I changed the vas 2sc1815/2sa with BD139-140, i get some more power, but even at 30mA on the vas , I can't get more than 14-15v rms without distortion, What's wrong I dont know... I think I could get 20v rms from DC 36.5-0-36.5 transformer on this schematic
I get 22v rms out but that distorts like hell, transofrmer doesnt drop more then 1-1.5v from 36.5v so I dont think that's the issue, bride rectifier is 15 Amps ( "square" package ) one 6800uF cap per rail.
Output stage bias is 100mA tried even higher bias , same results.
I get 22v rms out but that distorts like hell, transofrmer doesnt drop more then 1-1.5v from 36.5v so I dont think that's the issue, bride rectifier is 15 Amps ( "square" package ) one 6800uF cap per rail.
Output stage bias is 100mA tried even higher bias , same results.
You can´t ask for precision values when you measure by ear.
from your description, I would expect 16/17V RMS at clipping start.
With a test sinewave and driving the proper load resistor, of course.
Your
Do you mean DC or AC?
Transformer windings before bridge rectifier or rectified and filtered DC rails?
In any case, , which will make amp clip quite earlier than apparent.
Of course, you see that on a scope
from your description, I would expect 16/17V RMS at clipping start.
With a test sinewave and driving the proper load resistor, of course.
Your
statement is confusing.
Do you mean DC or AC?
Transformer windings before bridge rectifier or rectified and filtered DC rails?
In any case, , which will make amp clip quite earlier than apparent.
Of course, you see that on a scope
The peak output should be able to reach about 3V (possibly 2V) below the rail. So if you have 36.5V D.C I would expect to be able to achieve at least 33v output.
If it sounds distorted there may be other reasons. In a complementary feedback circuit like this oscillation can occur in the output stage because of the relatively high loop gain and phase shifts. A Miller capacitor, like your 100pF, normally prevents this but there might be some layout issues (wiring leads etc) that cause the problem.
If the transistors appear to be getting hotter than you expect, perhaps oscillation is the cause.
On the other hand do your loudspeakers handle the power (33V peak into 8 ohms is 80W)
Sometimes the cone excursions can hit the loudspeaker material in the cabinet and cause a similar sound to clipping.
If it sounds distorted there may be other reasons. In a complementary feedback circuit like this oscillation can occur in the output stage because of the relatively high loop gain and phase shifts. A Miller capacitor, like your 100pF, normally prevents this but there might be some layout issues (wiring leads etc) that cause the problem.
If the transistors appear to be getting hotter than you expect, perhaps oscillation is the cause.
On the other hand do your loudspeakers handle the power (33V peak into 8 ohms is 80W)
Sometimes the cone excursions can hit the loudspeaker material in the cabinet and cause a similar sound to clipping.
No , the speakers aren't clipping , fed them 30v rms ( they are 6 ohm ) from my Pioneer amp and they dont distort at all, they get a bit warm but dont distort . mencanically the voice coil doesnt hit the back magnet plate even at powers ( short for testing ) way over they rated power, they are not expensive at all, hi-fi , but they sound pretty good.
I know how a speaker that distorts or mecahnically hits the back plate of the magnet sounds like , and it's not the case.
The output transisotrs get just warm,
I know how a speaker that distorts or mecahnically hits the back plate of the magnet sounds like , and it's not the case.
The output transisotrs get just warm,
Hi John, looks like something is not quite right in this calculation.
33Vpeak = 23.33452378 Vrms
Pout = Vrms * Vrms / 8ohm = 68Wrms
Well, not a big difference - just to be precise
Cheers,
Valery
Sorry -approximating due to haste. Duly corrected.
Regarding the distortion problem - another point about the high capacitance of the high fT transistors is that they really can build up a high base charge. 100 ohm base resistors might not be enough to remove the charge. So at rest there appears to be no problem. For a signal, the bases stay on longer simply because the charge cannot be extracted. The average current begins to increase - which would cause premature clipping when the transistors cannot switch the currents quickly enough (e.g. 4A instead of 2A).
This was the old 2N3055 (RCA hometaxial base) transistor issue when handling frequencies of ~20kHz or so, but can also occur in these new high ft devices. I have not specifically looked into this in your circuit, but you could check by measuring the supply current and seeing if it increases with signal level. If it does, perhaps reducing the base resistors will help. As a suggestion I would make them 33 ohms, but check the drivers can handle the power! The basic design concept here is to make sure that -Ib=+Ib.
