I am confused regarding the use of a Pi ( CRC ) filter for a solid state class ab amplifier.
Looking at build guides for say the honey badger or Symasym it appears to be quite normal to use a CRC filter. However other articles say that using such a filter is not good for class ab due to the large variation of current from quiescent to loud music passages, which will obviously give a voltage variation across the R.
Any thoughts on these two opposing views would be appreciated.
Thanks
Alan
Looking at build guides for say the honey badger or Symasym it appears to be quite normal to use a CRC filter. However other articles say that using such a filter is not good for class ab due to the large variation of current from quiescent to loud music passages, which will obviously give a voltage variation across the R.
Any thoughts on these two opposing views would be appreciated.
Thanks
Alan
As long as you don't drive the amplifier into clipping, (may be less voltage available) there will be no difference if the amplifier uses complimentary or quasi complimentary output stage.
As always in audio electronics, there is not a simple 'recipe' answer.
Using the R will reduce hum, but also reduce the steady-state maximum power output. Which is most important to you?
Using the R will reduce hum, but also reduce the steady-state maximum power output. Which is most important to you?
Thanks. What I am trying to understand is with varying current the voltage drop across the R will also vary, meaning the supply voltage is constantly varying. Is this not a bad thing? Or are the voltage variations small enough to be ignored.
For example: R= 0.1 Ohm 25mA quiescent current =0.0025 V drop across resistor . When large currents are drawn say 2.5 A across the same R there will be a voltage drop of 0.25 V. Just asking
Also I note that it is common with CRC filters the R is usuall made of several low wattage resistors in parallel, why not use one higher wattage resistor?
Thanks
Alan
For example: R= 0.1 Ohm 25mA quiescent current =0.0025 V drop across resistor . When large currents are drawn say 2.5 A across the same R there will be a voltage drop of 0.25 V. Just asking
Also I note that it is common with CRC filters the R is usuall made of several low wattage resistors in parallel, why not use one higher wattage resistor?
Thanks
Alan
0.1 ohm will not make much difference to hum and ripple because it is the same order of magnitude as capacitor ESR. Anything larger in value will start to eat into your peak voltage capability. Maybe this is why traditionally power amps just have big reservoir caps?
the 25mAdc is a continuous current and gives rise to a continuous Vdrop of 2.5mVdc.
The 2.5A is a transient as a result of sending ~2.5Apk to the speaker. It's total period is half a cycle of the dominant frequency (1ms for a 500Hz transient) and it is only at this 2.5Apk for a very short part of that period (maybe 100us). For the remainder of the time (900us) the transient current in the supply rail is less and sometimes zero (500us) (i.e. back to the 25mA quiescent current).
Transient current demand is met by the local supply rail decoupling.
The smoothing capacitor simply recharges the decoupling.
The R of the CRC hardly sees any of the transient.
The Last C does most of the work in recharging the decoupling.
The first C does a bit to recharge the decoupling, but most of it's duty is in recharging the second C.
So your first job is to decide how much HF local decoupling is required IN the amplifier and where it needs to be located.
The second job is to decide how much MF local decoupling is required IN the amplifier and where it needs to be located.
The third job is to decide how much smoothing is required IN the last C
The fourth job is to decide how much smoothing is required IN the first C.
The final job is to decide what impedances to provide between these four different duty capacitances. Some will be discrete resistors, some will be resistances of cabling, some will be inductances of cabling and maybe some will be discrete inductances.
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