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Modifying a Labophot for use with Hi CRI LED
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glmory



Joined: 30 Dec 2016
Posts: 25

PostPosted: Sun Dec 24, 2017 10:59 am    Post subject: Reply with quote

mawyatt wrote:
Hello,

I've attached a schematic that uses the MT3608 as an LED "CURRENT MODE" controller. This requires just 2 resistors, one a current sense resistor (Rx) and another to scale the feedback potentiometer resistance to control the maximum allowed LED current (Ry).

The example shows the MT3608 with a BC-135 LED, a 5ohm 4watt resistor for Rx (can be a couple 10ohm 2watt resistors in parallel), and a 16Kohm resistor for Ry. Maximum LED current is limited to specification of 0.9amps, and minimum LED current is 0.12amps, with nominal at 0.6amps.



Is the idea here to leave the trimpot on the board in place, then wire a parallel resistor in? Or are you still removing the trimpot?

The idea I really liked about this modification over other modifications is that the wheel on the front of the microscope still controls the output. Would I be losing that feature with your retrofit?
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mawyatt



Joined: 22 Aug 2013
Posts: 1296
Location: Clearwater

PostPosted: Sun Dec 24, 2017 11:04 am    Post subject: Reply with quote

glmory wrote:
mawyatt wrote:
Hello,

I've attached a schematic that uses the MT3608 as an LED "CURRENT MODE" controller. This requires just 2 resistors, one a current sense resistor (Rx) and another to scale the feedback potentiometer resistance to control the maximum allowed LED current (Ry).

The example shows the MT3608 with a BC-135 LED, a 5ohm 4watt resistor for Rx (can be a couple 10ohm 2watt resistors in parallel), and a 16Kohm resistor for Ry. Maximum LED current is limited to specification of 0.9amps, and minimum LED current is 0.12amps, with nominal at 0.6amps.



Is the idea here to leave the trimpot on the board in place, then wire a parallel resistor in? Or are you still removing the trimpot?

The idea I really liked about this modification over other modifications is that the wheel on the front of the microscope still controls the output. Would I be losing that feature with your retrofit?


Either way should work. I don't know what the trim pot is on the microscope, but if you have the actual value you can recalculate the resistor values.
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mawyatt



Joined: 22 Aug 2013
Posts: 1296
Location: Clearwater

PostPosted: Sun Dec 24, 2017 11:36 am    Post subject: Reply with quote

glmory wrote:
mawyatt wrote:
Hello,

I've attached a schematic that uses the MT3608 as an LED "CURRENT MODE" controller. This requires just 2 resistors, one a current sense resistor (Rx) and another to scale the feedback potentiometer resistance to control the maximum allowed LED current (Ry).

The example shows the MT3608 with a BC-135 LED, a 5ohm 4watt resistor for Rx (can be a couple 10ohm 2watt resistors in parallel), and a 16Kohm resistor for Ry. Maximum LED current is limited to specification of 0.9amps, and minimum LED current is 0.12amps, with nominal at 0.6amps.



Is the idea here to leave the trimpot on the board in place, then wire a parallel resistor in? Or are you still removing the trimpot?

The idea I really liked about this modification over other modifications is that the wheel on the front of the microscope still controls the output. Would I be losing that feature with your retrofit?


Here is more information. Assume your pot is Rpot and the 2.2K resistor in the schematic is called Rz. A little circuit analysis will give you the following:

Rz=Rpot/[Iledmax*Rx/V3 - 1].

Assume your microscope pot is 10K ohms and your LED max current is 0.9a and Rx=5ohms. V3 from the MT3608 specification sheet is 0.6V.

So Rz is 10K/[0.9*5/0.6 -1] = 1.538K ohms, select 1.5K.

If Rpot is 1K, then Rz ~ 150 ohms, Rpot is 20K ohms then Rz is 3K ohms.
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glmory



Joined: 30 Dec 2016
Posts: 25

PostPosted: Sun Dec 24, 2017 11:40 am    Post subject: Reply with quote

The potentiometer on the front of the microscope ranges from 0 (on) to ~1300 ohms.
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mawyatt



Joined: 22 Aug 2013
Posts: 1296
Location: Clearwater

PostPosted: Sun Dec 24, 2017 12:35 pm    Post subject: Reply with quote

So for 1300 ohms, Rz should be 200 ohms if Rx is 5 ohms and LED max current is 0.9amps.
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glmory



Joined: 30 Dec 2016
Posts: 25

PostPosted: Sun Dec 24, 2017 1:16 pm    Post subject: Reply with quote

Just to make sure I follow.