Regarding the distortion problem - another point about the high capacitance of the high fT transistors is that they really can build up a high base charge. 100 ohm base resistors might not be enough to remove the charge. So at rest there appears to be no problem. For a signal, the bases stay on longer simply because the charge cannot be extracted. The average current begins to increase - which would cause premature clipping when the transistors cannot switch the currents quickly enough (e.g. 4A instead of 2A).
This was the old 2N3055 (RCA hometaxial base) transistor issue when handling frequencies of ~20kHz or so, but can also occur in these new high ft devices. I have not specifically looked into this in your circuit, but you could check by measuring the supply current and seeing if it increases with signal level. If it does, perhaps reducing the base resistors will help. As a suggestion I would make them 33 ohms, but check the drivers can handle the power! The basic design concept here is to make sure that -Ib=+Ib.
Lets do a quick calculation for maximum positive swing (negative being roughly the same since the topology is completely symetrical):
First let's assume ideal components, i.e. transistor beta is infinite, and power supplies are ideal. This will give us the maximum theoretical swing.
The tail current of Q1/Q2 is set by the 20k resistor to the negative supply, resulting in (V3-Vbe)/20k = 1.715mA for both transistors at a 35V power supply voltage.
Ideally, half of this current makes a voltage drop on R3, that equals 2.57V. The voltage on the emitter of Q5 is Vbe closer to the power rail than that, and it's collector may at best become equal to that. However, the base-emitter junction of Q9 is adds this Vbe right back, so in essence, with ideal components, the emitter of Q9 can swing up to 35 less 2.57V = 32.43V. At this voltage, the current through the load plus R14 equals 32.43V/(6+0.2)Ohm = 5.23A, producing a voltage drop of 5.23V x 0.2Ohms = 1.046V, so the maximum theoretically possible output voltage with a 35V power supply is about 31.4V peak, or 22.2Vrms.
A non-ideal power supply will further reduce this. If a regular DVM was used to measure the power supply voltage, it will actually measure the peak of the power supply sawtooth waveform (resulting from rectification and imperfect filtering) and not the bottom of it which is what determines the clipping point. So, measuring 1.5V reduction compared to no signal may well mean the lowest voltage in the ripple waveform may well be 3 or more V lower than the unloaded voltage, Translating to a reduction of the output by a further 2-3Vrms, let's optimistically say to 20Vrms.
Finally, we are using real components. At ~5A peak output current, and Q5 and Q9 near or at saturation, the actual current gain of these transistors will be far lower than expected, so base current will be required by Q9. What is the maximum we can get? This happens when all the tail current of the input pair flows through Q1, resulting in doubling of the voltage drop on R1, to about 5V. This would be input stage clipping. Under no signal conditions the current through Q5 is roughly 9mA, and when the input stage clips, it rises to 23mA, the difference being the maximum current available to the base of Q9, which is 14mA. At this point the added voltage drop on R1 has further reduced the output swing by 1.8V. 14mA into the base of Q9 is actually well in the ball-park of what can be expected at clipping. Under those conditions Q5 is also not going to be fully saturated, more likely with Vcb near zero rather than Vce. This will further reduce available output by 0.3Vrms or so. When it all adds up, we get around 18V RMS at most, and this can already be considered hard clipping, and audible distortion may occur before that.
Finally, most regular DVMs will only be accurate to about 400Hz or so, thereafter showing a lower than actual voltage due to a slow and very low power OPamp being used as a peak rectifier. They are based on measuring the peak voltage and showing it divided by square root of 2 assuming a sinewave. The only accurate measurement may well be had at the mains frequency.
In other words - even a cursory examination shows there is no real mystery to the 'low output voltage swing'.
Moral of the story: debugging amplifiers requires a scope.
First let's assume ideal components, i.e. transistor beta is infinite, and power supplies are ideal. This will give us the maximum theoretical swing.
The tail current of Q1/Q2 is set by the 20k resistor to the negative supply, resulting in (V3-Vbe)/20k = 1.715mA for both transistors at a 35V power supply voltage.
Ideally, half of this current makes a voltage drop on R3, that equals 2.57V. The voltage on the emitter of Q5 is Vbe closer to the power rail than that, and it's collector may at best become equal to that. However, the base-emitter junction of Q9 is adds this Vbe right back, so in essence, with ideal components, the emitter of Q9 can swing up to 35 less 2.57V = 32.43V. At this voltage, the current through the load plus R14 equals 32.43V/(6+0.2)Ohm = 5.23A, producing a voltage drop of 5.23V x 0.2Ohms = 1.046V, so the maximum theoretically possible output voltage with a 35V power supply is about 31.4V peak, or 22.2Vrms.
