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Old 15th July 2009, 08:54 PM   #11
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Yep, the diode makes the trick. (of dual supply working)
You've just created a buck regulator turned to current source.
This is clever, congratulations.
Note, that a dc voltage at a inductor is zero, which is the point here.
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Old 15th July 2009, 11:21 PM   #12
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And if you connect the load halfway into the sense resistor???
Does this make an active current source of equal impedance?

Hate to throw away double free power... If you are making it anyway...
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Old 16th July 2009, 09:01 AM   #13
AndrewT is offline AndrewT  Scotland
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Hi,
is the switched current sink ON for half the time?
Does the inductor smooth the switched current to effectively half the peak current?
Does the amplifying transistor see an effective constant current load ~=average smoothed current through the inductor?

Does any of that mean that the average bias has been reduced to ~50% of the peak bias?

Is it simpler to just turn down the bias to half? i.e. quarter (-6dB) of the maximum ClassA output power.
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Old 16th July 2009, 09:27 AM   #14
QED047 is offline QED047  United Kingdom
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Quote:
Originally posted by kenpeter
And if you connect the load halfway into the sense resistor???
Does this make an active current source of equal impedance?
Not sure what you mean here (but it sounds intriguing)... equal impedance to what?

I (simply) see everything below the junction between the follower source and load as a "black box" constant current sink. Its apparent DC impedance varies as a direct function of the follower voltage divided by the set current.

On the subject of "free power" there can, of course, be no such thing! As you know we're only re-distributing it here. The 2 Amps on the meter above is what's being drawn from the PSU. The current in the sense resistor (i.e. the actual bias current in the Class A stage is 1.9 x as much (1 + the efficiency factor of the switcher) thanks to the diode pumping the unwanted potential back into the reservoir capacitors of the PSU (albeit that it must be re-introduced carefully to avoid noise issues).

So on my breadboard setup shown above there's actually a fairly repectable 3.8 Amps bias flowing through the follower stage.
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Old 16th July 2009, 09:32 AM   #15
AndrewT is offline AndrewT  Scotland
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Quote:
Originally posted by QED047

On the subject of "free power" ..........................
...............The 2 Amps on the meter above is what's being drawn from the PSU. .............................there's actually a fairly respectable 3.8 Amps bias flowing through the follower stage.
2A (100% duty cycle) from the supply, but 3.8A (100% duty cycle) through the Follower.

How?
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Old 16th July 2009, 09:34 AM   #16
QED047 is offline QED047  United Kingdom
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Quote:
Originally posted by AndrewT
Hi,
is the switched current sink ON for half the time?
Does the inductor smooth the switched current to effectively half the peak current?
Does the amplifying transistor see an effective constant current load ~=average smoothed current through the inductor?

Does any of that mean that the average bias has been reduced to ~50% of the peak bias?

Is it simpler to just turn down the bias to half? i.e. quarter (-6dB) of the maximum ClassA output power.
No, hopefully my previous post will make it clearer how the efficiency is gained. What's happening in broad terms is that the unwanted energy that is normally lost in a a resistive element entirely as heat is temporarily accumulated in the inductor and then released back into the supply reservoir capacitor(s).
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Old 16th July 2009, 09:37 AM   #17
QED047 is offline QED047  United Kingdom
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Quote:
Originally posted by AndrewT

2A (100% duty cycle) from the supply, but 3.8A (100% duty cycle) through the Follower.

How?
By returning (recycling) the unwanted (heat) energy as electrical energy!
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Old 16th July 2009, 09:48 AM   #18
QED047 is offline QED047  United Kingdom
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I have to say here and now that the principles are absolutely sound ~ it's not a perpetual motion device

The fundamental principles are the same as those used in Class D amplification but their application is different. I really like single ended Class A. I like the simplicity of a ZEN style two-transistor amp (voltage gain in pre-amp, current gain in power follower). But the heat kills it for me.

Class D amps are complicated beasts that I doubt could be assembled on a breadboard - but keeping the switching out of the signal path and using it to manage the unwanted energy in the current sink is proving to be reasonably straight forward and is delivering a massive saving in energy.
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Old 16th July 2009, 09:55 AM   #19
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I believe some power LED driving chips would work here fine.
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Old 16th July 2009, 10:43 AM   #20
QED047 is offline QED047  United Kingdom
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Quote:
Originally posted by Wavebourn
I believe some power LED driving chips would work here fine.
Yes, although there aren't many (at the moment) that deliver more than a couple of Amps. Also they tend to operate at a fixed frequency ~500kHz with variable mark-space. This increases the peak-to-peak ripple voltage on the inductor which the self-oscillating design above keeps symmetric. Instead, the frequency shifts around with the audio frequency of the follower.

Fortunately the switching current is relatively low in the world of switch-mode designs so a small-ish transistor with low gate charge can be selected and the switching frequency can be kept high. The inductor I've used is also a lot higher in value (220uH) than most switchers so the ripple is kept small. I've gone for a toroidal with powdered core to store as much energy as possible while keeping stray field to a minimum. I think this is the best strategy.

The differential amp has to be high speed and have an excellent CMRR to handle the rail-to-rail swing of the sense voltage. I used an old OP37 from the junk box with a gain of 10 and it works fine.
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