Hello, If I use a 500ohm resistor instead of a choke 120ohm/20H.
My voltage will get lower than if I used the choke, and I will get more ripple. It will also get even lower when the amp is under "heavy load" (!?sag) with the R than with the L. Ok.
But will I get more ripple when the amp sag (under LOAD) ? or just a lower voltage but with the same ripple than idle ?
Thank you.
My voltage will get lower than if I used the choke, and I will get more ripple. It will also get even lower when the amp is under "heavy load" (!?sag) with the R than with the L. Ok.
But will I get more ripple when the amp sag (under LOAD) ? or just a lower voltage but with the same ripple than idle ?
Thank you.
Yeh power supplies interactions are complex.everything affects everything else. Theory and reality dont always match. Its called prototyping. Give it a go and see what happens.
Yeah, if you have all of the parts, the real thing will beat simulations any day.
But make sure the real input AC line voltage also remains steady under heavy loading.
But make sure the real input AC line voltage also remains steady under heavy loading.
By eyeball 20H looks like a Z of >15K at the ripple freq of 100 Hz.If I use a 500ohm resistor instead of a choke 120ohm/20H.
That is an order of magnitude more filtering than a 500R resister.
And much better regulation, only 120R in the choke. 👍
Agree with @jhstewart9. The 500 ohms has higher resistance at DC so will give lower voltage.
It has lower impedance for ripple so will give higher ripple.
Jan
It has lower impedance for ripple so will give higher ripple.
Jan
Of course the ripple in both situations is higher under load ...Hello, If I use a 500ohm resistor instead of a choke 120ohm/20H.
My voltage will get lower than if I used the choke, and I will get more ripple. It will also get even lower when the amp is under "heavy load" (!?sag) with the R than with the L. Ok.
But will I get more ripple when the amp sag (under LOAD) ? or just a lower voltage but with the same ripple than idle ?
Thank you.
which stuff is?Hello, If I use a 500ohm resistor instead of a choke 120ohm/20H.
My voltage will get lower than if I used the choke, and I will get more ripple. It will also get even lower when the amp is under "heavy load" (!?sag) with the R than with the L. Ok.
But will I get more ripple when the amp sag (under LOAD) ? or just a lower voltage but with the same ripple than idle ?
Thank you.
pre or amp.
If amp is a s,e or p-p?
For pre the coil is better for amp some consideration must be done
Walter
Hi guys, thank you for the answers. It is for a power amp (push-pull Of 2 kt88), if I could get rid of the choke It would make the base so much smaller 😉
If I make sure I don't load down the secondary winding. I will ONLY get the voltage drop from the R(choke), so ripple would stay the same as idle ? (I guess).
If I make sure I don't load down the secondary winding. I will ONLY get the voltage drop from the R(choke), so ripple would stay the same as idle ? (I guess).
For a p-p in class A/B with 88 is not a good idea to put the chocke. But also the resistor
If it is in class A can be resonable.
There are some vintage example of the use of the resistors in, p.e., Leak p-p amp but they are class A.
I used a chocke in p-p of 300 B class A/B biased at 40 mA each but with a GZ34 , 47 uF, chocke ( 10H 300mA) and 4 x 220 uF as filters.
Good results in terms of dynamic
Walter
If it is in class A can be resonable.
There are some vintage example of the use of the resistors in, p.e., Leak p-p amp but they are class A.
I used a chocke in p-p of 300 B class A/B biased at 40 mA each but with a GZ34 , 47 uF, chocke ( 10H 300mA) and 4 x 220 uF as filters.
Good results in terms of dynamic
Walter
I am not going to catch up with reading Posts # 1 - 9.
Instead, I will respond to the Title of this Thread:
CRC with more resistance than CLC
Example:
A CRC with R = 200 Ohms
A CLRC with DCR 100 Ohms plus R 100 Ohms
Ooops, the same total resistance, to illustrate a point.
Both have the same DCV output, at the same Load current
One of them has Far Less Ripple.
Guess which one?
Adjust the Total DCR plus R to get the B+ that you want.
Do you want better load regulation (more constant DCV versus load current)?
Then use a higher voltage primary, Choke input filter, and a bleeder resistor that actually draws a fair amount of bleeder current.
Engineering at one of its best implementations!
Yes, there are tradeoffs, Expen$e, space for the parts, magnetic field spray, weight.
Deal with those, and you will have something elegant.
Done!
Instead, I will respond to the Title of this Thread:
CRC with more resistance than CLC
Example:
A CRC with R = 200 Ohms
A CLRC with DCR 100 Ohms plus R 100 Ohms
Ooops, the same total resistance, to illustrate a point.
Both have the same DCV output, at the same Load current
One of them has Far Less Ripple.
