damaging amps with wrong impedance

Thread Starter

veenife

Joined Jan 16, 2014
42
hey... i know that if one plug an amp with high output impedance to and loudspeaker that has a lower input impedance than the amplifier, the amp can get damged...

but one i still dont understand, what is exactly happening there that the amplifier gets damaged....

...is that because of the high voltage of the amp that gets reflected back due to non matching impedances or someting like that.....???.... can someone proper explain it????

cheers!
 

wayneh

Joined Sep 9, 2010
17,496
Too much current flows into the load and something will get too hot.

Also, even if the load is protected against overheat and over current, power transfer is very poor. The load pulls the voltage too low. Current flows, but power is voltage times current.
 

Thread Starter

veenife

Joined Jan 16, 2014
42
but if the load (speakers) are sucking all the current and getting hot... why is the amp that gets burned out then?
 

Brevor

Joined Apr 9, 2011
297
[QUOTEbut if the load (speakers) are sucking all the current and getting hot... why is the amp that gets burned out then?][/QUOTE]

Its the amp that gets hot.
 

Thread Starter

veenife

Joined Jan 16, 2014
42
... okay... so the amp is generating power... current gets pumped by the low load impedance and no voltage can be delivered properly... so voltage stays at the amp and doenst have where to go and so start to accumulate in the amp till it burns the circuit down...???? am i on the right path?
 

richard.cs

Joined Mar 3, 2012
162
It's important to realise that the stated (lowest) impedance of an audio amplifier is the impedance of the load it can safely drive*, not it's output impedance in a transmission-line matching kind of way.

The output impedance of most audio amplifiers is designed to be very low so that the amplifier behaves as a voltage source. A lower load impedance then corresponds to more current being drawn from the amplifier for a given voltage and this results in much more power being dissipated in the amplifier.

*At reduced output levels you can drive lower impedances but inefficiently and there's a risk of damage if the source audio level is increased.
 

MrChips

Joined Oct 2, 2009
30,717
The power delivered to the speaker can be approximated by

P = V*V/R

Suppose your volume control is set so that the signal delivered to an 8Ω speaker is 16V rms.

Power = 16 * 16 / 8 = 32W

Now suppose you disconnected the speaker and replaced it with a 4Ω speaker. The power delivered to the speaker is doubled:

Power = 16 * 16 / 4 = 64W

The current draw from the amp is also doubled from 2A to 4A.
If the amp is not designed to deliver this extra power and current then the amp can be damaged.

Normally, one would turn the volume down but the amp is still in danger of being over driven should someone choose to pump up the volume.
 

studiot

Joined Nov 9, 2007
4,998
There have been several partial answers here, but reality is more complicated.

richard.cs has noted that the output impedence of amps is designed to be low - indeed very low often less than 0.1Ω and for a good amp down to 0.001Ω.

Mr Chips has pointed out that if you present a low load to the amp terminals it will attempt to deliver a greater current.
However it will only be partly successful since it will be likely limited by its transformer or power supply.

wayneh has commented on protection circuits. Very likely the amp is protected against connecting a short circuit - it will just turn off. In this instance a 'short circuit' is usually taken to be around 1Ω

So in the situation described by MrChips where a speaker or set of speakers that aggregate to say 4Ω or even 2Ω will attempt to draw excessive current.

You have asked why the amp, not the speakers suffer.
Well it depends. If you connect say one big 6Ω speaker it may well suffer as well as / instead of the amp.
If however you connect two 8Ω speakers in parallel, neither speaker will be overdriven, but the amp will be.

The mechanism of damage in speakers is ohmic overheating.

In the amp it may well be the power supply that suffers since it will be continuously outputting its maximum and beyond.
If the power supply in tha amp can actually supply the extra current then the output transistors may well be taken outside their SOA or safe operating area.
The SOA is a complicated graph that defines parameters within which a transistor may be run. It is complicated because the safe voltage and power levels depend upon the current so it is not a simple relationship and some products of voltage and current are safe for certain values of current, but not for others.
 

inwo

Joined Nov 7, 2013
2,419
My frame of reference is always mains powered circuits.

They can safely supply only appliances rated at a certain voltage or current draw. = impedance:)

Plug in a 24 volt lamp (low impedance load), poof.
Or if it's a large enough wattage to over load supply. Supply will fail.

Try a 240 volt lamp on a 120 volt supply. (high impedance load) Not so bad!
Power will not be delivered efficiently however, and light will be dim.

There will be a certain voltage/amps design (again, impedance) that will work best.
Impedance matching.
 

studiot

Joined Nov 9, 2007
4,998
My frame of reference is always mains powered circuits.

They can safely supply only appliances rated at a certain voltage or current draw. = impedance

Plug in a 24 volt lamp (low impedance load), poof.
Or if it's a large enough wattage to over load supply. Supply will fail.

Try a 240 volt lamp on a 120 volt supply. (high impedance load) Not so bad!
Power will not be delivered efficiently however, and light will be dim.

There will be a certain voltage/amps design (again, impedance) that will work best.
Impedance matching.
Have a care, neither amplifiers and loudspeakers nor mains necessarily work by impedance matching or follow the maximum power transfer theorem.
 

wayneh

Joined Sep 9, 2010
17,496
... so voltage stays at the amp and doenst have where to go and so start to accumulate in the amp till it burns the circuit down...???? am i on the right path?
Not really. The damaging factor is usually too much current through a component. This causes power to be dissipated as heat (in the amount of I^2•R). When this gets excessive, the component gets hot and either shuts down or blows up, like a fuse.

Some types of components can be damaged by excessive voltage, without excess current, but that's not what is happening with a low impedance load placed on a high-impedance amplifier. And voltage does not "accumulate", although charge can accumulate on a battery or a capacitor, leading to a higher voltage. Again, not relevant to the amplifier problem.
 
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