Need an electrical engineer

[Phil Thien]

I have a question in regards to passive crossovers (for loudspeakers) and their effects on audio amplifiers.

I've asked this on forums that cater to the topic over the years, my confidence level in the answers is not high.

If I take an audio transducer that can reproduce the entire audible spectrum (20-20k, and this is theoretical), and which has a fairly flat impedance curve, and connect it to an amplifier, I can observe the amount of work the amplifier does by (for example) measuring the current draw of the amplifier (from the mains).

Now, what happens if I use a fairly large cap in series with the transducer? Say 400uf or thereabouts. Something that will essentially roll-off anything going to the transducer below 200-Hz, at 6db per octave.

My thinking is that, because capacitors don't get hot, it is rolling-off the low end by increasing the impedance seen by the amplifier as the frequency decreases (because amplifiers deliver less power into lower impedances).

And that if the amplifier is seeing higher impedances at lower frequencies, it is "working less hard," and that eliminating the need to produce this low frequency energy would lighten its load (and tax on the power supply, etc.) and make it easier to produce cleaner output at higher frequencies.

I've been told this is incorrect, and that the audio amplifier is still "cranking out the power at low frequencies." I don't understand how this can be, as I'd expect the energy then to go somewhere (the caps don't get hot, and the transducer doesn't reproduce the sound). The energy has to be accounted for somewhere. Is the load more difficult for the amp, so it is generating more heat which needs to be dissipated by the heatsinks? I'd think this wouldn't be the case, as very low impedances are actually more difficult for amplifiers to drive.

I do not understand the relationship between frequency and current, to an audio amplifier. Is the output of an audio amplifier current-limited? If we aren't producing tons of LF output, can we generate more current into the higher frequencies?

Anyway, if you know your stuff, please enlighten me.

Answers from EE's or especially those with backgrounds in this field would be very welcome.

[Jim O'Dell]

Interesting question. I can't help with the answer, but I'm curious what comes out of this. I would "think" that the amp would be reproducing all the information all the time, because the source is sending it. Using a crossover can only change what is output from it. But I understand your question as to where that part of the spectrum goes. It is energy. Capacitors store energy, but they have a limit as to their capacity (sorry about that), so what happens when the capacitor reaches it's limit? and how does that affect the amp? Maybe we should all get electronic crossovers and multi-amp our speakers so that we only send to the amp what it is supposed to amplify. But then, the question is still there on a pre-amp level....I think I'll shut up and wait for the answer.....:rolleyes: Jim.

[Jamie Buxton]

You're correct that a capacitor in series with the speaker increases the impedance seen by the amp at low frequencies. The output voltage of the amp doesn't change much when the load impedance goes up, so less power gets from the amp to the speaker.

However, you might not be able to observe this effect by measuring the power that the amp draws from the wall. Amplifiers aren't super-efficient. That is, only some of the mains power goes to the speakers. The rest becomes heat. In measuring at the wall, you're measuring the heat-producing portion plus the speaker portion, and then hoping to see the change in the speaker portion when you add the series capacitor.

[Phil Thien]

You're correct that a capacitor in series with the speaker increases the impedance seen by the amp at low frequencies. The output voltage of the amp doesn't change much when the load impedance goes up, so less power gets from the amp to the speaker.

However, you might not be able to observe this effect by measuring the power that the amp draws from the wall. Amplifiers aren't super-efficient. That is, only some of the mains power goes to the speakers. The rest becomes heat. In measuring at the wall, you're measuring the heat-producing portion plus the speaker portion, and then hoping to see the change in the speaker portion when you add the series capacitor.

Thank you Jamie. The measuring at the wall was just to illustrate my point, I'm not actually going to try that.

You say "less power gets from the amp to the speaker." Is less power being generated by the amp by adding the capacitor, all other things being equal. Or is the amp still producing the same amount of power?

[Bob Coleman]

I don't really know what's what when it comes to audio equipment, but it sounds like you've connected a high pass filter to the output of a signal generator. The higher frequencies ignore the cap and drive the speakers and the low frequencies are shorted through the capacitor. The energy is stored and discharged by the capacitor out of phase with the driving signal.

To sum up - the amp puts out less power, with the capacitor changing the circuit impedance.

[Phil Thien]

The energy is stored and discharged by the capacitor out of phase with the driving signal.

So you're saying the LF energy goes through the cap, and some part of it exits the cap inverted in phase, cancelling some part of the signal that made if through the cap that did not invert in phase?