Guess which one?
Adjust the Total DCR plus R to get the B+ that you want.
Do you want better load regulation (more constant DCV versus load current)?
Then use a higher voltage primary, Choke input filter, and a bleeder resistor that actually draws a fair amount of bleeder current.
Engineering at one of its best implementations!
Yes, there are tradeoffs, Expen$e, space for the parts, magnetic field spray, weight.
Deal with those, and you will have something elegant.
Done!
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Frederico Acardi,
OK. My thick skull finally got it!
My guess about a CRC, and a CLC power supply . . .
The Ripple will change versus the load current.
Think of the capacitive reactance of the input cap.
Charge the cap to X volts DC.
Now discharge it with 30mA for 5 msec. (the rectifiers are not conducting).
The DC voltage will drop to an amount, Y volts.
Again, Charge the cap to X volts DC.
Now discharge it with 60mA for 5 msec. (the rectifiers are not conducting).
The DC voltage will drop to LESS than Y Volts (more ripple)
The drop of the DC Volts, IS the ripple.
Is your question answered now?
Have fun designing, building, and listening!
OK. My thick skull finally got it!
My guess about a CRC, and a CLC power supply . . .
The Ripple will change versus the load current.
Think of the capacitive reactance of the input cap.
Charge the cap to X volts DC.
Now discharge it with 30mA for 5 msec. (the rectifiers are not conducting).
The DC voltage will drop to an amount, Y volts.
Again, Charge the cap to X volts DC.
Now discharge it with 60mA for 5 msec. (the rectifiers are not conducting).
The DC voltage will drop to LESS than Y Volts (more ripple)
The drop of the DC Volts, IS the ripple.
Is your question answered now?
Have fun designing, building, and listening!
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It makes sense that if the system (not loaded) shows similar ripple from R to L comparison at filter location, then-
Perhaps the quiescent background 'hum' level mite not benefit from the iron. Sound floor 'Hum' being primary battle with Hi voltage / Hi Z systems like tube amps.
Then once the music starts and loads begin to fluctuate... that ripple cant be good for sound quality-imo
But can it be readily heard? Seems such could be easily seen on a 'scope as some rendition of distortion. I believe affecting lower freq's more than hi's-
That is if ripple is proportional to load.
I had always thought (perhaps wrongly) that a resistor instead of choke in Pi network was effect of production bean counters.
Hmm
Perhaps the quiescent background 'hum' level mite not benefit from the iron. Sound floor 'Hum' being primary battle with Hi voltage / Hi Z systems like tube amps.
Then once the music starts and loads begin to fluctuate... that ripple cant be good for sound quality-imo
But can it be readily heard? Seems such could be easily seen on a 'scope as some rendition of distortion. I believe affecting lower freq's more than hi's-
That is if ripple is proportional to load.
I had always thought (perhaps wrongly) that a resistor instead of choke in Pi network was effect of production bean counters.
Hmm
What is not loaded ? Idle current in tubes is not 0 even in stingy class B and the input tubes consume current .
If you don't use a choke the capacitors must have much higher values for the same ripple .
Or just make it worse and hope you won't hear the hum ...
If you don't use a choke the capacitors must have much higher values for the same ripple .
Or just make it worse and hope you won't hear the hum ...
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My description of 'not loaded' is per quiescent definition of 'inactivity'-
Within that measure are design parameters that are, essentially, driver and bias currents of which should be excused as 'quiescent' in my previous response.
Yes, there are other loads in the operating system. But my point was to compare inactive system (no signal) to loaded system (full signal)
Now if the PP was operating in class A, then that changes everything ;<)
Jim
Within that measure are design parameters that are, essentially, driver and bias currents of which should be excused as 'quiescent' in my previous response.
Yes, there are other loads in the operating system. But my point was to compare inactive system (no signal) to loaded system (full signal)
Now if the PP was operating in class A, then that changes everything ;<)
Jim
If the PP stage is in Class A the load current should change very little.Now if the PP was operating in class A, then that changes everything
No more than 10% from no signal to full signal. For most common audio power tubes you will find that in the operating specs. 👍
At any current the ripple is higher replacing the choke with a resistor ... so it is up to you what you want .
At 100 Hz the Z of inductor of 20H is 12.500 ohm
So the ripple is more attenuated respect to R
This in theory, then must be considered the quality of inductor but I think it is fine
Walter
So the ripple is more attenuated respect to R
This in theory, then must be considered the quality of inductor but I think it is fine
Walter
This is what I ask about the circuitIf the PP stage is in Class A the load current should change very little.
No more than 10% from no signal to full signal. For most common audio power tubes you will find that in the operating specs. 👍
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