Wow, talk about a train wreck. But I learned some stuff.
Does anyone have a link to an explanation of the difference between current and voltage amplifiers? Something at a good level for a guy with a good understanding of DC circuit, Ohm's law kinda stuff and an understanding of what impedance is would be especially appreciated. I had some classes on transistors and amps, 20 years ago, but I don't remember most of it and I don't seem to remember anything at all about voltage amps compared to current amps.
Does a current amp try to maintain a constant amperage by increasing voltage when impedance increases? Maybe I'm being too simplistic by thinking about it in such an Ohm's law kinda way?
Thanks,
Chris
Does anyone have a link to an explanation of the difference between current and voltage amplifiers? Something at a good level for a guy with a good understanding of DC circuit, Ohm's law kinda stuff and an understanding of what impedance is would be especially appreciated. I had some classes on transistors and amps, 20 years ago, but I don't remember most of it and I don't seem to remember anything at all about voltage amps compared to current amps.
Does a current amp try to maintain a constant amperage by increasing voltage when impedance increases? Maybe I'm being too simplistic by thinking about it in such an Ohm's law kinda way?
Thanks,
Chris
Hi Chris.
Yes, and so unnecessary.
No, not simplistic at all, since that is what happens. The voltage does indeed go up, but importantly, the current is maintained, so if we had 1 Amp RMS output, then with an 8 Ohm load, the voltage that will develop is 8V RMS across the load/Voice Coil. But if the impedance rises to 9 Ohm, you will see 9V RMS and so on. Of course, if voltage driven, then the current will in effect go down since 8/9 = 0.888 Amp.
We do know this: That it is the current through the VC that makes for motion and hence sound. Hence linear current and linear load would make for a linear motion. But so many things prevent that. We think that the voltage across the VC is what is being tracked, but it is not. With voltage drive, the voltage has to be converted to current - with current such a conversion does not take place. But we may end up with non-flat response due inductive EMF (above 100 Hertz typically) and motional EMF (below 100 Hertz typically) and also to a degree microphonic EMF. For the most (not always with vented boxes) the EMF forces the impedance up, the motor system wants to revert to becoming a voltage source - but with current send, that current always stays 1 Amp.
Indeed, as John K pointed out earlier, the response can be seen and predicted by looking at it as a voltage divider. If you keep the impedance flat across frequency, then you can drive it from any source impedance, both voltage (low) and current (high).
So the topic is really this: Some of us are hearing a distinct improvement in the clarity of the speaker when we start dealing with these things. There are a number of strategies, I have adopted one, others have adopted parallel resistance to flatten the Z response. Also, when doing this, we also see a flattening of the electrical current phase angle.
Again, there are pros and cons for the load on the amplifier, but then there is already with many commercial speakers that are not my cup of tea dropping below 2 Ohm, ouch.
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But Joe, John K gave a clear and factual explanation that the flattening does nothing to the speaker (no change in drive voltage) but asked more of the amp.
You say that you would like a good discussion. Why not debunk John's reasoning if you think it wrong? If you let that float, it'll comeback again and again because it undermines whatever you say.
Or say that you feel that it makes the speaker sound better despite the fact that the drive signal doesn't change.
It would be helpful to actually address the arguments and get them out of the way. Or not.
Jan
Sorry Jan, but I really don't want to go down that road again. Much of what I say is either ignored or being cherry picked, I realise belatedly that it is not worth the effort - especially when I am asked to debunk something I don't feel a need to debunk, in fact nothing to debunk.
And floating an idea is now a sin?
In fact, I don't even know what you guys want - maybe my IQ is not high enough, but I will float ideas if I feel like it. And I enjoy others floating theirs. I enjoy brainstorming ideas - the universe is still unfolding and let's have some fun.
"An expert is somebody who has made every mistake there is to be made in his field." Only very slightly paraphrasing Niels Bohr.
I am not yet an expert then. 🙂
And floating an idea is now a sin?
In fact, I don't even know what you guys want - maybe my IQ is not high enough, but I will float ideas if I feel like it. And I enjoy others floating theirs. I enjoy brainstorming ideas - the universe is still unfolding and let's have some fun.
"An expert is somebody who has made every mistake there is to be made in his field." Only very slightly paraphrasing Niels Bohr.
I am not yet an expert then. 🙂
So are you saying that John's analysis is correct? Or has he missed something, and if so, what?
What analysis? I agree with 80-90% of what John says.
But he has a habit of assuming that the rest of us know less than we actually do and lectures us on things we already know. He could ask first. Other than that, I have no problem whatsoever.
