Phototransistor Replacement Question - Page 1
Forum section: electronics / projects

author: bostonman
author_rank: Super Contributor
timestamp: October 22, 2023, 02:28:00 pm
content: A brief background: I recently purchased a Commodore 1351 Mouse (sold as is) and discovered it didn't have vertical movement. After researching, I learned the biasing resistor(s) need to be changed to satisfy the >4V to the logic.Upon taking measurements (refer to attached schematic - referencing the top right circuit for vertical movement), I found Vcc to be about 4.76v (slightly low but fine), 4.15v between the two diodes, and 0v between the cathode of the second diode and R2. My assumption is the diode is bad, possibly open.Since I had the unit disassembled, I figured it was a good opportunity to see if the diodes and/or phototransistors have part numbers on them, and unfortunately they do not.I'd like to replace both the diodes and transistors for two reasons: one is to install new ones that will last since it's required to increase current over time to boost the gain, and second to provide others with a good replacement should they have issues too.Both the diode and transistor are in the same size package, and I'm measuring: 0.228” height, 0.178” wide, and 0.06” thick.My questions are: what wavelength should I search for, do specific parameters exist that I need to concern myself with, and has anyone tinkered with this mouse or a similar circuit?I am uncertain if I need to find a diode/transistor combo that compliment each other or I can just buy them separately.

author: mjs
author_rank: Regular Contributor
timestamp: October 22, 2023, 02:41:38 pm
content: Almost any wavelength will be ok, if it's just normal slot interrupter type.As long as current transfer ratio (i.e. how much current is passed through phototransistor for given LED current) is roughly ok, it should work. And of course polarity must be the same.I would take a look at OP550 / OP240 or similar. You might also find these by taking apart a through hole photointerrupter.

author: bostonman
author_rank: Super Contributor
timestamp: October 22, 2023, 02:50:38 pm
content: QuotePolarity of what?Unfortunately I don't have much current to play with since Vcc is only 4.76v, so I'm uncertain if enough light will be emitted at such a low current.

author: bostonman
author_rank: Super Contributor
timestamp: October 23, 2023, 02:18:45 am
content: After some investigating, I think the LTR-301 and LTR-302 are a good fit, however, a few items on the datasheet I don't understand.The datasheet lists "BIN" A-F. After some reading, it seems this is something to do with selecting the current based on the BIN sent. The baffling part is what if I want one with lower current?

author: floobydust
author_rank: Super Contributor
timestamp: October 23, 2023, 05:19:44 am
content: I think you are confusing your voltage measurements - they are at the IR emitter LED's, not the phototransistors. It looks like one of the two series LED's is open? No voltage drop/current on R2.These optical switches can be hard to work on, sometimes ambient room/bench lighting can cause photocurrent. The mouse gets nailed with ESD as well.I find physical alignment is critical and sometimes a gentle bend while measuring output voltage can tell you if they are crooked and need to be re-soldered to a better angle.I think you can view the IR LED's with a cellphone cam and maybe find a weak LED.For phototransistors, I see a A0101930 or A101910 Rev. A with the photo-transistor in an emitter-follower config and 15k to GND. Is that what you have, or what's the mod? It's E-C pinout.That would be 0.17mA which is not much (assuming MCU is CMOS). It also depends on how wide the wheel slots are. Narrow ones can block (waste) a lot of light.Mouser LTR-301 spec. is BIN G sorta like OP550B, high sens. with min/max.Digi-Key, LCSC LTR-301 spec. is BIN E, like OP550C, middle sensitivity.LTE-302 IR emitter also ranked according to output.

author: mjs
author_rank: Regular Contributor
timestamp: October 23, 2023, 06:13:51 am
content: By polarities I mean pinout, i.e. anode/cathode for LED and emitter/collector for phototransistor must be the same way in the package. If they're swapped, the won't work.Bin means how sensitive the transistor is, i.e. how much current for given amount of IR radiation is passed. This is typically compendated for by selecting suitable load resistors, so that's not an important parameter. Usually more sensitive is better, but usually you've got to live with what you get from the store and can't pick the bin unless buying huge quantities.

