Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.
Sebi you are right...here is no buffer to help when CD-ROM is shaking. I don't think that is a good idea to build unles the car will be standing still.
I agree that a LDO will work fine. If the car is running, the voltage from the alternator is around 13.5V. You only get a little less than 12V if your car is not running.
HI there,
the circut for powering cd-rom from car instalation is very simple as you know(you need only two regulator for 5v and 12v plus some capacitors). CD-rom works very good on all kind of streets I checked it. If you have for example 8x CD, for audio disks it works with 1x speed. If something is wrong(holes in road ) it will speed up. But don't use CD-rom like samsung, btc, creative
Regards
Hi, I've the power suply who feed the CD room with +12 and +5 Volts, all
run well, its open and eject the cd, but don't play anything, I want to get
analog sound to inject it in a cassette player. Please somebody can help
me??
Thanks very much,
Fridel
I am facing the same problem. What I have done so far is that I used the power voltage regualtor 7805 - the one having the rounded metal hausing, and the result was that it worked for few minuets after that the regulator stops (and the player as well) due to overheating. I think there should be a mistake made in the circuit. If any one can offer a working circuit please post it and it would be very appreciated.
use a T0-220 package. and a heatsink or solder to a isolated plane on pcb.
Since the 5V regulator also drops 7 volts at the drawn power I suspect you do not have enuf thermal dissipation. That or you failed to consider the addidional power required to regulate the 12V to 5V.
Remember the power rating needs to be higher than required to prevent premature failure of the regulator at temperature.
Example you need 250ma @ 5V for the cdr.
Thats 1.25W for the 5V plus. 12V-5V= 7V ==> 7V @ 250 ma = 1.75W
Total = 1.25W + 1.75W = 3W add some overhead 4-5W dissipation is required.
What exacty 7805 are you using? complete part number please.
By your description it seems you are using a LM78L05. Metal T0-39 package.
that part is only rated at 100mA
Looking at the maximum power dissipation without a heatsink @25C max power is .9W
so if you needed 100ma, that'd be .5W for the CD, and .7 additional dropping from 12V to 5V for total 1.2W. Exceededs maximun dissipation.
If you simply place a 30c/W heatsink on the regulator you should be ok as that will up your power dissipation to just under 2W.
Without heatsink = .9W with 30c/W 1.9W.
good luck
thatnks for your interest.
One mentioning wothy point, I would like to direct your attention to, is thatmy CDR rating (written on specs sticker) is 1.5 Amp!!!
Yes I use the package you mentioned. ..T039, What do I do now?
If you have a completely different circuit that works with that case please post it and I can go for it directly.
Theres nothing wrong with the circuit. You just need to use a 7805 with the proper power rating. not a 78L05. 78L05 can only deliver 100mA!!!
Like I said get a 7805 in a TO-220 package. or a T03 package.
Then in order to dissipate the power YOU need attach the regulator to either an isolated section of the PCB ( not grounded ) or to the chassis ( metal ) using appropriate insulator to prevent shorting and allow thermal dissipation. Due to the high power requirement of the cd 7.5W ( 5v * 1.5A ) plus dropping the 12V to 5V = 10.5W for a total of 18W dissipation. Note the max dissipation to a "Infinite heatsink" is only 20W.
Note you can use regulators in parallel to boost current and reduce the dissipation needed using a single device. Devices in parallel will Equally share the load current and heat dissipation.
You could also look into a switching regulator that uses 12V in and delivers 5V out.
It will have a lower power dissipation for dropping the 12V to 5V. Its more efficient than a linear regulator. Notice that a linear regulator drops more power than the CDR requires. Thats due to the linear regulator needing to drop 7V at 1.5A to supply the CDR with 5V @ 1.5A using a 12V source.
good luck have fun
Can someone help me with the amplifier part please? I mean, the output of the cd drive line out is connected to the amplifier... Which one? Audio, Power or anything else and what should be the rating of the amplifier and now how do i get the output to the speakers... They need atleast 600Watts right, so how can i get it to deliver so much power? Can someone please explain it to me clearly as to what is happening exactly? I want to know how exactly this arrangement is working... Please... help me...
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.