I assume that the lower the frequency, the more equal the two parts (the part inverted vs. the part that isn't inverted), so more of the energy is canceled as the frequency drops?

[Phil Thien]

BTW, everyone is welcome to speculate.

If you're an actual degreed engineer, though, please mention that so I can identify those speculating vs. those that are not.

[Jamie Buxton]

You say "less power gets from the amp to the speaker." Is less power being generated by the amp by adding the capacitor, all other things being equal. Or is the amp still producing the same amount of power?

The amp's actual output power is measured at the load -- the loudspeaker. So adding the capacitor reduces the power output by the amp.

Consider this experiment... Connect a loudspeaker to the amp (without the capacitor), and play music. There's power reaching the loudspeaker. Now disconnect the speaker. The power delivered to the speaker is zero. That is, the amp is outputting zero power. Or here's another... The 115 VAC power outlet in your wall. With nothing plugged into that socket, there's no power being drawn out of the socket. There's still that 115 volt AC waveform happening at the socket. You can see it with an oscilloscope. But no current is coming out. Power is voltage times current. With no current, there's no power coming out. Plug in a lamp or the like, and now you get power transferred from the wall to the lamp.

[Phil Thien]

The amp's actual output power is measured at the load -- the loudspeaker. So adding the capacitor reduces the power output by the amp.

Consider this experiment... Connect a loudspeaker to the amp (without the capacitor), and play music. There's power reaching the loudspeaker. Now disconnect the speaker. The power delivered to the speaker is zero. That is, the amp is outputting zero power. Or here's another... The 115 VAC power outlet in your wall. With nothing plugged into that socket, there's no power being drawn out of the socket. There's still that 115 volt AC waveform happening at the socket. You can see it with an oscilloscope. But no current is coming out. Power is voltage times current. With no current, there's no power coming out. Plug in a lamp or the like, and now you get power transferred from the wall to the lamp.

This seems logical to me. Bob Coleman's post above about the phase shift in the capacitor causing some cancellation of part of the signal has me wondering, through.

[Chuck Wintle]

what is the impedance of the transducer you are using(theoretical value) as this will determine the 3db point of the roll off or 1/2 power point and the -6db point or the 1/4 power point, that is the power that goes to the transducer. i think a good understanding of Coulomb's will help in understanding how the "power" is dissipated at low frequencies .

[Phil Thien]

what is the impedance of the transducer you are using(theoretical value) as this will determine the 3db point of the roll off or 1/2 power point and the -6db point or the 1/4 power point, that is the power that goes to the transducer. i think a good understanding of Coulomb's will help in understanding how the "power" is dissipated at low frequencies .

This is a theoretical question, and isn't so much about the power that goes to the transducer, but how much power is generated by the amp.

You'd think those two #'s would be the same, but resisters in passive networks can convert power to heat and therefor the amp can generate power that isn't accounted for at the transducer.

If Bob Coleman is correct and capacitors cause a 180-degree phase shift that sort of destroys watts at descending frequencies, then the amp is still producing the watts and the capacitor is somehow eating them with the phase shift (without converting them to heat). But that doesn't sound right to me. But my intuition has screwed me before.

[Jamie Buxton]

but resisters in passive networks can convert power to heat and therefor the amp can generate power that isn't accounted for at the transducer.

True, but you're confusing yourself by changing the circuit diagram. You've added resistors to the network, when your original diagram only had a capacitor. With resistors in the network, the output power of the amp becomes the sum of the power delivered to the transducer, plus the power delivered to the resistors.

the amp is still producing the watts then the amp is still producing the watts and the capacitor is somehow eating them

This sentence doesn't make any technical sense. If a generator (like the amp) is "producing watts", the watts are going someplace, like the loudspeaker or those resistors. Those watts produce observable heating or sound. Capacitors don't "eat" power. In a network which includes reactive components -- capacitors or inductors -- there are some currents running around which are out of phase with the driving voltage, and aren't involved in the power calculation. Power is only what arrives at the resistive loads -- the loudspeaker or those resistors in the crossover.

[Jamie Buxton]

Let's say you connect a capacitor directly across the output terminals of a voltage generator like a music amp. (Let's also suppose the amp is a magic amp that will not be damaged by this. Real amps would be damaged, or more likely they will have protection circuitry built in to shut the amp down when presented with this situation.)