author: bostonman
author_rank: Super Contributor
timestamp: October 23, 2023, 12:50:47 pm
content: The diode and transistor both sit in a plastic housing keeping them in (hopefully) alignment.As mentioned, yes, it does appear the second diode is open. I could have tried changing the resistors (as discussed on repair sites) to try boosting current (assuming the diode is just very weak and not open), but felt this was a good opportunity to replace them rather than boost current, drive them harder, and then they wear out faster; although I don't see myself using the mouse much as it's more of a collectable item.The voltage needed for the IC is >4v. From reading, the diode/transistors degrade over time, the voltage to the IC dips below 4v, and movement fails in that direction. Changing the resistors to boost the voltage is the fix, but again, I felt boosting voltage will only increase the rate they degrade.Unless I missed something, I will buy the LTR301 and LTR302 and replace all eight (four and four).

author: floobydust
author_rank: Super Contributor
timestamp: October 23, 2023, 05:05:26 pm
content: Sounds like a good plan. The binning according to sensitivity I can't make much sense out of because these are run at low collector current anyway, so I doubt it matters.Don't forget to check the electrolytic caps, I had those go bad on a cheap chinese Razer mouse and it wasn't anywhere near as old as this.I'm also repairing an encoder that uses these "side looking" phototransistor and IR LED. I find the LED's age and go dim.If the alignment is not great, the mouse movement will be flaky. When one phototransistor/LED is blocked, the other should not be, for quadrature.So after you put them in, maybe check the emitter voltage as you slowly turn the wheel. They might need a solder reheat and nudge.Because the plastic frame I doubt is that great for alignment.Also, these parts are fragile, they don't take lead bending or they just break apart.

author: bostonman
author_rank: Super Contributor
timestamp: October 24, 2023, 01:08:32 pm
content: I ordered the LTR and LTE 301 and 302 (previously I accidentally put LTR for both parts and realized one is 'R' for receiver and 'E' for emitting.Hopefully the parts arrive before this weekend so I can install them and update this thread.

author: bostonman
author_rank: Super Contributor
timestamp: October 30, 2023, 12:53:33 am
content: I installed the replacement diodes and photo transistors, and the mouse seems to work. It was a quick test run before I cut the leads after soldering (just in case I needed to re-position the components) but I got movement in both directions.Just need to clean the flux residue and reassemble before testing again.

author: bostonman
author_rank: Super Contributor
timestamp: November 01, 2023, 01:25:21 am
content: Just an update, the mouse is reassembled and works great.Prior to reassembling, I took measurements (the emitters of the photo transistors should output >4v because that's the minimum 'high' for the IC they feed). I measured 4.5v on all four photo transistors.As always, thanks for the help.

author: bostonman
author_rank: Super Contributor
timestamp: November 01, 2023, 02:04:19 am
content: Actually, I have a question about the transfer of these two components.I'm measuring 19.4mA through one of the diodes and 1.16mA through the photo transistor that is in line with the diode (actually, all the voltages and currents are almost equal).Due to being unfamiliar with these circuits and devices, I'm trying to understand how to calculate the transfer ratio based on the datasheets. Actually, the datasheet for the LED is confusing. Figure 2 shows the graph for 50mA, how do I figure out what the graph is at the measured 20mA (rounding up from 19.4mA)?Also, figures 4, 5, and 6, are missing units on the axis, so I'm confused on what these are telling me.

author: bostonman
author_rank: Super Contributor
timestamp: November 01, 2023, 04:55:12 pm
content: Just realized, I didn't attach the datasheet:https://optoelectronics.liteon.com/upload/download/DS-50-93-0013/LTR-301.pdfAlso, just an edit. I stated all voltages and currents are almost equal. I meant that for reach diode pared with a photo transistor (there are x4, x2 for the vertical and x2 for the horizontal), the voltages and currents through each are almost all equal.

author: floobydust
author_rank: Super Contributor
timestamp: November 04, 2023, 07:23:47 pm
content: Light has a complicated measurement system, its own units some very old like candela, some 3D some 2D. I'm not an expert on them. I don't think you can really go much further into analysis without some way to measure that.The LTE-302 IR LED spec ref. is 20mA and radiant intensity given in (3D units) mW/sr, steradians. The graphs are relative - look at the 1.0 point and then to the other axis to see what is the reference. So Fig. 5 relative to IF20mA they show intensity as linear with IF. A "1.2" would be 120% for example.The LTR-301 IR phototransistor spec is referenced to 1mW/cm2(2D area). For that you can see the many different resulting collector currents depending on the part's bin #.Then there is the distance between the two, the slot width on the wheel etc. that would make a long physics exercise.