The usual suspects have turned up again. Hello Scott, you are next?
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Sorry Jan, but that just caught my eye. Answer would depend on what you mean by 'drive signal does not change.'
I mean, I am very willing to discuss this, but I don't want to misunderstand your position and I don't want you to misunderstand mine. Can we handle this rationally? I am willing to try, but the tone has to be a conducive one.
Just what John (and others in the past) have pointed out. If you flatten the impedance as seen by the amp, the signal that the speaker (and crossover) sees doesn't change a microvolt. So if it now sounds different, despite that the signal is exactly as before, what could cause that, apart from perception factors? Or if it is the perception factor, lets just say so.
BTW My 'float' comment was meant to mean 'if you (JR) let the comment from John K float, it will come back again and again'.
The fact that the impedance flattening does not change in any way the signal offered to the speaker is obvious to many so people will continue to comment that.
Jan
BTW Re-reading this and trying to find a possible misunderstanding that I missed, are you talking about current drive amplifiers?
In that case I can understand that flattening the impedance as seen by the amp does change things from the speaker perspective.
But then you also lose the advantage of current drive in that it sort of automagically compensates for speaker impedance variations. The combination of current drive with impedance flattening gets you in even more trouble. Seems to me that at points where the speaker impedance peaks, more current gets diverted through the flattening network and the peaks get even more pronounced than without the flattening network (assuming current drive).
Jan
In that case I can understand that flattening the impedance as seen by the amp does change things from the speaker perspective.
But then you also lose the advantage of current drive in that it sort of automagically compensates for speaker impedance variations. The combination of current drive with impedance flattening gets you in even more trouble. Seems to me that at points where the speaker impedance peaks, more current gets diverted through the flattening network and the peaks get even more pronounced than without the flattening network (assuming current drive).
Jan
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Well, John brought up a point or two that I hadn't considered. That's the whole point of technical discussion. So if there's anything you disagree with, let's get it out there and see if there's a hole in his argument.
Like all things, there are shades of grey.
Esa Merilainen is totally fixated on current drive. I have his book (maybe you could review it). He has some decent points to make and I do recommend a read of his book (even if it is written in Finglish) and that it is more crusading for my taste (but we still get on and I have no problems with him). But we are not going to convert the world to current drive, and there are many reasons for that. It's just not going to happen, period. But there is something to be said to tracking current as the method of motion in motors. There is something to be said about exploring techniques that are somewhere in the middle.
Now please, understand that sometimes you observe things and develop a gut feeling, one that may be true or just completely wrong. But having run so many 'what ifs' in SoundEasy (John is an expert on SoundEasy and I have spoken to Bohdan Raczynski and consider the earliest distributor Peter Stein a friend) that simulating a current source (high Z) always produced a zero degree phase angle. This gave me an idea worth exploring. This is only a tip of the iceberg, but earlier John said something I had already observed, that equalising the Z meant that Thevenin's predicted the response irrespective of the box alignment and so on. So, to cut a long story short, I took an existing design of mine, calculated (using SoundEasy) the values seen directly (not part of the crossover) by the amplifier directly and then used (within reason) only series inductors and series capacitors (tweeter) in the way that Esa preferred. SoundEasy showed that it should hang together nicely. I built the crossover and found that everything behaved as predicted. Then took a listen and was both surprised and very pleased with the result. There was a degree of obvious clarity not there before. It was also easier on the ear, less fatiguing - a guy I know with Ambience ribbon speakers (designed by Tony Moore) came over for three hours and said they sounded like ribbons. I considered that signficant praise.
Above is but an extremely brief recount of a much longer story. But is it so wrong to share this and see if other get similar results?
I have since done a number of designs this way - I like what it does. It makes a speaker that is able to be driven by any source Z (that alone cannot be bad).
For an 8 Ohm speaker, I aim for a flat 6 Ohm Z. With 4 Ohm it is somewhat more challenging, but aim as high and flat as I can. But sometimes you can't, then I simply won't do it. I have always aimed at easy-to-drive speakers and here we get something that looks as close as a resistor as possible..
It's 2 AM here. So will say goodnight.
Cheers, Joe
First, I don't assume others have less knowledge than I. But I recognize that everyone has a different level of understanding. Thus I try to address things in a manner that allows most people to cut through the spin and understand what is reality and what is fantasy.
The bottom line is simply this:
Case 1: With a voltage source, the current through the VCs will be affected by changes in the speaker Z due to temperature, or other factors, regardless of whether there is a compensation network or not. The voltage across the speaker will not be affected.
