Hi all,
After my sucessfull Class A headphone project (by the way I had some modifications on it and it works very good now)
http://www.diyaudio.com/forums/showthread.php?s=&threadid=104767&highlight=
I decided to make a speaker amplifier based on it.
I've been inspired from Rod Eliott's "Class A Op-Amp" project to make it a pure Class A design.
The bias current is 2 A or so. And it can give about 20W RMS into 8 ohm.
I havent tested it yet. But on 5spice software it works very very good. With 0 phase drift and 250Khz bandwidth. However I cannot make FFT and THD analyses.
Anyway, may you have a look and give some critics on it pls.
Thanks in advance a lot.
After my sucessfull Class A headphone project (by the way I had some modifications on it and it works very good now)
http://www.diyaudio.com/forums/showthread.php?s=&threadid=104767&highlight=
I decided to make a speaker amplifier based on it.
An externally hosted image should be here but it was not working when we last tested it.
I've been inspired from Rod Eliott's "Class A Op-Amp" project to make it a pure Class A design.
The bias current is 2 A or so. And it can give about 20W RMS into 8 ohm.
I havent tested it yet. But on 5spice software it works very very good. With 0 phase drift and 250Khz bandwidth. However I cannot make FFT and THD analyses.
Anyway, may you have a look and give some critics on it pls.
Thanks in advance a lot.
Hi,
the LTP input pair are not balanced.
R1 (22k) with near 24V will pass about 1.1mA.
Half of this should go through each of Q1 & 2.
Q5 passes about 9mA and it's base current will be about 20 to 40uA.
This leaves about 0.5mA through R4. It's value should be about 1k1 to 1k2.
You could add a trimming pot across a slightly higher value of R4. Try 1k3//(6k8+10kpot). Then set up the trimmer to balance the Vre on each side of the LTP. Oh, no Re, try adding emitter degeneration resistors of 22r to 200r. This will reduce the gain slightly but may sound better.
Why did you choose a & b grades of bc559/549?
I would have chosen c grade bc560/550 throughout.
the LTP input pair are not balanced.
R1 (22k) with near 24V will pass about 1.1mA.
Half of this should go through each of Q1 & 2.
Q5 passes about 9mA and it's base current will be about 20 to 40uA.
This leaves about 0.5mA through R4. It's value should be about 1k1 to 1k2.
You could add a trimming pot across a slightly higher value of R4. Try 1k3//(6k8+10kpot). Then set up the trimmer to balance the Vre on each side of the LTP. Oh, no Re, try adding emitter degeneration resistors of 22r to 200r. This will reduce the gain slightly but may sound better.
Why did you choose a & b grades of bc559/549?
I would have chosen c grade bc560/550 throughout.
Some thoughts
You don't want any possible DC from the signal input to Q1 base.
Solution: Put a capacitor between the input and Q1 base.
You want to minimize any DC offset that your amplifier could produce.
Solution: Connect a big series capacitor with R9. (I mean big capacitance.)
You want to maximize linearity of the differential amplifier, reduce gain and thus also reduce possible stability problems.
Solution: Put small equal emitter resistances to Q1 and Q2.
While the Zener diode D1 is a voltage source, it is also a noise generator. The noise is amplified in Q7 and goes to the speaker.
Solution: Connect a capacitor parallel with D1 to suck the produced noise.
If your amplifier has stability issues, you must compensate.
One possible solution: Try a small capacitor between Q5 collector and base.
Instead of the current source (R2, R3, Q3 and Q4) maybe you want to use something simple.
Solution: Use a bootsrap connection from the amplifier output. You need one capacitor and two resistors for that.
You don't want any possible DC from the signal input to Q1 base.
Solution: Put a capacitor between the input and Q1 base.
You want to minimize any DC offset that your amplifier could produce.
Solution: Connect a big series capacitor with R9. (I mean big capacitance.)
You want to maximize linearity of the differential amplifier, reduce gain and thus also reduce possible stability problems.
Solution: Put small equal emitter resistances to Q1 and Q2.
While the Zener diode D1 is a voltage source, it is also a noise generator. The noise is amplified in Q7 and goes to the speaker.
Solution: Connect a capacitor parallel with D1 to suck the produced noise.
If your amplifier has stability issues, you must compensate.
One possible solution: Try a small capacitor between Q5 collector and base.
Instead of the current source (R2, R3, Q3 and Q4) maybe you want to use something simple.
Solution: Use a bootsrap connection from the amplifier output. You need one capacitor and two resistors for that.
Regarding output stage, I have already seen something like this 😉
The input LTP might be improved, and the output stage should be driven from low impedance, rather than from Q5 collector.
The input LTP might be improved, and the output stage should be driven from low impedance, rather than from Q5 collector.
Input LTP now improved I guess...
The 15pF on NFB line to kill 2,2Mhz oscillation which I recently noticed.
An externally hosted image should be here but it was not working when we last tested it.
The 15pF on NFB line to kill 2,2Mhz oscillation which I recently noticed.
No there is no special reason to use darlington. In fact I use only a power resistor for this position in my first design. (Headphone amp). After then I though to put a 5A regulator IC (like LM338 or LT1083) as a constant current source but I faced that power dissipation limits exceeded. Then I found that solution on a web design. Its not special, and not my design.
Hi Dxv,
that mirror on the input LTP is tried by many and many more citicise it for poor sound quality.
In addition the current into the base of Q5 unbalances the LTP. Is that why is sounds bad?
Go back to resistor loading.
that mirror on the input LTP is tried by many and many more citicise it for poor sound quality.
In addition the current into the base of Q5 unbalances the LTP. Is that why is sounds bad?
Go back to resistor loading.
AndrewT said:
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.
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.Dxvideo said:
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?
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;
Is that modification makes balanced the LTP? Because you mentioned "the LTP input pair are not balanced." in your first comment.
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 was not working when we last tested it.
Is that modification makes balanced the LTP? Because you mentioned "the LTP input pair are not balanced." in your first comment.
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
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
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.
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.
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?
"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?
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).
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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