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Pure Class A Single End Amplifier Idea!
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Old 18th July 2007, 09:14 AM   #11
Gordy is offline Gordy  United Kingdom
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Quote:
Originally posted by AndrewT


Go back to resistor loading.
Yes, and add the small emitter resistors to the input pair as has previously been suggested. In this way the gain is effected by resistors, which are easy to match, and not by active devices, which are more difficult to match.
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Old 18th July 2007, 09:20 AM   #12
Dxvideo is offline Dxvideo  Turkey
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So you say, keep the first design and just add dedicated emitter resistor to each transistor on input like 270ohm or so. Is it?
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Old 18th July 2007, 09:40 AM   #13
AndrewT is offline AndrewT  Scotland
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Quote:
Originally posted by Dxvideo
So you say, keep the first design and just add dedicated emitter resistor to each transistor on input like 270ohm or so. Is it?
Re=270r might be a bit high.
The circuit will work and work well, but there are upsides/downsides to high Re values.

Re reduces stage gain, but makes gain more predictable.
Re increases stage noise, but this is not usually a problem in power amps if values are selected to minimise the audible effect.
Re enables measurements to be taken to check operation and for debugging.

For a low power amp probably using sensitive speakers I would keep Re<=100r but it will work with higher values.
The gain of the stage can be restored by increasing the LTP current. This also reduces the stage noise. This way you can get back to where you are with your present schematic. Have you noticed that very low noise BJT input opamps have high operating current? Why?
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Old 18th July 2007, 09:54 AM   #14
Dxvideo is offline Dxvideo  Turkey
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I will try all these alternatives. But first I must be sure;
I will increase the input stages current, its currently 1,1mA as you calculated. So I can increase it to 2mA by changing the resistor value with 12K.
I will add 82ohms to emitters. Mean;
An externally hosted image should be here but it no longer works. Please upload images instead of linking to them to prevent this.

Is that modification makes balanced the LTP? Because you mentioned "the LTP input pair are not balanced." in your first comment.
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Old 18th July 2007, 09:56 AM   #15
Gordy is offline Gordy  United Kingdom
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Re-listing the suggestins from the member contributions:

1. return to collector resistors in the input pair
2. add low value emitter resistors to the input pair (experiment with values in simulation; AndrewT suggests 22 to 200 Ohms)
3. add input capacitor
4. consider current source to replace R1
5. add capacitor across R8 (as you have)
6. consider compensation capacitor across B-C of Q5 (as you have)
7. add capacitor across D1 zener diode
8. although the Q3, Q4 current source will work, you may wish to experiment with a boot-strapped current source resistor as an alternative (see Note 1)
9. if you have a dc off-set at the output one solution is a large value capacitor under R9


Note 1: for inspiration for boot-strapped current source refer to http://sound.westhost.com/project03.htm
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Old 18th July 2007, 10:14 AM   #16
Dxvideo is offline Dxvideo  Turkey
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Ok thats the answers,
1. return to collector resistors in the input pair --> done
2. add low value emitter resistors to the input pair --> done
3. add input capacitor --> I have no DC offset in my source. I will use a class A bipolar preamp before it, and it already have an output cap.
4. consider current source to replace R1 --> done
5. add capacitor across R8 --> done
6. consider compensation capacitor across B-C of Q5 --> done
7. add capacitor across D1 zener diode --> It needs to have an explanation "why?"
8. although the Q3, Q4 current source will work, you may wish to experiment with a boot-strapped current source resistor as an alternative --> I examined Rod Eliotts circuit. However I am newbie on solid state designing. (In fact this is my first design) And sorry I couldnt understand. Is it for output power increasing? Because as I remember the bootstrap concept is used for that???
9. if you have a dc off-set at the output one solution is a large value capacitor under R9 --> I dont want to change my phase response, this is a bad idea I think. But may be I can add a op-amp dc servo between output and NI input.. Does it affect the phase response?
Thx.
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Old 18th July 2007, 11:01 AM   #17
Dxvideo is offline Dxvideo  Turkey
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But I am still confused on collector resistor issue.. This is a friends comment on it (which owner of current mirror idea) ;

"I study the distortion and find that much is IHF which is caused by the fact that the amplifier is not linear and positive and negative half cycle amplitude is different by 3V - VERY BAD! this is caused by the long tail pair that is operating harder in NFB for positive half cycle, so I remove R4 and replace it with a current mirror dropping to THD to a respectable 0.0057% which is very good."

What do you think?
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Old 18th July 2007, 11:16 AM   #18
AndrewT is offline AndrewT  Scotland
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Hi,
the voltage across the collector load (R12) is fixed by Vbe of the VAS (Q5).
This fixes the current through the collector load. Iload=Vbe/R12=0.6/2700=0.22mA.

The LTP tail current must be accurately set to twice the collector current, Ic = Iload+Ibase.
Ibase =Ic5/hFE=6mA/(200 to 400)~=0.03mA to 0.015mA.
Ic=0.22+0.02=0.24mA. Two times Ic=0.48mA.
Itail=[24-1]/12000~=1.9mA
You are miles away from balanced.
You must reduce R12 to balance the currents in the two halves of the LTP. You must do this accurately. That's why I suggested the trimmer option.

Add a cap across Zin (same as across R8). This is the RF filter. To make the RF filter effective you may need to add a series resistor to the input line (this adds noise so keep the value lowish 330r to 1k0).

You have confirmed that your source is DC blocked. Now look at the resistances on the two inputs.
The non-inverting sees 4k7 (R10)
The inverting sees 4k7//220r (R8 & 9)=210r.
This unbalances the DC conditions across the LTP and forces the output to offset. DO NOT fit a pot across the emitters of the LTP. Balance the input resistances and then fit a trimmer to the CCS feeding the VAS to fine tune the output offset.

Consider what happens to the current through R1 when signal changes force the +ve rail to change voltage. Signal induced modulation will cause the LTP currents to vary. This will impact on the sound (probably bad). I suggest the tail current be fixed. You have a number of solutions, select one.
Miller compensation cap (C1) will be audible. Consider alternative compensation methods.
Cap across D1 will lower the impedance seen by the base of the CCS (Q7). The lower the impedance, the more constant the CCS action with varying load currents. The cap will also sink some noise generated by the Zener. But this cap may be audible (good or bad).
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Old 18th July 2007, 11:18 AM   #19
AndrewT is offline AndrewT  Scotland
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Quote:
so I remove R4 and replace it with a current mirror
so must be a typing error!
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Old 18th July 2007, 11:20 AM   #20
Dxvideo is offline Dxvideo  Turkey
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No pls look at the first circuit. R4 is the collector resistor in first circuit.
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