6V LED driver - old microscope upgrade - Page 1
Forum section: electronics / beginners

author: Tsou
author_rank: Contributor
timestamp: November 04, 2018, 02:27:48 pm
content: Hello!I've an old microscope with an halogen bulb but I'd like to replace it with something more recent, a LED light.The halogen bulb is taking 6V and a potentiometer makes this voltage change between 0V and 6V in order to make the halogen bulb more or less bright. (Image of the microscope below)So, I read some schematics (my main field is Biomedical Sciences, not Electronics) and did my own (see LED.png). I also did the layout of the board (see LED-layout.png).I've used the parts I've laying around, they may not be the best components to use...The zener diode is a 3.3V, with the 1 ohm resistor it makes the current go up to 3.3A max. The LED takes 6V and max 20W, so I should stay within the limits (3.3A x 6V = 19.8W).My plan is to use the 6V going in the potentiometer of the microscope to drive my circuit, and use the in-frame potentiometer to vary the 6V going in the op-amp.I've added some component, in regards to the advices given in this post :https://www.eevblog.com/forum/projects/linear-led-driver-with-mosfet/msg1796558/#msg1796558Do you think my design is alright? Does it need something else?Thank you!

author: spec
author_rank: Frequent Contributor
timestamp: November 04, 2018, 03:59:17 pm
content: + TsouHell, this has been a long journey for you: I have just read through your other thread on this subject too.About your latest circuit, attached above, it is a classic constant current generator and as such is a good architecture which has been used countless times before.There are a few detailed issues with your implementation though. I will do an analysis and report back.But just for starters, the TIL072 is not suitable for negative rail input signals and the frequency stabilizing capacitor value is way to low for safety.

author: flynwill
author_rank: Regular Contributor
timestamp: November 04, 2018, 04:00:32 pm
content: Well it's not going to work off 6 volts.According to the CREE datasheet the LED typically drops ~6.1V at 3.3 A. Assuming your FET is fully saturated you still the 3.3V being dropped by your sense resistor you will need at least a 10V supply.Second problem: At 3.3A & 3.3V your sense resister is going to be dissipating ~10W. From your layout I would guess that you've not left space for more than a 1/4W resistor.With a 6V available supply you probably need to be looking at 3V LED modules, and a much much smaller value for your sense resistor (I'd suggest 0.051 ohm 1/2 Watt resistor). Also if you stick with the linear design your FET is going to need a heat sink. A switching driver will be a lot more efficient but also more design work.Also you probably don't need a 20W LED. How may Watts is the halogen bulb you are replacing? I would wager a 1 or 2 Watt LED will be plenty of light.

author: spec
author_rank: Frequent Contributor
timestamp: November 04, 2018, 06:55:38 pm
content: + TsouCould you build the attached circuit for feasibility proving.In your schematic you do not show the value of the potentiometer, so I have assumed 10K. The value is not that critical, but could you post the value you are using.If you build the feasibility circuit use low ESR capacitors. The values of the capacitors are not critical, but the leads should be as short as possible and connected as close as possible to the MOSFET. Likewise with the gate resistor, R1.Before you turn the feasibility circuit on, make sure that the wiper of the potentiometer is fully at the 0V end of its track.When you turn the circuit on, the LED should be off, but at about halfway travel of the wiper, with any luck, the LED should start illuminating.Carry on turning the potentiometer until the wiper is fully at the 6V end. The LED brightness should increase. With the wiper fully at the 6V end, note if the LED is bright enough for your purposes.Be interested to hear the results

author: spec
author_rank: Frequent Contributor
timestamp: November 04, 2018, 07:00:59 pm
content: Quote from: flynwill on November 04, 2018, 04:00:32 pmI had similar thoughts.

author: mariush
author_rank: Super Contributor
timestamp: November 04, 2018, 07:41:43 pm
content: Tsou, why don't you use a LED driver?Look for example at something like this, Monolithic Power MP24833:https://www.digikey.com/product-detail/en/monolithic-power-systems-inc/MP24833-AGN/1589-1896-ND/7361862It's a 3A max led driver chip - yes I know your LED is ~ 3.3A but really, 3A is still going to be bright enough and you're only gonna extend the life of your led.The chip also has an ENable pin, which supports analogue or PWM dimming. Basically, the chip turns on the LED once voltage on this pin is above 0.67v and you can control the brightness percentage by keeping the voltage between 0.67v and 1.35v ... so 0.67v is nearly off, 1.35v is full brighnessSo how about this... you have the transformer output, you have the bridge rectifier, add some capacitor to smooth out the output of that rectifier and you probably have around 6v or more.Now you can either use tiny linear regulator to get something like less than 1.5v (there are linear regulators with fixed output voltage) or you can use a couple of resistors (voltage dividers) to get something close to 1.35v - now you can connect ground and 1.35v+ to your potentiometer and the wiper of your potentiometer goes to the ENable pin of this led driver.So you have brightness adjustment with a max cap at 3A.The led driver works at 200kHz , so even if it uses PWM to adjust brightness, it will probably not flicker significantly.

