I was wondering if it is possible to increase the impedance of a speaker by adding a resistor in line (series) to one of the terminals and what the effects of doing so might be? Specifically- adding a 2 or 4 ohm resistor to a 4 ohm speaker for a 6 or 8 ohm load. Could it work or would it sound like butt?

It would have to be a high powered resistor on the order of 100 watts, and it would have to be connected in series (i.e., in-line, like you said) to increase impedance. It doesn’t increase the impedance of the speaker, only the total impedance load the amplifier will see. I assume you’re looking to get 8-ohms from a 4-ohm speaker? Not sure how it would sound, but it is a significant waste of amplifier power. If you’re using an 8-ohm speaker with an amp rated for 100 watts @ 8-ohms, the speaker gets the amp’s maximum power output. Connect the amp to the 4-ohm speaker/4-ohm resistor combo and the amp still puts out the 100 watts. However, the power is divided between the speaker and the resistor, so all the speaker gets is 50 watts. So to answer your question, it will work but you lose 3-dB headroom. Regards, Wayne A. Pflughaupt

Thanks! Thats pretty much what I was thinking would happen. What I have is a reciever rated for a 6-8 ohm load and a pair of 4 ohm speakers. I want a temporary fix (don't want to risk blowing my amp)untill I can get a separate power amp to run them. If I used a 2 ohm resistor, would the speaker still only see half of the amp's power or would the wattage be divided 2 to 1 to the speaker? Another thing I was thinking is that I could add a 4 ohm subwoofer in series to each speaker so that the power loss would at least be used for something, but how big of a pain would it be to build a crossover for a sub in series with a full range speaker?

I did some research and answered my first question. According to termpro.com/articles/spkrz.html it seems that the speaker would get about twice the power of the 2 ohm resistor. amplifier output=rated power X (rated impedance [ohms]/total impedance[ohms]) 146 = 110 X (8/6) The amplifier is rated at 110 watts into an 8 ohm load,so theoretically, the amp will produce 146 watts into a 6 ohm load. watts per driver(or resistor)= output X (speaker impedance/total impedance) speaker 97 = 146 watts X (4/6) resistor 49 = 146 watts X (2/6) The resistor will use approx. 49 watts leaving 97 watts to drive the speaker. This will only equal an approximately 13 watt drop in power and a 1 db or less derease in performance vs. an 8 ohm speaker powered by the same amp. My second question still stands- how would I make a crossover for a 4 ohm sub in series with a 4 ohm full range speaker?

Again, it must be reiterated, if you are determined to place a resistor in series with a speaker, then that resistor needs to be able to handle the extra wattage. That usually means paralleling a bunch of resistors to achieve the desired resistance value and the wattage rating. For example, to get a 2 ohm resistor rated at 80 Watts, you'd want to use eight 16 ohm resistors rated at 10 Watts, and parallel all of them together to get a 2 ohm resistance that could handle 80 watts. What you are really doing is burning power across the inserted resistor. This is just a stop gap and shouldn't be used for a long-term solution. Another option is just just not play the audio system too loudly until the new amps are in place. You'd be better off using a receiver that had bass management to handle the X-over duties, and then using a separate amp to power the subwoofer (if it's a passive model). There are also the Paradigm X-30, and the Marchand XM9, both are active X-overs.

You also seem to be doing power calc's at max power per channel ratings. First you are not going to have the amp cranked all the time (if you do, you are limiting its life) and second, that rating is not going to be accurate for all channels driven anyway. You will notice a significant reduction in power because you are burning 1/3 of the power as raw heat disapation. The correct way to do this is to use an impedance matcher.

Grab onto a 50W light bulb that's been on for a few minutes to get an idea of what 50W of power feels like. I've had resistors melt carpet at normal listening levels, and they weren't carrying nearly as much percentage of the load as we're talking about here. The short answer is, don't do it.

A more elegant solution is to simply use a single high-powered aluminum housed resistor, like the ones on the back row here: Just to give you an idea of the size of these things, those nuts on the connection terminals take a 7/16” wrench. They are designed to dissipate heat, so there’s no danger of melting things as long as the proper wattage is used. You should be able to find or order them at a local electronics hobbyist shop. Common brands are Dale and Ohmite. This would work even as a long-term solution, but as everyone has noted, it’s not ideal because you’re wasting the amplifier’s power. EDIT: Rob's suggestion for an impedance matcher sounds like a better solution, but I'm not familar with them. Perhaps Rob can elaborate. Regards, Wayne A. Pflughaupt

Thanks to everyone for the replies. I now have 2 questiions- 1- What is an impedance matcher and how would I use/hook it up? 2- Could a low pass filter be easily made for a 4 ohm sub attatched in series to my 4 ohm speakers to give me an 8 ohm load?

Aaron, the sooner you rid yourself of the notion that you can simply add the impedance of a "speaker" and a "subwoofer" to arrive at a larger impedance just to make it a "gentler" load for your current receiver that shouldn't be driven at high volume levels, the better off you will be. In audio terms, impedance is the resistance profile of the load (speaker) over its frequency spectrum, mainly from zero to 20,000 Hz. If the subwoofer is a 4 ohm load, its spectrum of operation is usually under 200 Hz (usually lower, unless you have small satellite speakers), and having the speaker in series with the subwoofer isn't going to present an 8 ohm load to the receiver because its impedance profile will look more like a 4 ohm load over the course of the total audio frequency spectrum.

Thank you- an answer and an explaination is what I wanted. I could not because I did not know if it was even possible- which is why I asked the question. Thank you for the information. I now know it is not possible AND why.

An impedance matching transformer will present an 8ohm load from the primary to the reciver while driving a lower impedance load on the secondary. The problem with this is that its more costly then a resistor.............but it's really the only "right" way to do if you must do it to begin with.