DEA K-15A Active Speaker Repair Number 3 - I have three of these to fix, here is the THIRD one

[@stevejagger8602]

Merry Xmas Richard Completeness in repairing is so important because nobody wants come backs.

[@greengrayradio1394]

Two equal zeners in series will double the watt rating (just look at the current, which is equal in a series circuit, and multiply I x V). And Merry Christmas, Richard and Julie!

[@danblankenship5744]

I found "Freeze-Spray" (or invert a can of compressed air "Keyboard duster") will freeze the glue, making it easier to be removed in large chunks. The dark brown glue is hard and brittle, but the tan is pliable, making it difficult to flake off.

[@pieteboe]

Merry Christmas Richard, also to your wife. And thank you for another video

[@pauldery7875]

Nice repair Richard, Paul, USA!!!

[@toner5709]

Merry Christmas . Good detective work and explanations.

[@Dutch_off_grid_homesteading]

Heya, nice wen you have repair these kind of speakers you already know them and repairing will be easier

[@TheEmbeddedHobbyist]

can't beat doing repairs to the same equipment. I used to repair CD players and i might have 50 waiting, but they were all the same make and with the sane electronics inside. so you know the fault as soon as you hit the power button. made it quite an easy evening job after work.

[@LearnElectronicsRepair]

Yes I wanted to put that across in this video 🙂

[@izzzzzz6]

So in the end you simply removed the glue and checked some iffy joints and then it worked again? What happened to those amp modules where one was producing bass or full range and one was producing only trebble or mid I think. Were those the other two of these?

[@JustInspiredKent]

There was a burnt (open circuit) resistor that needed changing too.

[@LearnElectronicsRepair]

@@JustInspiredKent Yes, and I also fitted a higher wattage resistor, explained why and upgraded the zener diode as well.

[@LearnElectronicsRepair]

The Biema Alpha Line Array Amplifier with no bass, only treble, I repaired it here : kzbin.info/www/bejne/aGPPaKKOhKuZqs0 And in the end it was not any of the suggestions which I discussed in this video. It's working fine now and the customer paid me. All is good.

[@andymouse]

@@LearnElectronicsRepair "And in the end it was not any of the suggestions which I discussed in this video" please remind me of what that one was ...cheers.

[@LearnElectronicsRepair]

@@andymouse I was referring to the Biema Alpha repair that 1zzzzzz6 asked about. In that video I discuss some viewers suggestions and it turned out to be none of them.

[@bulwinkle]

Comment for the algorithm.

[@bulwinkle]

DEA or DOA?

[@TheEmbeddedHobbyist]

EAD Every Amplifier Dead.

[@LearnElectronicsRepair]

hehe

[@andymouse]

@@TheEmbeddedHobbyist SQUEAK!!!!!!!!!!!

[@edmguru9056]

Hi Richard it was me who commented about your zener diodes in series wattage calculation previously, and I'm afraid to say that you're still wrong or the diode manufacturers at Toshiba and other websites are lol. I would have a look online and check who's right here.

[@LearnElectronicsRepair]

@edmguru Hi, sorry I couldn't remember who it was, thanks for posting. Let's see who is right here. This webpage shows a simple circuit which is equivalent to the one in this amplifier www.hnhcart.com/blogs/active-components/how-to-calculate-wattage-of-a-zener-diode It has different voltages to ours but the principle and maths is exactly the same. So you will see that in step 3 you calculate the wattage of series resistor Pr=VxI where V= the voltage (6.9V) dropped across the resistor which in the example is a 62R so Pr=6.9Vx110mA = 759mW. Now if we used two 31R resistors in series to form R, then for each resistor the voltage dropped across it would be half of 6.9V = 3.45V but the current would be the same so each resistor is dissapating 3.45x110mA - 0.3795mW. This means we could safely use two 0.5W 31R resistors instead of one 1W 62R in the circuit. I hope we can agree on that? *And the same applies to the zener diodes* So onto the zener diode, in the example on the link the zener is rated at 5.1V so the power dissapated by the zener Pz=5.1Vx110mA = 561mW. In this case we would use a 1W zener. Now if we use 2x 2.5V zeners in series then (because the current still remains the same) Pz for each is 2.5Vx110mA = 275mW so we can see that each zener could be replaced with a 400mW zener. Now imagine we halved the value of R from 62R to 31R. The voltage across R remains the same so this would double the current to 220mA. Now our single 5.1V 1W zener would be dissapating Pz=5.1Vx220mA = 1.122W and burn up but if we used two 2.5V 1W zeners in series each would dissapate 2.5Vx220mA = 0.55W and be just fine. Therefore I stand my case. Replacing a 24V 1W zener with two 12V 1W zeners in series doubles the maximum wattage that the circuit can dissapate to 2W. Your turn 🙂

[@andymouse]

@@LearnElectronicsRepair Using two 12V Zener diodes in series is a practical solution when a 24V Zener diode is unavailable. Just verify tolerances, check power ratings, and test the configuration to ensure proper functionality in your circuit. 😊