author: Tsou
author_rank: Contributor
timestamp: November 04, 2018, 07:56:45 pm
content: Quote from: flynwill on November 04, 2018, 04:00:32 pmWhat a big mistake I did there, forgetting the drop voltage in the resistor...For the 10W, I was planning to use a 20W 1 ohm resistor with aluminum shell.The halogen bulb is a 20W 460lm one, and unfortunately I tried some conversions between lumen and candella to see if small LED would be enough, and it seems that it wont.So I check the datasheet, and I think that I don't need more than 2.5A, which is already ~120% total flux (and the total flux is ~1500lm, so 3 times more than my actual one).With 2.5A and a 0.05ohm 1% resistor I'd have 0.125V drop in the resistor, which is not a lot, plus at this current the voltage across the LED is 5.9V. This make 6V total, so it should be good.Given the fact that I'd be mostly at around ~33% total flux (like the actual bulb), this is 0.03V and 0.6A, so no problem with the 6V.The only problem I've is : I need 0.125V between the resistor, so I need 0.125V in the positive input of the opamp. To do so, is it possible to put a resistor divider in parallel with a zener? If it is, I'd need a 1:20 conversion for a 3V zener and a 1:24 conversion for a 3.3V zener.The other option is to take off all the electrical parts of the microscope, put a new 9V power supply and use the same schematic, with a 1ohm 20W resistor (3V x 2.5A = 7.5W to dissipate).I don't know if the LED is very sensible to the voltage, because I'm not sure the power supply of the microscope gives a stable 6V... Enough for the 6V halogen bulb to work at least...

author: Tsou
author_rank: Contributor
timestamp: November 04, 2018, 08:05:55 pm
content: Mariush, that's a nice way of doing things! Probably more efficient, simpler, ...I didn't think about it because I was afraid of flickering (I'll be using my camera to take pictures and videos) and I searched about solutions and only found the one I posted here.So with your solution, the 'smooth' 6V I get from the rectifier goes on a 6V line on my board, to drive the LED and also goes in a linear rectifier on the board to get 1.5V, for the potentiometer? Almost like I did on my schematic?Edit : The only problem I see is that the design is more difficult and need more parts, I think I'll try both solutions if they work both and design this one a little bit later, 'cause it'll take more time to do and I don't have the parts needed for now

author: Zero999
author_rank: Super Contributor
timestamp: November 04, 2018, 08:30:46 pm
content: What voltage does the 6V microscope power supply actually give? Under light loads, it's likely to be higher than it would if it were fully loaded.Another issue is the TL072 will not work with its inputs that close to the negative rail, so when the potentiometer is adjusted to a low power setting it will start to behave unpredictably. Swap the TL072 for a single supply op-amp such as the LM358.Reduce the voltage reference and R3 to compensate.

author: Tsou
author_rank: Contributor
timestamp: November 04, 2018, 09:27:34 pm
content: @spec I'll try your circuit when I can, but I don't have the LED for now, neither my microscope near me@Hero999 I don't know the voltage with no load, I didn't try that and I don't have the microscope to try it, but I read somewhere that "The 0 - 6V output of the Nikon's internal PSU is DC so could be the basis of an LED supply, but a suitable dropping resistor is needed as it is 10V open circuit and would immediately blow an LED typically rated for 4.5V max" (http://www.microscopy-uk.org.uk/mag/indexmag.html?http://www.microscopy-uk.org.uk/mag/artaug07/dw-labophot.html).So I guess we can say that the voltage under light/no loads is 10V, I don't know if this could have any effect on the circuit...I changed the reference of the opamp, the value of R3 and C1 and I add a resistor divider to have ~0.125V on the positive input of the opamp (see LED2.png)I'll also need to add some capacitors to smooth the 6V supply

author: spec
author_rank: Frequent Contributor
timestamp: November 04, 2018, 09:48:24 pm
content: Quote from: Tsou on November 04, 2018, 09:27:34 pmOK

author: Zero999
author_rank: Super Contributor
timestamp: November 04, 2018, 09:55:35 pm
content: Quote from: spec on November 04, 2018, 06:55:38 pmBe careful with that circuit. It might blow up the LED with the power cranked up to the maximum, as there's nothing to limit the LED current, other than the on resistance of the MOSFET, especially if the 6V supply is higher than that.Quote from: Tsou on November 04, 2018, 09:27:34 pmDon't worry about the open circuit voltage, the LED needs the forward current to be regulated. not the voltage, which is what this circuit does.When you calculated the values of R5 and R6, you missed that RV1 is in parallel with R6, so the output voltage will be lower than that. If you don't know the value of RV1, then you could use the spare op-amp on the LM358 IC as a unity gain buffer between R6 and RV1.

