Axiom Home Page Forums General Discussion Technical Questions Question for physics whizzes

Hey,

I have a question for people who didn't sleep through physics lectures like I did way back when. There's story to this, and I'm just curious as to how this stuff works.

I'd been having a problem with my center channel (VP150) sounding somewhat distorted when turned up loud. My receiver is the Arcam AVR100, which is rated at 70Wx5 (in "real" watts). My other speakers are the QS4s and M40s. I found myself often needing to set the center channel to +4 dB for poorly-mixed soundtracks, and I've always understood that a receiver driving 8 ohm loads wouldn't have problems with a sensitive 6 ohm load like the VP150.

Thinking I needed to buy a mono amp for the center channel, I talked a guy at the store where I bought the Arcam. He suggested switching the receiver to 4-6 ohm mode, setting all three front channels at -2 dB, then turning up the gain as need be. It was explained to me that the VP150 is "hungry for current rather than voltage", and the distortion I was hearing was due to an overworked amp.

The suggestions worked magnificently, and even playing the worst-mixed soudtracks (read: Star Wars), dialogue is now crystal clear with the volume turned up.

Could somebody explain to me why and how this worked, or recommend a good resource to help make all this clear to me?

Thanks

To keep this short, power supplies are designed to operate with some sort of a load. They are designed to maintain a given voltage into a given load. For example take a 12 volt power supply that's been designed to deliver 1 amp of current. If we connect a load that has a resistance of 12 ohms, this will allow the power supply's max rated current to flow through the load. If we connect a second 12 ohm load across the power supply's output terminals, the output current will be 2 times as much as the supply can safely deliver (assuming that the supply can maintain it's rated voltage under the heavy load) and the supply will soon blow a fuse or fail. The same thing happens when you connect too low of an ohm load to an amplifier. Lower ohm loads can allow the amplifier to produce more output current which results in more output power, but too low of an ohm load will cause the amplifier to fail or trip its protection circuitry . The amplifier expects to 'see' a certain minimum resistance (ohms) to assure a limited maximum current flow at its maximum output. The best example I can give you would be one of a faucet. The load represents a tap on the faucet. An 8 ohm load would be like having the tap half closed. A 4 ohm load would be like then opening the faucet fully. If you had a pump behind the faucet trying to produce a constant pressure in the system it would have to work much harder to maintain the same pressure with the faucet fully open.

Lets say we have a 100 watt amplifier and it can drive a minimum ohm load of 8 ohms. To produce 100 watts, the amplifier will have to deliver 3.538 amps of current. To produce a current flow of 3.5 (3.538 actually but lets round off the numbers to make it simpler) amps into a 8 ohm load, it will have to develop 28 volts across the load (the voltage at the speaker terminals at full power will be 28.284 volts).If you reduce the ohm load to 4 ohms, the current flow will double. Since the max safe current output is only 3.538 amps and the lower ohm load causes more than 3.538 amps to flow, the amplifier may well be damaged by the lower ohm load. In most cases on cheap units the circuit protection will kick in. To protect the system and to be able to pass safety concerns due to overheating problems in amplifiers with cheaper power supplies the manufactures put these 4/6 ohm switches. What this does is cut the power (voltage) available instead of producing 28 volts you may find the you may only have half the voltage available. This will reduce the current delivery into the 4 ohm load to a level that the amplifier can handle.

Less Resistance = More Current

Here are some base formulas to work this out for yourself.

Where:
P = Power in Watts
E = Electromotive Force in Volts
I = Electrical Current in Amps
R = Electrical Resistance in Ohms
SQR = Square Root
I use 'E' to represent voltage most of the time but sometimes you'll see 'V' used for voltage.

BASE FORMULAS
P=IxE
E=IxR

TO FIND VOLTAGE
E=P/I
E=IxR
E=SQR(PxR)

TO FIND CURRENT
I=P/E
I=E/R
I=SQR(P/R)

TO FIND POWER
P=IxE
P=E2/R
P=I2xR

TO FIND RESISTANCE
R=E2/P
R=E/I
R=P/I2


Hope this helps

I had totally forgotten those formulas. I'll play with this in my head and get it straightened out.

Thanks.