Mooly,
you have just stated the opposite to what I am thinking.
My last post referred us back to the similarity of the two diagrams in post 25.
But it is not up yet or has gone "missing in action". (edit - the missing post now appears as post39)
you have just stated the opposite to what I am thinking.
My last post referred us back to the similarity of the two diagrams in post 25.
But it is not up yet or has gone "missing in action". (edit - the missing post now appears as post39)
I claim the reservoir caps do directly influence the pass band of the amplifier.
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look at post25's two diagrams.
In both versions the speaker current is fed through one or both capacitors when the charging circuit is OFF.
The capacitors are in series with the speaker when current flows around the output circuit.
But it does.
Your post showing the effect ripple has on the output is exactly that.
During the moment that rail voltage has dropped, the gain for the output has dropped, a lot. That is the dip in the output voltage.
When the capacitor gets recharged, the rail voltage rises and the gain rises to give the near full output voltage when the big differential has been restored.
The gain has changed downwards during the rail voltage dip !
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Definition of pass band.
"An amplifier designed to pass a definite band of frequencies with essentially uniform response"
To me the reservoir caps play no direct part in that statement and yet you are claiming that they do. Whether the amp has 100uf, 1000uf or 100,000uf reservoir caps, the pass band is the same. There is no "filtering", no roll off, no modification of the pass band. Only when we define what the output may be (in terms of current and power delivery) can we start to put minimum values on the reservoir caps. If we undersize them we get distortion as the rails drop below the expected output but we have not changed the "pass band of the amp".
I would say the gain doesn't alter too.
When the ripple encroaches on the output signal, well I would call that distortion I suppose. The gain of the amplifier is unchanged. It all comes down to how you interpret things. You could say an amplifier that is turned off has a gain of zero, but that isn't what we would mean of course.
When the ripple encroaches on the output signal, well I would call that distortion I suppose. The gain of the amplifier is unchanged. It all comes down to how you interpret things. You could say an amplifier that is turned off has a gain of zero, but that isn't what we would mean of course.
The gain is set with Rfb and R3. I don't see how this changes with rail voltage. I also expected a dip in gain.
The max output voltage is limited by cap value.
Its pointles to have a 30 V supply that can only supply 10 V.
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Could it be that the network (cap-amp-spkr) changes as the R of the amp lowers and/or that the network (transformer-cap) changes? More current from the transformer than the cap.
While playing with spice it seemed that transformers are to big and had no problem charging the caps within one half cycle.
While playing with spice it seemed that transformers are to big and had no problem charging the caps within one half cycle.
Yup I got everything I need
I'm gonna keep the 4700uF I got on at the moment.. just gonna have to mount them side ways. As for those Nichicons 1000uF, I'll use them to upgrade my on amp 1000uF G-Luxons. Also where can I get horizontally mounted LM1875's?? digikey doesnt have them it seems.
hehe weather here... well lets just say we get plenty of liquid sunshine
only if there is adequate margin. As the margins/overheads reduce the gain drops.
Yes I agree, it is pointless having a supply that sags badly when current is delivered to the load.
That is why bass frequency amplifiers need capacitance for smoothing. That smoothing becomes the LF supply.
As the frequency required from the amplifier gets lower the capacitance requirement goes higher.
Other wise the gain drops.
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I think this is the formula that shows gain varies with the margin left when Vout approaches Vrail:
Vout = Vin / [R1/{R1+R2} + 1/Aol]
from:
Operational amplifier - Wikipedia, the free encyclopedia
about 2/3rds down under Negative feedback applications
Then it follows that:
Gain = 1 / [R1/{R1+R2} + 1/Aol]
It's the effect of Aol not being extremely large that most ignore. When the margin between Vrail and Vout is high and the open loop gain is also high then the simplified gain formula holds reasonably well. Well enough, that in general we can all ignore the error.
I hope I got it right, but I'm sure someone can correct me.
Vout = Vin / [R1/{R1+R2} + 1/Aol]
from:
Operational amplifier - Wikipedia, the free encyclopedia
about 2/3rds down under Negative feedback applications
Then it follows that:
Gain = 1 / [R1/{R1+R2} + 1/Aol]
It's the effect of Aol not being extremely large that most ignore. When the margin between Vrail and Vout is high and the open loop gain is also high then the simplified gain formula holds reasonably well. Well enough, that in general we can all ignore the error.
I hope I got it right, but I'm sure someone can correct me.
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