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AB amplifier, first try
AB amplifier, first try
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Old 24th September 2017, 05:28 PM   #1
DIYnovice is offline DIYnovice
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Default AB amplifier, first try

Hi.
I try modify some net circuit. Any obvious problem here?
Any stability problem?
Thermal stability not include. I try thermistor over R12, R13.
Thx to all advice.
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Old 25th September 2017, 12:47 AM   #2
HI-FI PRO is offline HI-FI PRO  China
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Dosen‘t seem like theres any plorbem
Thermal stability can be fixed by placing q11 onto same heat sink as q14 q15 are mounted on
However main concern for you is over biasing for your voltage gain stage 2sb 649.
Maximum power through transistor approximated to be half of vcc divided by emitter resistance times half of vcc
which means (15/47)(15)= 4.785W of power disiplatio. unless your vas is heat sinked it will die .

For your first design, highly recomended that you play at lower power for you gain stages, then step up the bias after you tested the first one to be working.
I would also recomend the collector resistors of the diffrential pair to all be the same values.

Approximation of idle bias for your vas is about 30-(3.3/2)(820)= voltage appearing at base of q7.
Voltage at emitter = vb-0.6 = approx0.765v
0.765/47*29v= approx470mW of quisent. This is half of maximum disipation without heat sink for vas. it will run boling hot.

Last edited by HI-FI PRO; 25th September 2017 at 01:02 AM.
 
Old 25th September 2017, 01:03 AM   #3
Bigun is offline Bigun  Canada
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AB amplifier, first try
The challenge with this circuit is keeping a stable current flow through the VAS, Q7,9, as temperature variations will affect their operating points. The use of R7 and R10 helps quite a bit, I made a circuit like this but with much higher degeneration on the VAS for better stability. With current variations through the VAS you also get variations in bias for such a simple Vbe multiplier (Q11). This too can be improved using the Hagerman circuit. What I did is shown here:TGM5 - all-BJT Simple Symmetric Amplifier

I was never that happy with the sound.

I had significantly better results with this one:TGM7 - an amplifier based on Greg Ball SKA
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Old 25th September 2017, 04:15 AM   #4
CBS240 is offline CBS240  United States
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pretty straight forward double diff, push pull VAS. degeneration of the LTP might be good. Med power VAS is not always a bad thing if designed properly but I believe it limits the components as far as specs go, that are available for such an operation. It all depends on the output stage for that.......which is a different subject.
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Old 25th September 2017, 02:56 PM   #5
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Ok. Thx. What value R7,R10 is good and max bias?
 
Old 25th September 2017, 04:46 PM   #6
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This is DC situation. Power is near 560mW.
With 47ohm, Q9 1.136W Biger resistor = biger THD. With same bias.
In simulation.
Any solution?
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Old 25th September 2017, 04:51 PM   #7
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I use heatsink to Q7-Q15. Q7 and Q9 solo.
Or not?
 
Old 25th September 2017, 10:20 PM   #8
HI-FI PRO is offline HI-FI PRO  China
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I recomend that R7 R10 stays at 470 ohms for the first prototype build. Everything else can be kept same and should work
You'd also want to adjust vbe multipleier at lowest current first, something like equal emitter and collector resistance for lowest biasing, once working and sound, you then adjust bias and play arround.
 
Old 26th September 2017, 08:01 AM   #9
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for d1 to d4 you can replace them with zener diodes, typically 2.4v to 4.7v is recomended. Larger voltage means biasing of diffrential pair changes less due to missmatching of transistors and resistors.
Voltage through r1 r2 can be estimated as zener voltage minus transistor be voltage assumed to be approx 0.6 to 0.7v.
this is the total current through q2 q3.
The current through just q2 estimated to be half of total current.

example 4.7v zener, voltage accross r1 estimated to be 4.7-0.6= 4.1v.
V=IR, because voltage is constant through r1r, the current will be popotional to current.

if r1 is 1k ohms this means the total current through q5 q6 is 4.1/1000= 4.1mA
The current through individual transistor q5 estimated to be half of 4.1mA = 2.05mA.

Because current is fixed you can then estimate the voltage drop from collector resistor.
Lets say you use a 2k2 resistor for Collector.(r5 r41)
the voltage accross 2k2 would be 2.05mA times 2200ohms=4.51V

This means the voltage accross R10 is estimated to be 4.5v-vbe drop of q9.
vbe approx 0.6v
this means 3.9v will be droped accrosed r10.

From there you can estimate idle current of q9 and detrimine heat sinks and resistor value of r10.

Maximum power consumpotion of resistor and q9 can be estimated
this is half of supply voltage divided by the value of r10. This tells you the current through r10 and q9.
From there you know current and volt drop accross q9 r10 is approx half of supply voltage. you can calculate maximum power disipation.
When you output approx 15v pk-pk sine wave the maximum power accross r9 would be this number

EDIT: Large zener diode values means if your diffrnetial pair needs lots to output a large voltage, your maximum voltage swing will be reduced. However mostcases this voltage swing is low because the following stage does most or all of the amplification.
Estimation of maixmum power example r10 is 470ohms, current would be 15/470=0.0319A
maximum power through q9 approx 0.0319*15 = 0.4785w
By the way this assumes r38 r39 are short circuited and all diffrential pair are directly suppled by vcc

Last edited by HI-FI PRO; 26th September 2017 at 08:21 AM.
 
Old 26th September 2017, 11:15 AM   #10
mcd99 is offline mcd99  United Kingdom
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Just wondering about the position of the shunt compensation. In this amp it is on the base of the VAS transistors. Normally the shunt compensation is the collectors of the VAS transistors. Not tried simulating the stability of this design so may work ok. But at present the VAS may be excessively affected by the varying load of the driver transistors as this is the only loading. So just asking the question.

Paul
 

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