It was a Dr. Crankenstein m-80 2 channel amp with a sub woofer cross over. I think I picked it up for 15 bux but do to a loose wire it caught fire and that was that. Was getting ready to toss it in the trash when I got interested in building my first gain clone (potentially one to replace this burnt amp). Please forgive me for my ignorant question, I've been trying to read and learn, but I'm no electronic guru. If a car's electrical system puts out 12-14.4 volts, how was my little amp able to step it up high enough to run the two lm3886's. On the national spec sheet it says they need 28 volts right? Also while I'm asking newbie questions, how do I get the -v for the voltage rails using battery power? Thanks in advance.
I dunno about your specific amp's circuit to raise the voltage, but the V- rail is an easy one.
Say, for example, that you had 12v+ and ground to start with. First, you have to understand that ground is relative. Repeat that until you believe it.
You would simply set up a voltage divider. And example of which would be putting two large value resistors across the 12 volts in series. Now, you would choose the node between the two resistors as ground.
Since you have divided the 12 volts across the resistors and ground is in the middle (remembering that voltages, and the concept of ground, are relative) you have a "0" volts at the center of the two resistors. The 12 volts, "straddling" that node, becomes a (relative) 6v+ and 6v-.
There are your positive and negative rails and ground, from one voltage supply.
Now you can see that if, for example, your chip requires 28 volts you will need a total rail-to-rail voltage of 56 volts to split it into a +-28v plus ground.
Hope you are not more confused that you were before you asked... 🙂
GnD
Say, for example, that you had 12v+ and ground to start with. First, you have to understand that ground is relative. Repeat that until you believe it.
You would simply set up a voltage divider. And example of which would be putting two large value resistors across the 12 volts in series. Now, you would choose the node between the two resistors as ground.
Since you have divided the 12 volts across the resistors and ground is in the middle (remembering that voltages, and the concept of ground, are relative) you have a "0" volts at the center of the two resistors. The 12 volts, "straddling" that node, becomes a (relative) 6v+ and 6v-.
There are your positive and negative rails and ground, from one voltage supply.
Now you can see that if, for example, your chip requires 28 volts you will need a total rail-to-rail voltage of 56 volts to split it into a +-28v plus ground.
Hope you are not more confused that you were before you asked... 🙂
GnD
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