Case 2: With a current source, and the appropriate compensation network in place, both the current through the VCs and the voltage across the speaker will be affected by changes in the speaker Z.
If you want to say that unequivocally results in greater clarity, that's fine with me, though it seems a bit of a stretch.
Examples:
Case 1:Voltage source, Rs =0. Z increases, V is constant, I2 goes down.
Case 2:Current source, Rs finite. Z increases, Thus Z in parallel with Z*, (Z||Z*) increases. As a result of (Z||Z*) in series with Rs, V increase, Since V increases, Vs must increase to keep Is constant. But since V increased and Z* remains constant, I1 increases. Since I2 = Is - I1, I2 must decrease. So, voltage across the speaker increase but current decreases. That is, while the voltage across the speaker increases it does not increase sufficiently to counter the increase in Z, hence the current through the speaker decreases.
What this says to me is that, and remembering that the speaker Z includes a crossover network as well, the variation of the frequency response of the speaker as the VCs heat up will be different for a voltage and current driven system. Does that, in general, equate to better clarity?
In reality Rs of a voltage source isn't zero, but much less than Z or Z||Z*. With a current source, Rs is much greater that Z or Z||Z*. Thus, both V and I2 change for either source. But for a voltage source V changes much less than it does for a current source.
The bottom line is simply this:
Case 1: With a voltage source, the current through the VCs will be affected by changes in the speaker Z due to temperature, or other factors, regardless of whether there is a compensation network or not. The voltage across the speaker will not be affected.
Case 2: With a current source, and the appropriate compensation network in place, both the current through the VCs and the voltage across the speaker will be affected by changes in the speaker Z.
If you want to say that unequivocally results in greater clarity, that's fine with me, though it seems a bit of a stretch.
Examples:
An externally hosted image should be here but it was not working when we last tested it.
Case 1:Voltage source, Rs =0. Z increases, V is constant, I2 goes down.
Case 2:Current source, Rs finite. Z increases, Thus Z in parallel with Z*, (Z||Z*) increases. As a result of (Z||Z*) in series with Rs, V increase, Since V increases, Vs must increase to keep Is constant. But since V increased and Z* remains constant, I1 increases. Since I2 = Is - I1, I2 must decrease. So, voltage across the speaker increase but current decreases. That is, while the voltage across the speaker increases it does not increase sufficiently to counter the increase in Z, hence the current through the speaker decreases.
What this says to me is that, and remembering that the speaker Z includes a crossover network as well, the variation of the frequency response of the speaker as the VCs heat up will be different for a voltage and current driven system. Does that, in general, equate to better clarity?
In reality Rs of a voltage source isn't zero, but much less than Z or Z||Z*. With a current source, Rs is much greater that Z or Z||Z*. Thus, both V and I2 change for either source. But for a voltage source V changes much less than it does for a current source.
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Thanks for the reply, Joe. That makes complete sense to me.
I promise I won't keep mucking up your thread with my beginner questions but you lost me with the statement above. Voltage is the force that causes current to flow, while resistance/impedance controls how much current flows for a given voltage.
So I don't understand what you mean by "voltage has to be converted to current".
-Chris
Own experience adding resonance and inductance compensation for a woofer is that visual cone decrease low frq oscillation and especially on recordings with low frq rumble and also subjective think there is a change in sound. But my compensation trials sits close to driver and run active XO so driver look directly into power amp via speaker cable's RCL values.
Schematic in post 1 show a different compensation in it sits before a passive XO giving a high impedance route to driver via XO components and more cable RCL values.
Objective it looks like in sim for a power amp that if one zoom into uS/nS in group delay plot that it reflect a plot of speaker impedance load in that frq at highest impedance load comes first and lowest impedance load later, think it makes sense group delay plot replicate load impedance plot but if such small timing difference as uS/nS are is audio able may be a question.
Will try some tests in future if resonance compensation is visual in waterfall plot zooming into first slices around resonance frq with low rise time settings.
Schematic in post 1 show a different compensation in it sits before a passive XO giving a high impedance route to driver via XO components and more cable RCL values.
Objective it looks like in sim for a power amp that if one zoom into uS/nS in group delay plot that it reflect a plot of speaker impedance load in that frq at highest impedance load comes first and lowest impedance load later, think it makes sense group delay plot replicate load impedance plot but if such small timing difference as uS/nS are is audio able may be a question.
Will try some tests in future if resonance compensation is visual in waterfall plot zooming into first slices around resonance frq with low rise time settings.
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