Question concerning Delta Chip 66032

Welcome Guest! To enable all features please Login or Register.

Notification

Error

Question concerning Delta Chip 66032

Options

View

Previous Topic Next Topic

Chris6382chris		#1 Posted : 02 March 2011 07:34:48(UTC) Retweet
Joined: 27/11/2009(UTC) Posts: 1,218 Location: Middle of the US		I am installing a delta 66032 chip into an engine and wanted to replace the light bulb that came with the engine with an LED. What is the voltage coming out of the gray and yellow wires that feed the lights? I was thinking of going to Radio Shack and grabbing a couple LED's and was wondering what I should get. Also will I need to add a resistor as well, between the chip and the LED? If so what would be a good combination of LED and Resistor, looking for the lowest draw while still being somewhat bright. Thanks, Chris


User Profile View All Posts by User View Thanks


	Wish to join the discussion?! Login to your marklin-users.net community \| Forum account or Register a new forum account. Or Connect via Facebook

perz		#2 Posted : 02 March 2011 08:35:41(UTC) Retweet
Joined: 12/01/2002(UTC) Posts: 2,578 Location: Sweden		I think this newer version of the Delta decoder has a steady 20 V lamp voltage (older versions had a voltage that varied with motor speed). You will always need a resistor in series with the LED. A LED typically draws 10 - 20 mA and the voltage over it is approximately 3 V (in case of a white LED). This would give 17 V over the resistor, and if you aim at 10 mA this means a 1.7 kOhm resistor. Nearest standard values are 1.5 kOhm and 1.8 kOhm. The power rating of the resistor can be calculated as UU/R where U is the voltage and R is the resistance. Assume you have a 1.8 kOhm resistor: 17 V 17 V / 1.8 kOhm gives 160 mW, so 160 mW is the minimum power rating for the resistor. Add some margin to that to be safe.
		WWW

User Profile View All Posts by User View Thanks

H0		#3 Posted : 02 March 2011 09:44:31(UTC) Retweet
Joined: 16/02/2004(UTC) Posts: 15,463 Location: DE-NW		Hi, Chris! Originally Posted by: Chris6382chris Also will I need to add a resistor as well, between the chip and the LED? You need something that limits the current through the LED - and a resistor is a simple way to achieve that. You can have several LEDs (of the same type) in series. You'll need a smaller resistor and you'll get more light (and less heating) frome the same current.
		Regards Tom --- "In all of the gauges, we particularly emphasize a high level of quality, the best possible fidelity to the prototype, and absolute precision. You will see that in all of our products." (from Märklin New Items Brochure 2015, page 1) ROFLBTCUTS


User Profile View All Posts by User View Thanks

Chris6382chris		#4 Posted : 02 March 2011 17:26:47(UTC) Retweet
Joined: 27/11/2009(UTC) Posts: 1,218 Location: Middle of the US		Thanks to both of you for your answer. So I will look for a 160-170 mW rated resistor and wire it in series to my 3 volt LED. Chris


User Profile View All Posts by User View Thanks

H0		#5 Posted : 02 March 2011 17:38:25(UTC) Retweet
Joined: 16/02/2004(UTC) Posts: 15,463 Location: DE-NW		The standard type of resistors has 0.25 W (250 mW)
		Regards Tom --- "In all of the gauges, we particularly emphasize a high level of quality, the best possible fidelity to the prototype, and absolute precision. You will see that in all of our products." (from Märklin New Items Brochure 2015, page 1) ROFLBTCUTS


User Profile View All Posts by User View Thanks

frankie		#6 Posted : 02 March 2011 17:53:29(UTC) Retweet
Joined: 27/10/2006(UTC) Posts: 692 Location: Italy		I'm noy into electronics, but I have already used this solution, the resistor must be soldered on one of the pin of the LED, is there a way to figure out on which one? Because it works only on one of the two...
		Alessandro I have a CS1 Reloaded!


User Profile View All Posts by User View Thanks

perz		#7 Posted : 02 March 2011 19:29:46(UTC) Retweet
Joined: 12/01/2002(UTC) Posts: 2,578 Location: Sweden		Originally Posted by: frankie I'm noy into electronics, but I have already used this solution, the resistor must be soldered on one of the pin of the LED, is there a way to figure out on which one? Because it works only on one of the two... The resistor can be on either side of the LED, but the LED always needs to have the correct polarity. Probably when you only got it to work one way, you had the LED flipped as well.
		WWW

User Profile View All Posts by User View Thanks

H0		#8 Posted : 02 March 2011 21:53:32(UTC) Retweet
Joined: 16/02/2004(UTC) Posts: 15,463 Location: DE-NW		Originally Posted by: perz The resistor can be on either side of the LED, but the LED always needs to have the correct polarity. Might be worth to mention that the decoder polarity is not what many expect: the common wire (orange with M* decoders, blue with NMRA standard (e. g. ESU)) is plus, decoder function outputs (yellow and grey with M* decoders) are minus.
		Regards Tom --- "In all of the gauges, we particularly emphasize a high level of quality, the best possible fidelity to the prototype, and absolute precision. You will see that in all of our products." (from Märklin New Items Brochure 2015, page 1) ROFLBTCUTS


User Profile View All Posts by User View Thanks

Dave Banks		#9 Posted : 02 March 2011 22:57:26(UTC) Retweet
Joined: 08/03/2006(UTC) Posts: 1,047 Location: Gold Coast, Australia.		Chris also you can modify this chip to have a function for your light. http://www.bogobit.de/decoder/66032.html
		D.A.Banks


User Profile View All Posts by User View Thanks

Users browsing this topic
Guest

Forum Jump

You cannot post new topics in this forum.
You cannot reply to topics in this forum.
You cannot delete your posts in this forum.
You cannot edit your posts in this forum.
You cannot create polls in this forum.
You cannot vote in polls in this forum.