author: flynwill
author_rank: Regular Contributor
timestamp: November 04, 2018, 10:41:57 pm
content: Indeed this is much closer. I would strongly recommend characterizing that "6V power supply" in the existing scope if you can. I may be just a small transformer and rectifier with no filter capacitor. In that case you may want either add a filter cap to the design, or a diode in series with VCC to the opamp which will let you get away with a smaller capacitor as it now only has to filter the op amp power supply.While you are characterizing the PS, do check out the resistance of the existing pot. It might be quite low, and if you know it accurately you can eliminate R6.

author: spec
author_rank: Frequent Contributor
timestamp: November 04, 2018, 11:07:25 pm
content: Quote from: Hero999 on November 04, 2018, 09:55:35 pmYes, I did say to ensure that the pot is at the 0V end before switching the supply on and to gradually move the wiper towards the 6V supply line end. The current will be limited by five factors: the slope resistance of the LED which is significant as shown on the LED data sheet, the Rdss of the MOSFET which I think has a positive temperature coefficient, temperature effect on VGDth, which has a positive temperature coefficient, the impedance of the 6V supply line, and the current limitations of the 6V power supply.By the way, the purpose of the feasibility circuit is to establish, one way or another, if the approach of using the 6V Cree LED on the microscope 6V power supply is feasible. That is the fundamental enabling factor that needs to be established before building more complex circuits which will just cloud the issue at this stage.On a similar theme, it seems, from the above posts that the microscope 6V supply itself is suspect and at the very least will need extra smoothing/reservoir capacitors.

author: spec
author_rank: Frequent Contributor
timestamp: November 04, 2018, 11:42:30 pm
content: Quote from: Tsou on November 04, 2018, 09:27:34 pm100nF rather than the previous 100p for C1 looks more like it, especially with the relatively slow LM358.But I suspect that a bit more work will be required to compensate the circuit.

author: ratboy
author_rank: Newbie
timestamp: November 04, 2018, 11:58:22 pm
content: I retro fitted my old Reichert MicroStar IV with Led lighting using something likeLed:https://www.ebay.com/sch/sis.html?_nkw=6V%2F12V+LG+3535++4Chip+LED+Emitter+instead+of+CREE+XPG++LED+with+20MM+Copper+Star&_id=262224123950&&_trksid=p2057872.m2749.l2658Driver:https://www.ebay.com/itm/5A-Constant-Current-Constant-LED-Driver-Lithium-ion-Batteries-Charging-Module-/321905738478?hash=item4af313c2eeIf i remember correctly, i mounted the Led on a heat sink and hacked away the existing halogen base and squeezed it it.The led driver board, one of the fine tuning pot was removed and replaced using the wiring/pot of the existing controls.You can play with it while outside and see which pot is the one.The range of the existing pot + fix resistor, is sufficient to go from very dim to very bright.can't remember how i powered the board though, but the scope still uses the existing 240v in to power the whole thing, so i guess i used the existing transformer and switch.The above links are for similar items as the original ebay links are obsolete.

author: Zero999
author_rank: Super Contributor
timestamp: November 05, 2018, 09:46:21 am
content: Quote from: spec on November 04, 2018, 11:07:25 pmAnother issue is the forward voltage of the LED has a negative temperature coefficient which will cause the current to increase, as it warms up. The LED current should be measured using a meter and the potentiometer adjusted downwards if it starts to get close to the LED's maximum rating.

author: Tsou
author_rank: Contributor
timestamp: November 05, 2018, 04:43:20 pm
content: Quote from: Hero999 on November 04, 2018, 09:55:35 pmI did some improvements with your advices, I put an unity gain buffer, I also changed the zener diode for a voltage regulator and a resistor divider. I also add some capacitors to filter the supply voltage, because there's clearly no capacitors in the microscope, the supply voltage is coming in the potentiometer right from the rectifier.I'm also thinking, wouldn't it be more correct to just put away all the electrical parts and take a new 6V power supply, so I have something more recent and so I'm sure what's inside the microscope?I plan to make an epi-fluorescence module, so it could be interesting to get 6V from the power supply to power the fluorescent diodes, without the need to have multiple power supplies. A good 5A 6V power supply could be just enough I guess.Or I should get a more powerful one, like a 9V? But this mean I've to dissipate the extra 3V in the resistor, so this seems not to be a better solutionQuote from: ratboy on November 04, 2018, 11:58:22 pmThank you for your advice, I think I'll do my own circuit and also buy the circuit you showed, in case it doesn't work as good...

