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Answer #1Yes. 4 ohm speakers, when driven to the same voltage level (same volume control setting), try to draw four times the power (watts) of an 8 ohm speaker. This can overload the amp and cause it to fail. Answer #2It's a myth. There is really no 8 Ohm amplifier on the market. And there never was.

The amplifier will have an output impedance of around 0.04 ohms. In hi-fi we have always impedance bridging. Zout << Zin. That means the output impedance of the amplifier is much less than the input impedance of the loud speaker.

The damping factor Df = Zin / Zout tells you what Zout is.

Zout = Zin/Df.

If the damping factor Df = 200 and the loudspeaker impedance is Zin = 8 ohms, the output impedance of the amplifier is Zout = 8 / 200 = 0.04 ohms.

You see, there is no "8 ohm amplifier" on the market with a 8 ohm output impedance.

Scroll down to related links and look at "Voltage Bridging or impedance bridging - Zout < Zin".

Answer #3Actually, it's not a myth. Just a simple misunderstanding. An "8 ohm" amp does not mean it has an 8 ohm output impedance, it means the amp is designed to drive an 8 ohm load. Let me explain (warning, this is long, with math!).

Let's say I want to design an amp to provide 50 watts RMS into an eight ohm load with an input of 0dbv (which is 0.775VRMS).

How much voltage and current will I need to produce 50 watts? According to Ohm's Law, P=I^2R, so:

50watts=I^2 X 8ohms

I^2 = 50 / 8 = 6.25

I = sqrt 6.25 = 2.5

We will need 2.5 amps current capacity. Now that we know amps, let's figure volts. E=IR, so:

E = 2.5 X 8 = 20

We will need 20 volts RMS.

Now, how about voltage gain? Our output is 20V, our input is 0.775V, so the necessary gain is:

20 / 0.775 = 25.8

As you can see, this amp will need to supply 2.5A and 20V with a voltage gain of about 26. The design engineer will have to build the power supply with enough voltage to reach 20V at the output, and use output transistors capable of 2.5A.

I won't write out the calculations for a 4 ohm amp, you can do them yourself just like I did above, substituting 4 for 8. If you do, you will find the amp designed for a 4 ohm load will need to supply about 3.5A and 14V, with a voltage gain of 18. Note the voltage is lower, but the current is higher!

If I design this amp for a 4 ohm load, I will need bigger output transistors (3.5A vs 2.5A), but can get away with a lower voltage power supply (14V vs 20V).

See how the design is VERY dependant on which speaker impedance I intend on driving? Now lets see the ramifications of using the "wrong" speaker on an amp.

First, let's put an 8 ohm speaker on an amp designed for a 4 ohm load. We already know from our design calcs above that the amp will be designed to supply 14V. This is less than the 20V needed to drive the 8 ohm speaker to 50 watts, so it is obvious we will not reach full power. We will only get (I = E / R) 14/8 = 1.75 amps, and therefore (P = E I) 14 X 1.75 = 24.5 watts output.

So, if we put an 8 ohm speaker on an amp designed to drive an 4 ohm load, we will get less power, in this case about half. If we try to compensate by turning up the gain or increasing the input signal, the signal will clip and distort because the power supply can't reach the required output voltage. Note that no damage occurs to the amp (although some speakers can be damaged by driving them with a clipped signal), and the setup will sound fine as long as we keep the volume low enough so that our power is less than 24.5 watts.

Ah, but what about the reverse, putting a 4 ohm speaker on an amp designed for an 8 ohm load?

Let's do this: first install a correct 8 ohm speaker on the amp, feed an audio signal in, and adjust the gain so that 50 watts is produced. We already know from our calcs that at this point the amp will be producing 20V at about 2.5A.

Now, remove the 8 ohm speaker and without changing anything else, install a 4 ohm speaker. What happens? Well, if you feed a 4 ohm speaker with 20V, what is the current drawn? I = E / R, so:

I = 20 / 4 = 5A

Oops, this is a problem. The speaker will try to draw 5A, but our output transistors are designed for only 2.5A, I see smoke!

What is the power at this level? P = I^2 R, so:

P = 5^2 X 4 = 100

Oh, another problem. Our power supply is designed for 50 watts, but we are now supplying 100! More smoke.

Now, what if we turn the gain down until the current drops to 2.5A? well, at least no more smoke, since we are no longer overloading the amp beyond it's rated limit. What will the power output be? P = I^2 R, so:

P = 2.5^2 X 4 = 25

Again, if we turn down the gain such that 25 watts or less are delivered, everything will be fine, but if we try to drive the speaker to 50 watts, we may overload the output transistors. And since this amp has higher gain (26 vs 18), and it has a higher voltage power supply, we can drive the speaker to power levels approaching 100 watts, which will certainly pop the output transistors, probably burn up the power supply, and most likely ruin the speaker too.

So, answer #1 is correct. (Yes, I am a design engineer. And, yes, I wrote answer #1...)

Answer #2 is correct as well. An 8 ohm amp does not have an output impedance of 8 ohms. It is indeed much lower. But the question asker meant an amp designed for 8 ohm speakers when he asked the question, and these definitely do exist. And using a 4 ohm speaker on such an amp can cause it to fail. Ask me how I know this... I am no stranger to smoke, believe me.

Answer #4

According to legend and conventional wisdom, solid state (transistor) amplifiers can drive higher impedances than their rating demands. So a transistor amp with a label saying "4-ohm" will drive 4, 8, and 16 ohms with no troubles. You can go up but not down. Don't try 2 ohms.

Older legends say that tube amplifiers can drive lower impedances than their rating demands. So a tube amp with a label saying "8-ohm" may drive 4 and 8 ohms. You can go down but not up. Don't try 16 ohms. And never run a tube amplifier without speakers attached. Other legends of the tube age say that any attempt to use a tube amp with speakers of different impedance may cause damage to the amplifier.

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βˆ™ 14y ago
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βˆ™ 15y ago

Yes, but it will not be as loud as a 4-ohm cabinet. It's a myth. There is really no 4Ohm amplifier on the market. And there never was. The amplifier will have an output impedance of around 0.04 ohms. In hi-fi we have always impedance bridging. Zout << Zin. That means the output impedance of the amplifier is much less than the input impedance of the loud speaker. The damping factor Df = Zin / Zout tells you what Zout is. Zout = Zin/Df. If the damping factor Df = 200 and the loudspeaker impedance is Zin = 4 ohms, the output impedance of the amplifier is Zout = 4 / 200 = 0.02 ohms. You see, there is no "4 ohm amplifier" on the market with a 4 ohm output impedance. Scroll down to related links and look at "Voltage Bridging or impedance bridging - Zout < Zin".

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Q: Can 4 ohm speakers cause an 8 ohm amp to fail?
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