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Questions about connecting 8 ohm amp to a 4 ohm speaker.

Hacdrag

Member
I'm going to be connecting a low-voltage 8 ohm amp module to a 4 ohm loudspeaker. Using the same amp and a different speaker, I've done this before without issues. I'm not expecting high-fidelity, just having fun with a project. A vendor suggested putting a 4 ohm resistor between the amp and the speaker which I haven't done in the past.

I have a question regarding the pure mathematics of the setups with and without the resistor. Will I get more or less power out of the amp and more or less volume out of the speaker with the resistor in series? The amp will be powered by a 9-volt battery. Will a 4 ohm resistor in series drain the battery faster?
 
Will I get more or less power out of the amp and more or less volume out of the speaker with the resistor in series?
You will likely get less power from the amp with the resistor, but more dissipation in the amp without the resistor at higher volume levels.
Will a 4 ohm resistor in series drain the battery faster?
At the same volume level, no, if it's a linear power amp.
If it's a Class D switching amp, then yes.
 
Will a 4 ohm resistor in series drain the battery faster?
If you keep the volume knob in the same position in both cases, then the battery will drain slower with the resistor in place.

If you turn up the volume knob when the resistor is in place to make both options play at the same audible noise level, you will likely drain the battery faster with the resistor.
 
If you turn up the volume knob when the resistor is in place to make both options play at the same audible noise level, you will likely drain the battery faster with the resistor.
Since the current through the speaker is the same in both scenarios, I don't see how there can be much difference in the battery drain with a linear amp.
 
It's a class AB amp, if this makes a difference to the answer.

It would help if you specified exactly what the amplifier was - the issue of too low a speaker impedance is simply overloading the amplifier. For example, if you have a 100W 8 ohm amplifier, connect a 4 ohm speaker to it, and turn it flat out, then the amplifier will try and deliver 200W to the speaker, and most likely destroy itself. However, if you keep the volume to reasonable levels, it will be perfectly fine - the issue then is someone else possibly turning the volume up!.

However, if your amp is a little tiny module, 0.5W or whatever, it will most likely be perfectly fine driving 4 ohms rather than 8 ohms, it doesn't really have enough power to destroy itself, and the IC's used are well protected against such abuse.
 
Audio amplifiers in the the vacuum tube age used output transformers to match the impedance of the output tubes (high ~ 1000 to 3000 ohms) to the speakers. These transformers were carefully designed (hopefully) ,and worked best at a specific impedance transfer ratio ,with some having multipule secondaries for 4, 8, 16 ohms . Modern AB transistor amps bipolar and mosfet , don't care as long as the power dissipation and current limits are not approached. Clase E amps I don't know.
 
However, if you keep the volume to reasonable levels, it will be perfectly fine
The amp will dissipate more power at the same volume level, since it will be delivering twice the current at 1/2 the voltage, but that probably wouldn't be a problem.
 
Twice the current at half the voltage, is the same wattage - W=VI.
In the speaker, yes, but double the current will increase amp dissipation, which is what I was referring to.
 
It would help if you specified exactly what the amplifier was - the issue of too low a speaker impedance is simply overloading the amplifier. For example, if you have a 100W 8 ohm amplifier, connect a 4 ohm speaker to it, and turn it flat out, then the amplifier will try and deliver 200W to the speaker, and most likely destroy itself. However, if you keep the volume to reasonable levels, it will be perfectly fine - the issue then is someone else possibly turning the volume up!.

However, if your amp is a little tiny module, 0.5W or whatever, it will most likely be perfectly fine driving 4 ohms rather than 8 ohms, it doesn't really have enough power to destroy itself, and the IC's used are well protected against such abuse.
This is cheap module I like to play around with:


I'm not worried about it blowing up. I was just curious as to the specific power, volume, and battery drainage issues due to the impedance mismatch, and what affect the resistor in series would bring to the table.
 
This is cheap module I like to play around with:


I'm not worried about it blowing up. I was just curious as to the specific power, volume, and battery drainage issues due to the impedance mismatch, and what affect the resistor in series would bring to the table.
In that case, try it. If it sounds somewhere between reasonably good and acceptably bad and nothing gets too hot or buzzy, I think you'll have your answer.
 
I'm going to be connecting a low-voltage 8 ohm amp module to a 4 ohm loudspeaker. Using the same amp and a different speaker, I've done this before without issues. I'm not expecting high-fidelity, just having fun with a project. A vendor suggested putting a 4 ohm resistor between the amp and the speaker which I haven't done in the past.