author: bostonman
author_rank: Super Contributor
timestamp: November 05, 2023, 02:20:36 pm
content: QuoteHow do you know "1.2" is 120% and 120% of what?Articles online state I can use my camera phone to see the LEDs lighting. I've tried multiple times and my phone camera isn't picking up the light. Am I doing something wrong?

author: floobydust
author_rank: Super Contributor
timestamp: November 05, 2023, 06:34:04 pm
content: For the IR LED emitter, they mention the reference forward current on the chart axis label (20mA).For the phototransistor, they only mention it's "relative collector current" because of the binned range the datasheet gives. So it's relative to the part (bin#) and range for it.You could see what an IR remote control does as far as the cellphone camera, if it can see it. Try a dim room.The IR LED's used in the mouse have a very narrow beam and off-axis it might be seen as too dim or not seen at all.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: November 06, 2023, 05:43:53 am
content: Quote from: bostonman on November 01, 2023, 04:55:12 pmThat's what one would hopeQuote from: bostonman on November 01, 2023, 02:04:19 amThe CTR here is simply 1.16/19.4 = ~ 6% - that's quite good.Opto datasheets can be confusing, and some only give end to end CTRs for paired devices.If you want to see end-to-end specs, look at the slotted interrupter data sheets.

author: bostonman
author_rank: Super Contributor
timestamp: November 07, 2023, 02:40:49 pm
content: Quote6% is considered good? Does this mean 94% of the light is lost in the air?I'm still unable to see the LEDs with a phone camera. I'm thinking of buying one of those cards to see 940nm, however, I'm finding Thorlabs to be extremely expensive (up around $90) and the ones on Amazon have poor reviews.Does anyone have a recommendation?

author: mjs
author_rank: Regular Contributor
timestamp: November 07, 2023, 03:50:58 pm
content: Yes, most of the light is lost and 6% is ok..To see IR up to maybe 940-1000nm, get any USB/wifi 'night' camera with IR LEDs from Amazon/eBay and block or disconnect the LEDs. These cameras typically have either removed or switchable IR block filter in front of the sensor.Note that the IR light is directed by the small lens structure and may not be well visible from the side.

author: bostonman
author_rank: Super Contributor
timestamp: November 08, 2023, 12:26:17 am
content: Not sure it's worth spending this much just to see the light. I thought those cards would be a few bucks and maybe I could purchase an assortment for various wavelengths for possible future needs.Should these parts have had the 'BIN' number on the label? I confirmed that neither the diode or photo transistor list the 'BIN', so one wouldn't know the radiant intensity based on the datasheet.I have a silly question. Should the phone camera even pick up the light from the diode? In the attached spectrum, it shows that 940nm is in the black area, not the red. Does this mean by seeing black on my phone camera that I'm seeing the actual "light"?

author: mjs
author_rank: Regular Contributor
timestamp: November 08, 2023, 02:47:27 pm
content: Depends on the IR filter on the camera, there are big differences. I would not expect very good sensitivity for 930nm in any case - try shining the LED directly to the camera - from side the light is much dimmer.I've seen bin code in small quantity LEDs very rarely. Usually it's in the reel and distributors don't usually bother with it.

author: bostonman
author_rank: Super Contributor
timestamp: November 09, 2023, 03:25:07 am
content: I've tried aiming my phone at all directions without seeing anything. Most likely I'll just give up at this point.

author: floobydust
author_rank: Super Contributor
timestamp: November 09, 2023, 04:01:35 am
content: As I mentioned, try another IR source like a TV remote control and see what your phone cam shows.

author: bostonman
author_rank: Super Contributor
timestamp: November 10, 2023, 03:32:17 am
content: I should have included that I did try your advice.Funny enough is that my cable remote lights one of the buttons whenever any button is pressed. When I saw the bright shiny red LED in my phone camera, I was blown away; until I realized it wasn't the IR.After I tried another remote without any visible light and saw the IR light up on the phone camera.