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Old 25th October 2006, 07:44 PM   #1
Baldin is offline Baldin  Denmark
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Default Discrete Class-D anno 1986

Just thought I would share my first encounter with class-d amplification.

This artichle is from 1986, from the now closed Danish magazin "Ny Elektronik" ...... so I've had it in my drawer for 20 years

It's unfortunatly in danish, but the diagram and component list is universal

We're talking about a total construction with only 5 transistors ....

5W from a single 12 VDC ... distortion .... not known .... switch freq 100 kHz

Maybe it could be adopted to some more power and split supply.

I know it isn't high end, but I think it's quite intriguing anyway

UcD = 14 transistors ....... this one 5!
http://home20.inet.tele.dk/audio/Cla...ronik_1986.PDF
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Old 25th October 2006, 08:28 PM   #2
BWRX is offline BWRX  United States
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That's a very cool circuit! Clever implementation of a prefilter RC feedback loop. What is the purpose of R11? Why not move C3 after the output filter?
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Old 25th October 2006, 08:38 PM   #3
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It says in the text that R11 is for biasing the output transistors.

Yes I would guess C3 could be moved to the output, but that would change the effect of R11.

T1 and T2 forms a schmit-trigger and T3 is an inverter.
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Old 26th October 2006, 08:04 PM   #4
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cute little amp... I wish to READ more but I cant(understand it)...

anyways, how many watts could it output?? maybe 15W?

EDIT: oooppps 5W...

thanks Baldin
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Old 26th October 2006, 08:12 PM   #5
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A single 12V rail with a half bridge output stage will only give you 4.5W peak (2.25Wrms max) into 8ohms and 9W peak (4.5Wrms max) into 4ohms. And that's assuming a full output voltage swing and the output stage can provide enough current into those impedances.
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Old 26th October 2006, 08:16 PM   #6
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Quote:
Originally posted by BWRX
A single 12V rail with a half bridge output stage will only give you 4.5W peak (2.25Wrms max) into 8ohms and 9W peak (4.5Wrms max) into 4ohms. And that's assuming a full output voltage swing and the output stage can provide enough current into those impedances.

Brian,

could you give me the maths here... I cant seem to remember WHERE I saved those equations.. also, how do you calculate power output with dual polarity supply (+/-)...

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Old 26th October 2006, 08:18 PM   #7
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RX5
There is not that muct to read about the workings in the article.
I think you would need to simulate to get a better understanding and to find ways to improve. .... could be quite usefull to have a very simple desig for, say 20W .... would be enough to drive a small active speaker for the PC or something.

Then we'll of course also need a similar simple SMPS
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Old 26th October 2006, 08:24 PM   #8
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P = ((Vp/sqrt(2))^2) / R

So as Pafi says:
Vp = 12/2 = 6 V

P = ((6/1.414)^2)/8 = 2.25 W in 8 ohm

For a +- supply Vp equals the transformer voltage
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Old 26th October 2006, 08:25 PM   #9
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Quote:
Originally posted by Baldin
RX5
Thee is not that muct to read about the workings in the article.
I think you would need to simulate to get a better understanding and to find ways to improve. .... could be quite usefull to have a very simple desig for, say 20W .... would be enough to drive a small active speaker for the PC or something.

Then we'll of course also need a similar simple SMPS

hhhmmmmmm WHY didnt I think of it?? hahahaha sure would be nice to simulate and improve...

in a few days....
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Old 26th October 2006, 08:25 PM   #10
BWRX is offline BWRX  United States
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To calculate peak theoretical power output you just need to know the maximum output voltage across the load and the load impedance.
Peak power = (V^2)/R
RMS power = ((V/sqrt(2))^2)/R = (V^2)/(2*R)

For a single ended output the maximum theoretical output voltage is half of the total rail voltage. For a bridged output the maximum theoretical output voltage is the rail voltage.

Edit: Baldin beat me to it
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