000Z

To measure such low resistance you need 4 wire measurement eqiupment then cancels the leads resistance

A high sense Buck has nothing in parallel to the resistors so unsolder is not required, the Buck controller sense inputs are high ohmic

Barkuti · 2019-04-28T13:43:15.000Z

Lexel:

…

A high sense Buck has nothing in parallel to the resistors so unsolder is not required, the Buck controller sense inputs are high ohmic

Well mmmkay and so do low side buck or high/low side boost drivers, after all the sense resistor stack is in series with the output. Felt the need to pinpoint the general method.

Leads resistance could be “tared” or taken into account if the meter were to provide accuracy and range to measure ultra low resistance values, although contact resistances would still play a certain role.

Anyway, what sort of equipment?

:-)

Sirstinky · 2019-04-28T16:32:35.000Z

Barkuti:

Sirstinky:

… I finally got around to looking under the inductor! There are 3 of the resistors underneath. It looks like they are the same value in parallel? I measured one with my multimeter and got 0.1 ohm. I’m not sure how accurate that is. How would I figure the current value?
…

3 resistors in parallel, that is. Combined sense resistor value is 40mΩ.

What do you mean when you say I measured one with my multimeter and got 0.1 ohm? In order to properly measure an onboard resistor you need to desolder at least one of its terminals, if you don’t you will also be measuring any other component resistances in parallel simultaneously. A standard multimeter cannot measure such low value resistors accurately.

To determine the output current value use your multimeter to measure the voltage drop at the sense resistor stack terminals (any of them will do) while the driver is operating, milivolt scale of course. Then use Ohm’s law: I = V / R.

Thanks for the tip. I figured the reading wouldn’t be accurate, but thought I’d try it for fun. If I do that measurement of the voltage drop, what exactly am I looking for? Do I just measure voltage across the resistors while in operation?

Thanks again