A non-ideal power supply will further reduce this. If a regular DVM was used to measure the power supply voltage, it will actually measure the peak of the power supply sawtooth waveform (resulting from rectification and imperfect filtering) and not the bottom of it which is what determines the clipping point. So, measuring 1.5V reduction compared to no signal may well mean the lowest voltage in the ripple waveform may well be 3 or more V lower than the unloaded voltage, Translating to a reduction of the output by a further 2-3Vrms, let's optimistically say to 20Vrms.
Finally, we are using real components. At ~5A peak output current, and Q5 and Q9 near or at saturation, the actual current gain of these transistors will be far lower than expected, so base current will be required by Q9. What is the maximum we can get? This happens when all the tail current of the input pair flows through Q1, resulting in doubling of the voltage drop on R1, to about 5V. This would be input stage clipping. Under no signal conditions the current through Q5 is roughly 9mA, and when the input stage clips, it rises to 23mA, the difference being the maximum current available to the base of Q9, which is 14mA. At this point the added voltage drop on R1 has further reduced the output swing by 1.8V. 14mA into the base of Q9 is actually well in the ball-park of what can be expected at clipping. Under those conditions Q5 is also not going to be fully saturated, more likely with Vcb near zero rather than Vce. This will further reduce available output by 0.3Vrms or so. When it all adds up, we get around 18V RMS at most, and this can already be considered hard clipping, and audible distortion may occur before that.
Finally, most regular DVMs will only be accurate to about 400Hz or so, thereafter showing a lower than actual voltage due to a slow and very low power OPamp being used as a peak rectifier. They are based on measuring the peak voltage and showing it divided by square root of 2 assuming a sinewave. The only accurate measurement may well be had at the mains frequency.
In other words - even a cursory examination shows there is no real mystery to the 'low output voltage swing'.
Moral of the story: debugging amplifiers requires a scope.
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Hi ilimzn,
That was an excellent "back of envelope" conclusion. I think that coupled with a voltmeter that has poor HF response, you are exactly right.
Hi xXBrunoXx,
You can't trust a simulator for voltage and current readings comparing it to a real world circuit. Without instruments, you are really stuck at a guessing game. You need a good meter (>$100 USD) and some kind of oscilloscope. A used 5 MHz model would be a lot better than nothing. They sell really cheaply. The better your instruments are, the more you can trust them.
-Chris
That was an excellent "back of envelope" conclusion. I think that coupled with a voltmeter that has poor HF response, you are exactly right.
Hi xXBrunoXx,
You can't trust a simulator for voltage and current readings comparing it to a real world circuit. Without instruments, you are really stuck at a guessing game. You need a good meter (>$100 USD) and some kind of oscilloscope. A used 5 MHz model would be a lot better than nothing. They sell really cheaply. The better your instruments are, the more you can trust them.
-Chris
Hi xXBrunoXx,
We all started out like you as amateurs. It's a lifetime hobby and you will enjoy it more when you have at least some basic instruments. Upgrade them as you can afford it. I started out with a tube oscilloscope made by Stark, a 500 KHz single trace deal. Of course I upgraded it and sold it for what I paid for it. I don't expect that these days. The scope I upgraded to was a 10 MHz, dual trace Philips. It was amazing compared to the old Stark.
So pick up a good meter (you will have it for life) like a Keysight ora Fluke. Get one that is "true RMS" and has a higher frequency response. The cheaper meters don't even meet their own specs right out of the box.
-Chris
That may be, but at some point in time you will. Either that, or you will see a really good deal and know enough to take advantage of it.
We all started out like you as amateurs. It's a lifetime hobby and you will enjoy it more when you have at least some basic instruments. Upgrade them as you can afford it. I started out with a tube oscilloscope made by Stark, a 500 KHz single trace deal. Of course I upgraded it and sold it for what I paid for it. I don't expect that these days. The scope I upgraded to was a 10 MHz, dual trace Philips. It was amazing compared to the old Stark.
So pick up a good meter (you will have it for life) like a Keysight ora Fluke. Get one that is "true RMS" and has a higher frequency response. The cheaper meters don't even meet their own specs right out of the box.
-Chris
Fully agree.
"It's difficult to find a black cat in a dark room. Especially if it's not there."
When I was a student, I've got a pretty old second-hand scope as a "payment" for a small consulting work I did at that time (well, almost nothing). It was big, it had low-impedance inputs, so I had to arrange an attenuator, but it was great for the purpose - I could see what's going on inside the circuit.
Recommendation to xXBrunoXx - try to look at the secondary market, especially if this is not a one-off project (you plan some builds in the future).
Cheers,
Valery
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