author: Zero999
author_rank: Super Contributor
timestamp: November 07, 2018, 10:19:40 am
content: I think a new 6V power supply is a sensible option.If there are no smoothing capacitors, i.e. the 6V PSU is just a rectifier on a transformer, then a 220µF capacitor will be far too small to provide adequate smoothing. You probably need 10mF, i.e. 10000µF or even more.

author: Tsou
author_rank: Contributor
timestamp: November 07, 2018, 07:30:01 pm
content: Quote from: Hero999 on November 07, 2018, 10:19:40 amOuf, 10mF, that's a lot... I guess a new PSU would be better...I think I've found a good solution, I can put a new 12V PSU, get a 12V LED and use my board with 12V. So, R1 = 22k, I get 0.65V between R1 and R2, this means 1.3A across R5. We have 0.05 * 1.3 = 0.65V across R5 and 0.845W to dissipate, with a 1W resistor it should be ok. If it's too high I can lower the max current, these LEDs seem to be powerful enough. Someone on an microscopy forum told me that 10W LED are powerful enough for microscopy, this mean less than one amp, so I can lower the current if needed.LM358 and P16NF06L can handle 12V, but the AMS1117 is rated to 15V max, and 12V in usual conditions, so it's at its limits, but it should be ok if I'm right?

author: jmp76
author_rank: Newbie
timestamp: November 08, 2018, 12:04:36 pm
content: Hi,THE solution to drive a LED till 12.5 v à 3A is here:http://pichotjm.free.fr/Docu/Microscopie/Eclairage/_Modif_BHS.phpWith this solution you will be able to drive LEDs with low, medium and high currents as necessary. Solutions are given for auto fluorescence with green light (red vision) or with Royal Blue light (green vision). Solutions are given for epi scopy (reflection) and dia scopie (pass thru).Currents are delivered with 1024 step values. And are OK for LPA (polarisation) or DIC (differential contast Nomarski) in bio or metal options. You can do DIC photos @800 ISO and 1/800 s !!! (these functions are in research microscopes, very expensive ones: > $5000)Text is in french, but you can use Google Translate. May be no need to translate...Regards

author: Tsou
author_rank: Contributor
timestamp: November 09, 2018, 03:40:32 pm
content: I've made some improvements in the design of my PCB. The trace between R4 and R5 was too long and not large enough, so it has almost 20% of R5's resistance.Here I did a copper filled zone which include R5, R4 and the source of the transistor. I also shorten and thicken the trace between the opamp output and R4 but I'm not sure if it's necessary...Here's the new design, if it can help anyone else...

author: Zero999
author_rank: Super Contributor
timestamp: November 10, 2018, 10:57:59 pm
content: Quote from: Tsou on November 09, 2018, 03:40:32 pmWhat provision have you make for heat singing the MOSFET? It will dissipate a considerable amount of power and I don't see any room for a heat sink.The suggestion made by the previous poster, of replacing the 6V power supply with an LED driver was a good one. I understand you want to use the original potentiometer to control the brightness, which rules lots of options out. Do you know the value of the resistor?Have you considered using a switched mode power supply? It's fairly easy to make an LED driver with a comparator IC, such as the LM393. That would eliminate the heat sink issue.Here's an example of am :LED driver kit which uses the LM393. It should be fairly easy to modify it to suit your needs.https://www.velleman.eu/products/view/?id=366712https://www.velleman.eu/downloads/0/illustrated/illustrated_assembly_manual_k8071_rev1.pdf

author: Tsou
author_rank: Contributor
timestamp: November 10, 2018, 11:48:46 pm
content: I'll buy a ready-made circuit to power the LED in case my circuit does not work, but I'd also like to make this circuit, for education purpose, to go from design to final product.For the MOSFET, if I have 1.3A, a Rds_on of 0.1ohm max, I have 1.3A*0.1ohm = 0.13V then 0.13V * 1.3A = 0.169W. The 'Thermal resistance junction-ambient max' is 62.5°C/W so I have 10.6°C more than the ambient temperature, no? I can also go for a IRF3205 which has 0.008ohm of Rds_on to lower the power dissipation in the mosfet

author: Zero999
author_rank: Super Contributor
timestamp: November 11, 2018, 07:37:42 pm
content: Quote from: Tsou on November 10, 2018, 11:48:46 pmYou've got the power calculation completely wrong.The voltage across the MOSFET is not determined by the on resistance given on the data sheet, but the supply voltage and voltage drop across the LED.