The idea is to build a circuit which sets the current through the 5 ohm series resistor to 0.9 amperes. It does this by monitoring the output voltage of the LED and feeding it into a voltage divider.

This is done by the following wiring into the board:

Remove the existing trimpot. Connect one side of the existing potentiometer to the low voltage side of the LED, between the LED and the resistor. The other side goes to what I labeled as pin 2.



Connect a 200 ohm resistor from pin 2 to pin 1.

I believe the 2.2k resistor must be removed from the board (or shorted across) as well.

The details of how this circuit works are still giving me trouble. Ideally I would have a circuit which has a dim LED when the potentiometer is set to 1300 ohms, and a bright LED when the potentiometer is set to 0 ohms. I will have to think a bit about whether this design would do that.

I may dig through my pile of resistors and build it to see what happens. The circuit boards are only ~$1 each on Amazon so no great loss if it burns up a few more boards.
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mawyatt



Joined: 22 Aug 2013
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Location: Clearwater

PostPosted: Sun Dec 24, 2017 1:45 pm    Post subject: Reply with quote

The general idea is to have the MAXIMUM LED current, not the nominal LED Current, when the potentiometer wiper is at the Bottom end as shown on the schematic. The MINIMUM LED current is when the wiper is at the Top of the potentiometer. The NOMINAL LED current will be somewhere in the middle of the wiper range. Potentiometers have 3 terminals, wiper and bottom and top. A Variable Resistor has 2 terminals just like a regular resistor.

For your case with the 1300 ohm potentiometer on the microscope, remove the MT3608 potentiometer.

So whatever number on you image (probably 2) connects to the feedback pin on the chip (pin 3) which I call V3, this is where the wiper of your pot connects. Whichever end of the pot you prefer for maximum LED brightness, set the wiper to this end and measure the wiper to each end resistance. The end reading near zero ohms to the wiper is the end that connects to the 200 ohm resistor which has it's other end connected to ground. Now the last end of the pot goes to the end of Rx (5 ohms) that has the LED connected as shown, the other end of Rx goes to ground. The remaining end of the LED goes to the MT3608 output as shown. Note that I refer to VIN- and VOUT- as ground, or Gnd. They are all the same connection.

You can remove the 2.2K resistor (labeled 222) or just leave it in and calculate what value Rz would be in parallel to yield 200 ohms (220 ohms).

Just make sure the LED is wired correctly as the potentiometer and Rz.

The operation is that the negative feedback pin (3 on the chip) senses the current through Rx which is the LED current by way of Ohm's Law (Iled*Rx), I call Vx. The pot provides a variable ratio less than 1 of Vx. So the voltage Vx is reduced by the pot and this reduced voltage is compared within the chip to an internal 0.6 volt reference. As the ratio is reduced the required voltage Vx is increased (pin 3 must remain at 0.6V for the negative feedback loop to remain engaged), thus increasing the LED current & brightness since Vx=Iled*Rx to keep pin 3 at 0.6 volts and remaining in regulation. This is the concept of negative feedback with Current Mode operation.
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glmory



Joined: 30 Dec 2016
Posts: 25

PostPosted: Sun Dec 24, 2017 10:14 pm    Post subject: Reply with quote

That makes more sense. When I wire it up I will either find that it works correctly, or it works backward with high power at what was previously low power. If it is the later than I just need to rewire the potentiometer. I had noted earlier that two of the three outputs are currently shorted together. So I would just need to undo that, and use the two shorted terminals.


I made a shot at sketching out my understanding of your circuit design. Is there anything I missed, or was this what you had in mind?



I still don't understand the workings of the board well enough to be convinced this wiring works(hence all the blown boards). It seems simple enough though so I may try it anyways.
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mawyatt



Joined: 22 Aug 2013
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Location: Clearwater

PostPosted: Mon Dec 25, 2017 7:18 am    Post subject: Reply with quote

Please look at the schematic diagram I posted above again, and the first schematics I posted.

Potentiometers have 3 terminals, not 2. If your microscope control only has 2 terminals available (variable resistor not a potentiometer), then you may not be able to have the control work as you want, it may work backwards. Maybe you can flip the control over if that's the case.

CAUTION: Are 1 & 2 shorted together on your PC board (looks like they are shorted by PC traces from your image)??? This is different than the schematic I posted from the MT3608 data sheet earlier (see above). If this is the case then your wiring will not work (probably burn up the board as it will force maximum output voltage). BTW this might be why you are burning up boards when you were originally testing these.

