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Tweakers' Asylum Tweaks for systems, rooms and Do It Yourself (DIY) help. FAQ. |
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Tweakers' Asylum Tweaks for systems, rooms and Do It Yourself (DIY) help. FAQ. |
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In Reply to: Re: MOSFETs work fine too... posted by Steve Eddy on December 14, 2001 at 16:06:40:
Regarding ‘gained momentum’.There is no such a term mentioned in any of my undergrad physics text books. The momentum of a mass is given by its velocity and mass alone. There are no other variables to add. The force required to bring a mass that is travelling at some velocity to a halt is the same as the force required to accelerate that mass in the first place.
Introducing friction complicates things. In particular, the environment in which a speaker cone moves can be thought of as a mass, a force and a spring. The mass is made up of the speaker’s cone, the coil and any other moving parts, and the air mass that is moving with the cone.
The spring can be thought of as having two components - compression and extension. The compression spring can be thought of as all the forces that impact on the cone and prevent it moving. The compressibility of the air in front of the cone would be the main component (for a speaker moving outwardly).
Extension springs can be thought of as pulling against the speaker cone and so hindering its motion. The speakers spider and the surround at the top of the cone, electrical damping, and the damping that the box may provide can all be added together with the compression spring to give one figure for the spring (resistance to the cones motion).
This is not the same as friction. Friction prevents motion. Your hockey puck does not spring back. Speaker cones do. If a voltage were applied to a speaker from an infinite impedance source in one polarity only, say zero to 9 volts, then the driver returns to its rest position of its own accord. All piston-cone drivers will overshoot (decaying oscillations).
As it returns, the speaker produces a voltage.
If there is a low impedance source then the driver will return to its original position slower and will overshoot less.
Regardless of the tension of the spring, a driver will still have to slow down before it can move off in the other direction. That is physics. Whether or not the period in which this slowdown occurs is significant concerns Audio Engineers.
You claim that the ‘friction’ nulls the overshoot condition is erroneous. Leave aside the overshoot and just look at the conditions at the time the reverse voltage occurs (for the low frequency sine wave and drum beat with opposite phase to the sine wave at the instant the drum is struck). If it were a friction effect, like the hocky puck analogy, and ignoring any overshoot then you are 100% correct.
But if we replace friction with spring then at the moment the drum beat arrives there is a spring effect. This spring is going to pull the speaker in the same direction the amplifier was wanting to push it anyway. So the impedance will be higher that it would be if the speaker was stationary (for the overshoot condition, it would be momentarily lower then higher once it started moving in the right direction).
In an extreme case, the voltage produced by the speaker would be significant. A MOSFET amp (with feedback)can take the voltage produced by the speaker into account when calculating the correct voltage over the load. It does this by sensing the voltage at the output and comparing that to the input voltage.
It is because of the intrinsically higher impedance of the MOSFET amp that they are able to do this. An intrinsically lower impedance amp effectively blinds the feedback from any activity beyond the output devices.
And since the amplifier's output impedance is in series with the load, all of the current flowing through the load is also flowing through the amplifier's output impedance. And Ohm's Law says that you must be dropping 1 volt across the amplifier's output impedance.
No, the amplifiers output is in parallel with the load. The only series circuit is between the amplifier’s output and the supply rails via the output devices.
The amplifiers impedance is in parallel with the speaker. If the amplifier’s output impedance was zero ohms then this would be equivalent to a dead short across the speaker (assuming no loss on the leads).
The voltage drop across the amplifiers output is always the same as the voltage drop across the speaker (assuming no lead loss).
Your calculations are erroneous. When calculating the current drawn by the circuit you only need the voltage output of the amplifier and the impedance of the speaker at the test frequency. I’ve done this many times with an accurate current probe and it works out exactly as expected. The amplifier’s output impedance never enters the calculation. (I have done this with high impedance non-feedback MOSFET amps as well - no difference).
Output impedance disappears when the amplifier is switched off. A push-pull transistor amplifier has each pair of devices biased so that they push against each other (electrically). The output resists being changed from the value set by the bias (when the amplifier is idling).
The reason why amplifiers using Hitachi MOSFETs and a feedback circuit have effectively low output impedance is that any voltage appearing at the output, whether generated by the amplifier itself or the load, is compared with the input. An error voltage is generated to correct for any voltages that don’t match.
Apart from output impedance, the feedback corrects for non-linearities in the voltage amplifier stage. All the voltage amplification in most MOSFET amps is done by a transistor voltage amplifier stage.
The speed at which this can be done is roughly equivalent to a 3 megahertz frequency (depending on the stability of the amplifier and the placement and value of various capacitors in the voltage amplifier and feedback circuits).
Keeping all else equal, drop the load impedance down to 4 ohms. 17 volts into 5 ohms (4 ohms for the load, 1 ohm for the amplifier's output impedance) gives you 3.4 amps. Now you've got 13.6 volts across the load and 3.4 volts dropped across the amplifier's output impedance.
No, if the load is 4 ohms and the voltage is 17VRMS then the current drawn is 4.25ARMS. Note that this is not the case if you are measuring the amps voltage at the gates rather than at the output.
Just let me reiterate: the amplifier’s output impedance is measured between the amplifier’s two output terminals (usually load and Earth/zero volt). This can be done by using a constant current sine wave voltage generator to pass a voltage to the amplifier’s output via a resistor. By measuring the voltage across the resistor you can use Ohm’s law to calculate the amplifier’s output impedance. If the voltage is higher, then the impedance is lower etc.
If you want lower output impedance without feedback, then use devices in the output stage which have an inherently lower output impedance. I don't know that you'll be able to reach the level you'd like using MOSFETs and it appears that MOSFETs are the only devices you'd consider.
I took one of my amplifiers around to a certain advocate of Plinius amps and substituted one of his giant class A amps for mine. Compared to the transistor monster, my amp had a clearer midrange (the Plinius was ‘thicker’) and sweeter highs (driving 10’ high electrostatics).
He bought the amp. :)
Kind Regards,
Robert Karl Stonjek.
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- Re: MOSFETs work fine too... - Robert Karl Stonjek 00:20:28 12/15/01 (0)
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