I have a question regarding the pure mathematics of the setups with and without the resistor. Will I get more or less power out of the amp and more or less volume out of the speaker with the resistor in series? The amp will be powered by a 9-volt battery. Will a 4 ohm resistor in series drain the battery faster?

Here's what it looks like with various loads at various output voltages.

The original load 8 Ohms, with original 2v output:
2^2/8=1/2 watt (0.5 watts)

The new load 4 Ohms, with 1.4142v output:
sqrt(2)^2/4=1/2 watt (0.5 watts)

The new load with extra 4 Ohm resistor, with original 2v output:
4+4 Ohms:
2^2/8=1/2 watt, (1/2)/2=1/4 watt (0.25 watts)

The new load with extra 4 Ohm resistor, with 1.4142 times the original voltage of 2v output:
4+4 Ohms:
(sqrt(2)*2)^2/8=1 watt, (1)/2=1/2 watt (0.5 watts)

This shows that it looks like you have to turn up the output voltage (with the volume knob) to 1.4 times the original output voltage in order to get the same output power. That means that you use twice the power to get the same volume output when using an extra 4 Ohm passive resistor in series with the new 4 Ohm speaker.
If you can get away with just a 4 Ohm speaker and no extra resistor, you do not have to turn the voltage (volume) up you can actually turn it down, but you get the same output power.

With 2v output and 8 Ohms the current would be 1/4 amp.
With 1.4142v output and 4 Ohms the current would be 0.3536 amps.
With 1.4142 times 2v output and 8 Ohms the current would be 0.3436 amps.

With these results and the results farther above, the output power with a 4 Ohm speaker would be the same as with a passive series resistor or not. The input power with the extra passive resistor would be twice of that without the extra resistor. However, since the current would be the same and batteries tend to run down based on current flow, the run time may be the same or nearly the same for either setup. If the amplifier could handle the 4 Ohm load though, I'd probably go with that except I would not turn up the volume all the way.

Of course amplifier distortion is dependent on load values so we would expect a change in the distortion level. There's a good chance it would go up, but since we can turn the volume up less than with the 8 Ohm load it depends on how the distortion changes with volume setting also. Some amplifiers are worse at low volume and some worse at higher volume.

We could do some simulations to verify these results, I did not do that yet due to time constraints.
 
The 4 Ohm speaker will take 1.4 times as much current to get the same power as an 8 Ohm speaker. That will run the battery down faster, and also the larger current will cause more voltage drop on the battery.

A alkaline PP3 battery will have an internal resistance around 5 Ohms, which is quite low if you are driving a 4 Ohm speaker.

A resistor in series with the speaker will move some voltage drop from the amplifier to the resistor, but it won't do anything about the voltage drop of the battery when current is taken.
 
This is cheap module I like to play around with:


I'm not worried about it blowing up. I was just curious as to the specific power, volume, and battery drainage issues due to the impedance mismatch, and what affect the resistor in series would bring to the table.

The LM386 is rated for 4 ohms, you have no real issues using it with the 4 ohm speaker, despite the over complicated suggestions given. Also, there is NO MISMATCH, it's rated for 4 ohms, and transistor amplifiers don't use highly inefficient speaker matching.
 
The LM386 is rated for 4 ohms, you have no real issues using it with the 4 ohm speaker, despite the over complicated suggestions given. Also, there is NO MISMATCH, it's rated for 4 ohms, and transistor amplifiers don't use highly inefficient speaker matching.
I wasn't aware of this. All vendor documentation states:

  • 500mW maximum power output @ 8 ohms, without any mention of 4 ohms.
 
I wasn't aware of this. All vendor documentation states:

  • 500mW maximum power output @ 8 ohms, without any mention of 4 ohms.
From the Texas Instruments datasheet...

IMG_5871.jpeg
 
From the Texas Instruments datasheet...

View attachment 148504
I do seem to remember, though, that there are different LM386 versions, and TI licenses this to New Japan Radio who puts their NMJ 386 D in their amp boards, so am not certain whether this one is rated at 4 ohms. Here is that datasheet.


In any event, if I were wrong in the OP, and my chip is also rated for 4 ohms, none of the very informative replies, such as that from MrAl were a waste, as I learned a lot from them.
 

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