led, surface mount resistors.

oldspark wrote: The main risk with this operation (other than driving a short or low resistance) is probably the soldering. Rather than "replacing" the SMD resistors, you may want to add leaded resistors to the new LEDs. These may include the SMD value, else with a shorting link across the SMD/SMR. Or just add a dimmer ... (LOL)

Yes, this was part of my original question. To remove the current surface mount resistor or add another leaded resistor to the LED. Definitely quite a bit of soldering and solder removal. 20 LEDs and 20 resistors equals 80 solders/removals. Using the pot/dimmer inside the vehicle is not an option. The board is actually powered through a UART B serial data line. Granted, I do have the extra circuit board and I am really in no way worried about ruining the board, I would still like to the see the project be a success. I could also use the little silicon cap condom things to change the color, but what is the fun in that? I was thinking about what the first two posters had claimed that the resistors do not matter that would be okay using the current 147 ohm surface mount resistor.So, I did the math... I came out with I = (14.4v-3.4v)/147ohms which gave me .07 or 70 mA. Is this not way too much compared to the suggested 20mA. Will this dramatically reduce the life of the LEDs? I am new to this so, if the math is wrong or theory, please improve the learning curve. TIA

sallc5 wrote: I came out with I = (14.4v-3.4v)/147ohms which gave me .07 or 70 mA.

EXCELLENT! That is done correctly - at least "by the book". But I'll say more on that later.... But look at this the other way.... You are saying the 147R (R means Ohm - it's used instead of the Omega symbol) resistors mean 70mA through a 3.4V LED. (That is correct). Therefore your existing LEDs are also taking 70mA unless they are (say) 1.7V LEDs hence (14.4v-1.7v)/147R = 12.7/147 = 86mA. Do you see the issue - LEDs do not take that much (except for v.high intensity etc). So there is some "circuit issue" - it may be multiplexed, use series LEDs, or use a lower supply voltage (maybe 5V regulated - 5V/147R means 34mA MAXIMUM - less if you add LEDS - eg 22mA & 10mA for 1.7 & 3.4V LEDS). But yet again - is it important? Let's consider it this way.... Your new LEDs are higher voltage than the originals - or so I presume. They are unlikely to be higher voltage. If your new LEDs are the same voltage (as the original), they will draw/take the same current. If they are higher, they will take LESS current. Hence it should be safe simply to substitute the new LEDs. This assumes each LED replaces each original LED. (Otherwise you will need to figure out how the LEDS are connected - is it one resister per LED etc. If each LED is an indicator (bar graph etc) then yes - probably. If it's for illumination, then some LEDs are probably in series with a single resistor.) Oh geez! Too much? In summary - the circuit is probably more complex than you think, but it shouldn't matter. The new LEDs may not work or be bright enough, but since they shouldn't take MORE current than the old LEDs, the circuit should not blow up. The blues may not be bright enough because they may need the higher voltage and higher current. Besides - we are less sensitive to blue than green & red (& hence white). (Yes - aren't those blue headlights so much brighter than white - rotflmfho!!) The filters may work if the originals are white LEDs. For colored LEDs they usually don't (ie, red, green, yellow etc LEDs generally put out ONLY those colors so any "other color" filter blocks their light output. Regarding the 14.4V LED supply - that is unlikely - unless they dim & brighten with vehicle RPM. For various reasons - including the need for circuits to operate from (say) an 8V-10V to 16V vehicle supply - their supply will be protected and regulated. 5V is a common voltage for LED indicators/meters etc for historical reasons, plus reliable regulation at 5V even if the vehicle dips to 8V (eg, during cranking). Not that this voltage variation applies to all systems, but since equipment tends to be made to suit "all" applications.... Anyhow, there are various ways of checking the above, but it can get complicated. Especially if the LEDs are muliplexed (hence very high currents - maybe 70mA - but only for 10% to 30% of the time etc). And I was dumb enough to say above "But I'll say more on that later... (referring to your calcs "by the book"). So solely due to commitment (and not my need to ramble).... You will probably find that - especially for LEDs - near enough is close enough. I used to do my (12-1.7)V/20mA for cars etc. That became (12-2)V/mA = 10V/mA etc. Then like others I ignored the LEDs ~2V drop (for 12V). Why? Because: : 2V in 12V is under 20% error - not much (for LEDs). : 12V is really 13.8V typically; often 14.4V or more (ie, ~20% error!) : resistors came in "preferred values" bands of ~20% gap - ie, 10, 12, 15, 18, 22, 27 ec Ohms. Your 147R would have been 150R (else 120R or 180R) : resistors has a tolerance of 10% (so 150R could be 135-165R) etc etc etc. Although much has changed - like although "preferred values" may still be the E10 scale as above, their (manufacturing) tolerance is now 1% (not 10% or 5% etc), the approximations haven't. Why? Because LEDs tolerate and operate under a wide voltage variation. [This is what that other-site's "discussion" was about - someone claiming a "non-tolerance" (at the risk of thermal runaway) but the supplying all the evidence to prove me correct - though they tried to provide a "temperature measuring" document as a "proof of thermal runaway" for "LEDs without resistors".]       Anyhow, because LEDs can now be 3.4V, maybe their voltage can't be ignored. (Though you see the relative similarity between 10, 20 & 30mA when ignoring 1.7V & 3.4V in a mere 5V supply!). But I'd do your 14.4v-3.4v)/147ohms as (15-3)/150 = 15/150 = 100mA (else maybe 12/150 = 80mA). A better example may be calculating the resistor assuming 14.4V etc: R = (14.4-3.4)V/20mA = 550R => 560R (preferred E10 value); or R = ~10/0.2 = 500 => 470R or 560R (preferred E10 values). Rats - that last shows a typical problem. One might normally chose the 479R cos it's closer. Or would one chose the 560R anyhow because it's "safer" (less current), or because we know we rounded the voltage down hence getting a lower resistance than reality, or we allow for higher than 14.4V, etc etc. It is an experience thing.... If in doubt, do it by the book. But do "rough checks" with simple figures as I have shown. You may find you are out by orders of magnitude (ie, 56R instead of 560R), or that rough is good enough. Damn I hate commitments.    Maybe I should have breakfast eh? The I'll re-read by ramble. (Why can't we post-edit after later replies....). If it makes sense, maybe you can confirm the power (wattage) rating of the resistors? If the original LEDs are 1.7V from a 14.4V supply, those 147R resistors must be 1W (if not multiplexed) - and I doubt that they are! But you said it was powered from a UART serial line.... Do you mean the board is signalled via the UART (or the LEDs are controlled via the UART... nah) - because AFAIK UARTs are not for powering - only signalling (and using a serial signal for multiplexing or powering ... ) Got it! The UART may have a regulated output - eg - a 5V supply (ie serial-in to digital-logic out - eg, RS232 to CPU etc). Is that what is done? That makes sense with the existing "low" 147R. If the 5V is only powering the LEDs, a dimmer is possible (though probably undesirable).