For a safer solution in case your labeled 1 & 2 are shorted by PC board, do not short out the 2.2K resistor. Remove this resistor off the board and connect the end of the 200 ohm resistor to VOUT- not to your label 1, leave the other 200 ohm resistor end connected to label 2 as you have with the wire from your 1300 ohm variable resistor. If you don't want to remove the 2.2K resistor, then change the added 200 ohm resistor to 220 ohms (the parallel combination will be 200 ohms).

Bets & Happy Holidays,

Mike
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glmory



Joined: 30 Dec 2016
Posts: 25

PostPosted: Mon Dec 25, 2017 8:02 am    Post subject: Reply with quote

In the out of factory design two terminals of the potentiometer are shorted and it is thus being used as a variable resistor. I was figuring that if I just switch the shorted terminals from the center to left to connecting the center to right it would still function. That would swap it from 0 ohms for 6 on the wheel to maximum ohms at 6 on the wheel. So I left it as a resistor on the diagram.

The shorted 1 and 2 terminals observation is more interesting though.

THIS IS WHY I KEEP BLOWING UP BOARDS!

I noted earlier when poking at the board that I measured 0 ohms between pins 1 and 2. I concluded that something, maybe some capacitance, was interfering with the measurement and proceeded without proper understanding.

Looking at it again though I see your point. They are tied together! That means that instead of the board measuring 50k off the trimpot as R1 and 50k off the trimpot as R2 it is using 50k off the trimpot as R1 and 2.2k as R2 off the stationary resistor.

Checking the math:

Vout=0.6V(1+50k/2.2k)=14.2 V

Which is reasonably close to what I see out of the box

Boards therefore blow up not because I put too small a resistor between one of those pins, but because I put too large a resistor between pins 2 and 3. The 100k resistor is right at the upper limit of what it can take. Had I used a 50k resistor I probably would have failed less boards.

With the other design this means that I need to be pulling the potentiometer and connecting R1 between pins 1/2 and 3. Then I need to pull the 2.2k resistor(or connect in parallel/series while including it in my math)
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glmory



Joined: 30 Dec 2016
Posts: 25

PostPosted: Mon Dec 25, 2017 8:24 am    Post subject: Reply with quote

Based on my current understanding of the board, these are the two circuits I am considering when the new boards come.





I need to poke through what I have and see if I have a resistor which makes the second circuit work. If the first design does not perform satisfactorily I will certainly go to it.
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mawyatt



Joined: 22 Aug 2013
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Location: Clearwater

PostPosted: Mon Dec 25, 2017 9:21 am    Post subject: Reply with quote

Your understanding of the board behavior seems to be flawed regarding the current mode use. The voltage mode using the suggested 5 ohms resistor will work, but nowhere as good as the current mode. Follow my directions and you should have a really good solution (your current mode schematic is incorrect the 200 ohm resistor is not wired correctly), otherwise good luck.

Sorry I can't spend any more time on this, I feel I've pointed you in the right direction, if you chose to deviate, well that's your decision.

Best & Happy Holidays to everyone,

Mike
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glmory



Joined: 30 Dec 2016
Posts: 25

PostPosted: Mon Dec 25, 2017 9:42 am    Post subject: Reply with quote

Looking back at your circuit it seems that pin 3 should have been left unused. Hopefully this circuit creates the correct voltage divider:
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mawyatt



Joined: 22 Aug 2013
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PostPosted: Mon Dec 25, 2017 1:37 pm    Post subject: Reply with quote

This lower schematic probably won't work properly, you will just get about 0.12a LED current regardless of where the variable resistor is set because the right side of the 200 ohm resistor needs to be connected to what I call Ground, or VOUT-. Just connect the right side of the 200 ohm resistor to VOUT- and the junction of your 1300 ohm variable resistor and the left side of the 200 ohm resistor to whatever contact that connects to the feedback pin on the chip (pin 3 on the chip, not your label #3, this looks as if it's your #1 and #2 which are apparently shorted by the PC board). Then this should work.
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glmory



Joined: 30 Dec 2016
Posts: 25

PostPosted: Tue Dec 26, 2017 11:01 pm    Post subject: Reply with quote

Thanks!

For some reason I had it in my head I was connecting to the grounded side of that 2.2k resistor. Even if I was right in that guess, it is easier to connect to Vout- so hopefully this is my last revision:



I get the boards tomorrow so I should know by the weekend whether this works. At some point I am going to need to spend some time working out how the board works. The voltage divider circuit makes sense, but I am still completely lost as to the workings of this circuit.

Conceptually I see that controlling current would work better, and it removes the need of a trimpot which would need to be tuned so it is worth a shot.
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