There will be current running through the capacitor. It will be ninety degrees out of phase with the output voltage of the amp. The exact definition of power is the time integral of the current and voltage waveforms. Because of the ninety degree shift between the two waveforms, the time integral turns out to be zero. That is, zero power is being put out by the amp. Another way of looking at this is that there are no resistors in this simple output network. There's nowhere for the electricity to turn to heat. That is, there's no power being output by the amp.

[Bob Coleman]

The capacitor stores energy based on a time constant (it can only do things so fast). For high frequency signals, the capacitor doesn't really have time to respond and you can think of it as a short to this input - ie, no effect. But a low frequency signal has time to charge up the capacitor - which then acts like an open to this signal, effectively filtering the low frequency.

As to power consumption, you effectively have two loads in series now, so for a given input signal, the amp must deliver more current to get the equivalent power at the speaker. Scroll nearly to the end here:

http://www.kean.edu/~asetoode/home/tech1504/circuit1/srcc.htm for the relevant equations.

Without the capacitor, Xc is zero. With the capacitor, current is smaller, so the voltage drop at each component is smaller (one small quibble with Jamie: power at a load is either voltage drop at the load or current through the load, power output of a source is the sum of the power consumption of the loads, source output voltage times output current) So to get the same power out of the speaker, you would need to get more out of the amp. (Note - capacitors do not consume real power but because they change the overall impedance of the circuit, they have an effect on the output power)

[Phil Thien]

The capacitor stores energy based on a time constant (it can only do things so fast). For high frequency signals, the capacitor doesn't really have time to respond and you can think of it as a short to this input - ie, no effect. But a low frequency signal has time to charge up the capacitor - which then acts like an open to this signal, effectively filtering the low frequency.

As to power consumption, you effectively have two loads in series now, so for a given input signal, the amp must deliver more current to get the equivalent power at the speaker. Scroll nearly to the end here:

http://www.kean.edu/~asetoode/home/tech1504/circuit1/srcc.htm for the relevant equations.

Without the capacitor, Xc is zero. With the capacitor, current is smaller, so the voltage drop at each component is smaller (one small quibble with Jamie: power at a load is either voltage drop at the load or current through the load, power output of a source is the sum of the power consumption of the loads, source output voltage times output current) So to get the same power out of the speaker, you would need to get more out of the amp. (Note - capacitors do not consume real power but because they change the overall impedance of the circuit, they have an effect on the output power)

This makes more sense to me than your original post, which I may not have fully contemplated.

So I think my original thought, and Jamie's post, and you are in agreement that as the frequency decreases, the cap causes an increase in impedance, which prevents the amp not just from delivering power but from making the power in the first place (just talking the amp, not the amp's power supply)?

Edit to add: I think I just confirmed this at another forum that caters to amateur speaker builders. I didn't want to post over there to begin with because some of the answers immediately go into the esoteric and beyond my understanding. I took a chance and got further confirmation of what you guys are telling me.

Thank you for all your help.

[Gary Chester]

Sounds like you've got your answer, but here's another way to look at it...

Let say your amp is putting out 8 volts at "whatever" frequency into an 8 ohm load, that's 8 watts of power at the load (1 amp x 8 volts).

Now let's add that cap in series with the load and inject a 200hz sine wave, let's pick a cap value that happens to have an impedance of 72 ohms at 200hz.

The amp is still putting out 8 volts, but the load it sees is 80 ohms (8 ohms + 72 ohms). That gives a current of .1 amps, that means the amp is only putting out .8 watts (.1amp x 8 volts).

So your amp is not "cranking out the power at low frequencies", but it is "cranking out the voltage" at those frequency's.

Hope that helps...

[Phil Thien]

Sounds like you've got your answer, but here's another way to look at it...

Let say your amp is putting out 8 volts at "whatever" frequency into an 8 ohm load, that's 8 watts of power at the load (1 amp x 8 volts).

Now let's add that cap in series with the load and inject a 200hz sine wave, let's pick a cap value that happens to have an impedance of 72 ohms at 200hz.

The amp is still putting out 8 volts, but the load it sees is 80 ohms (8 ohms + 72 ohms). That gives a current of .1 amps, that means the amp is only putting out .8 watts (.1amp x 8 volts).

So your amp is not "cranking out the power at low frequencies", but it is "cranking out the voltage" at those frequency's.

Hope that helps...

Yep, good clarification, thank you.