I read through all of that and yes i do understand most of it. Thanks! Now I do think you are onto something with UART. Maybe it is regulated. Would a snippet from the FSM help you help me, haha? Here is a description of the hvac controls/driver info center: The instrument cluster converts the trip computer information to the driver information center (DIC) on the universal asynchronous receiving/transmitting (UART) B serial data line. A discrete input sends the information to the instrument cluster for calculation. The instrument cluster processes the raw data. The instrument cluster uses an internal time keeping function in order to determine the selected information. Schematics: (but not for the board itself...) All of the LEDs have an individual surface mount resistor. The LEDs themselves also never dim. There are other twist-lock style bulbs which are NOT leds. I am able to dim these bulbs in conjunction with the gauge cluster. This really has nothing to do with this though I guess. How would you suggest I test to see how many volts the factory LEDs are getting. So that I can confirm the voltage. I am guessing multi-meter with the car on...

The LED must be lit when testing the voltage.  Simply place one lead on one connection and the other lead on the other connection.  What is the DC voltage reading.  Also check the voltag drop across the resistor.  One lead on one end of the resistor, and the other lead on the other end of the resistor.  The LED must be lit up.

Your assumption is SPOT ON. I found the correct diagram. This definitely changes the game. So, I need to redo some math. Let me know what you think in the mean time.

My apologies this one is correct diagram. I need to postmoreto be able to edit, darn rookie. :( 5V as expected. So Would I also be right in assuming I can just pop in some twist-lock leds? Let's not go there quite yet though...

So, with this diagram clearing stating that the circuit board receives 5v, it is basically a waste of time to remove the 147R resistors. https://i45.tinypic.com/2lawv0k.png R = (5v-3.4)/.02 R = 82 So, I would "by the book" need an 82R 1/8W resistor. If anything the LEDS will not be as bright as they could be but more than likely bright enough for this application and not hard to look at. Is this a good assumption?

Don't ya hate "Rookies" that aren't rookies?! You are spot on (IMO!). I get 80R (not 82R), but big deal. 80/82 means 82R in preferred values. P = iiR = .02 x .02 x 82 = .0328W (check VI = 2.6x.02 =0.052 wth? P-vv/R = 2.6x2.6/82=.082 wth#2?) Oh well, seems under 1/10thWatt etc. Maybe a good way to change the SMD 147R to 82R is to parallel a 180R (185R) resistor with the 147R. Just an option. Otherwise I'd either short the 147Rs with a link, else take a flying lead from the "start" of the 147R to the 82R & LED. But just substitute the new LEDs for now. They might be bright enough (though probably not???). If not, a temp "parallel" 180R or "bypassing" 82R can be used to see the difference. (Maybe blue is too dark?) But I'm happy.... You seem to agree that you "can't blow the PCB" etc by merely substituting the LEDs. And it seems the new LEDs are NOT higher current than the originals. As to bluddy MLAs, FSM is Full Service Manual? No - Field...? (I knew my TLAs once upon a time - until I cross-technology'd) HVAC is not HiVoltage AC - it's Heinous Voltage Air-Con etc. It's important to know exactly what they mean..... LOL! (Save the FSM until AFTER we blow it up. That's what engineers & real blokes do!) Thanks for confirming my thoughts/suggestions. If it now blows up, it means YOU didn't confirm enough! LOL! (BTW - most boards can be repaired assuming non-proprietry etc. I assume to UARTs are still available. Only older stuff can be a problem - like trying to decode EFI/ECUs from the 1980s - you can't get the chip information!)

Yes, I just rounded up to the next value. My theory is to test it with a 6v battery with pot set to 150R.This should give me the same output of light as if using the 5v source at 82R. Then I can see whether or not it is as bright as I would like. These LEDs are pretty much staring me in the face while driving. Peripherally but that is your best night vision source. Even though the circuit boards are dealer expensive 1.5k range they can be had on ebay for fairly cheap. I already have a tested working board just for this project in case anything goes awry,I have the original to put back in. So,no worries there. Thanks for the help thus far. Let me know if pot theory is correct.

Oh yeah - FSM as in Factory Service Manual. You were close!