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Power supply for AMP in CAR


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HI THERE.
we all know that we can build amps for home use .
I would like to construct an amp for use in my car , with 12V supply .
Inorder to achieve many Watts , we should raise the voltage of the car battery to 40 or ever 70 volts .
As i know we should use a power converter .
I would like to have a diagramm or an idea of how i can do it .

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Another way would be to take the power inverter diagram from the projects section of this site and modify it.
I am referring to this project-->
http://www.electronics-lab.com/projects/power/018/index.html

Method 1:
Instead of using a tansformer that steps the voltage up by 10, use one which will step it up by 3 or 4. This is accomplished by the number of windings on the primary vs the number of windings on the secondary. You do not have to modify a transformer, they come in different types. Once you have the correct voltage out (or close), rectify it for DC and regulate it or use a voltage divider to tweak it in to the exact voltage you want.

Method 2:
Another way to accomplish what you want with this circuit is to use a voltage divider on the 12VDC to make it only 4 volts. Since this is a 10 times step up circuit, you will have 40 volts AC.
At the output of the transformer, you just run it through a diode to rectify it and add a large cap to smooth out the remaining AC.

Method 3:
Use the circuit "as is", run the output through the diode rectifier and add a cap for ripple rejection (DC voltage), then use a voltage divider or regulator circuit to trim it down to the voltage you desire. This could be done with zener diodes and resistors.

If you need a bipolar supply, you can run the output of the transformer through a bridge and get -40 as well as the +40 volts.

So you see, there are some choices. ;)

MP

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The problem with this design is Current handling capability.
The project you are looking at is only about 5 amps and a 100 watt Amplifier needs about 20 amps. You should make sure the transformer is 20 amps or better and use about 8 of the 2N3055 transistors 4 and 4 hooking three in parallel with each original.
Also see my post on the subject.

http://www.electronics-lab.com/forum/index.php?board=22;action=display;threadid=222

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You will need some current limiting power resistors on the emitter side of the power transistors unless you are going to use MosFets. The reason this is done is because power transistors like the 2N3055 are not matched. Current will flow through them differently from one to the next. Because of this, when you parallel them together like this, you will have one that will be less resistive than the others and will take all of the current load and overheat itself. The power resistors are a safeguard. Something in the 0.1 to 0.22 ohm range with a fairly good watt rating will work.

MP

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Cut the connection of the emitter to ground. Put 1 each between the emitter leg of the transistor to ground. You will not even know they are there as far as amplification of your circuit. But if they are not there and the system draws a lot of current, you could have a melt down. :D

MP

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I would not leave them out.
Why should the current run through these 2N3055s equally? There is nothing there to cause it. There is no balance here. The resistors will cause current to flow equally through all of them. If 1 2N3055 is doing all the work because it has the least resistance, you have a problem, don't you agree?

MP

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Yes, I work in the consumer electronics industry. I have also seen the amazing things that the industry gets away with. ::)Consumer electronics is not based on good design. It is based on how much money the company can make. If you can leave out several 20 cent parts and sell several hundred thousand units that will still make it past the warranty period, then it is considered a good thing. The goal is met. One should never base good design on this. One can always base good profit on this.
The only thing the resistors will do is control the amount of current flowing and insure that all paths to ground have the same resistance. When you have several paths to ground all open at the same time, the flow is going to take the least resistive path. That is a built in law with electronics.
I appreciate your stubborness. In fact, I respect you for standing on your opinion. If you want to research this further, please show me an engineering article which backs up your claim.
By the way, the MOSFETs are different. You do not need resistors on them.
MP

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OK, suppose we have four transistors in parallel and we like to switch say10 Amp@12 Volts.
The transistors are not exactly the same (not from the same batch) but we want to use them anyway.
The first has 0.2 V drop the second 0.3 the third 0.25 and the last 0.25.
This does not mean that all the current runs through the first one whit the lowest resistance sins they are all in parallel.
If we add resistors to each one of the transistors we also add further drop (0R1 X 2.5A = 0.25V) if the transistors had been equal.
Can you explain the reason for the resistors by adding info on the following drawings, please?
Sorry for being a pain in your neck, but I think debating things like this is very stimulating.
Just stop me when I become too much! ;D
The (Swedish) mule!

Ante ::)

post-929-142791415921_thumb.gif

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Ante: I do not think this debate really belongs in this thread. Others are probably tiring from it. I will have to stop now. Please check out the link above for more on transistors.

In answer to the question...since all of your transistors are on at the same time, your resistances are all in parallel and thus follow the rules of parallel resistors. You do not want one of them acting as a direct short because the others will no longer factor into the equation. In order to have 12 volts at 10 amp, you must have a total resistance of your circuit of 1.2 ohms when the transistors are turned on. (12/1.2 =10) otherwise you do not have 10 amps. If a transistor draws your total resistance down to 0.1 ohm, for example, then you are running 12/0.1 = 100 amps through this one transistor. The others will not react because all the current is flowing through the one. Adding small resistances to each leg of the circuit to balance the load will insure that the flow is equal. This is because of the parallel resistance formula.
This is only one small area of the design math. There are many other factors also, like hfe of the transistor, DC current gain, MMP, etc.

I hope that this explanation is helpful in resolving your questions.
Please, back to the inverter. I am interested in the outcome.

MP

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MP,
OK, there are some things I still not agree on. One example: how can a 10Amp load suddenly pull 100Amps through one transistor even if the transistor should be shorted?? I must have a lot to learn about this. But because this is not the right forum for this I rest my case. I am also interested in the outcome of this inverter but I will not question the construction again. By the way, the link did not contain any info on switching transistors.
Thanks.

Ante ::)

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