11-27-2023, 02:58 PM
Hi Guys
If you browse through a lot of schematics for audio gear, you see certain common values for the main filter cap immediately after the rectifier. How much if this is by design? How much is just copied from similar equipment? How do you know what the "right" value is?
Let's use our ubiquitous 50W push-pull amp model with these specs:
output transformer: 4kaa
100pk requires 316Vpk at 316mApk
tube saturation voltage is 60V
minimum plate voltage at full load: 316Vpk + 60V = 376V, call it 380Vdc
The effective load is then 380V / 316mA = 1k2
Assuming the plate winding on the power transformer has the usual 20% regulation, unloaded Va is then about 1.25 x 380V = 475Vdc. This would be derived from 330Vac. At full load, the winding voltage has fallen by 20% to about 270Vac and its 380Vpk is the relevant value to use.
We always use full-wave rectification for ALL windings, so the ripple frequency is twice the mains frequency, making the ripple 100Hz or 120Hz depending on where we live.
The approximate ripple voltage from a rectifier can be calculated using the ripple frequency, the peak voltage from the rectifier, and the filter cap value. Say we use 100Hz ripple for 50Hz mains and a 100uF cap to begin with.
Vripple = (Vpk x Tau) / (C x R-load)
where Tau is the reciprocal of frequency >> 1 / 100Hz = 0.01s, or 10ms,
then
Vripple = (380Vpk x 10ms) / (100uF x 1k2)
= 31V7ac
Truly, this is going to be marginal with respect to not having the output signal being modulated by AC ripple.
If we double the filter cap value the ripple will be half - 16Vac - a straight linear relationship BUT with diminishing returns if we go too large. 1mf (1,000uF) would cut the load ripple to just over 3V and this is what hifi guys do in their tube amps, often followed by a massive choke into another 1mF cap Great if you do not want to add solid-state hum filters.
In the past, high-voltage capacitors of this value range were not available and series-parallel banks would be needed. The cost would be prohibitive and the bulk of components needed to be managed. Now that snap-mount caps are affordable and of tremendous quality, there is no excuse for under-filtered power supplies EXCEPT for equipment that is never expected to be driven to its maximum capability. A music amp used to be such a device, although nowadays there is so much compression used in recording that peak-versus-average levels have narrowed. In a guitar amp, under-filtering is historic and thus copied widely. This means that any amp driven to clipping is also likely to exhibit a lot of hum - witness Komet amps, Tranwrecks, some plexis, and all their clones.
A very reasonable observation is to use a PT with better regulation, or at the very least use one rated for a bit higher voltage so that the sagged voltage will not modulate the output even at clipping. In TUT5's Stentorian chapter, you may recall that polypropylene caps were used as supply filter. The caps were 47uF each and that with three caps the clipped signal output was hum-modulated but at four caps it was clean - 141uF changed to 188uF. This result would have been the same were the caps electrolytic.
When you go to lower voltages as found in solid-state amps, the cap values are necessarily much higher.
Say we have a 100W amp at 8-ohms, which requires a peak signal of 40V at 5A. The saturation voltage of BJTs and mosfets is quite low and we may only need a few volts. Let's say, this is 5V. We now have an effective load of 45V / 5A = 9-ohms. We will use the same 50Hz mains >> 100Hz ripple >> Tau = 10ms.
We are being thrifty and try 1,000uF first (1mF):
Vripple = (45V x 10ms) / (1mF x 9)
= 450mV / 0.009
= 50Vac ???
This is more ripple than AC voltage coming in to the rectifier, so this value must be ridiculous or we have a wonky equation? If we go to a more common value of 10mF, at least the ripple is more reasonable at 5Vac.
Since the solid-state amplifier is likely to use symmetric supplies, it is a common assumption that half the power comes from either side of the PT center-tap - indeed it does over the complete signal cycle, but each peak is carried fully by just one side at a time. The "averaging" assumption would have the effect of making the load seen by each side twice as high in value and potentially reduces the ripple voltage to 2V5. This assumption is often used in assessing the ripple current rating required for the filter caps.
In both the tube and solid-state examples, it is clear that it is highly beneficial to have an idle DC voltage quite a bit higher than the expected loaded value, not just to accommodate supply and mains regulation, but to accommodate insufficient ripple reduction.
In any case, the first filter capacitor is the main element in fighting ripple through the supply line, so skimping on its value is never a good idea.
If you browse through a lot of schematics for audio gear, you see certain common values for the main filter cap immediately after the rectifier. How much if this is by design? How much is just copied from similar equipment? How do you know what the "right" value is?
Let's use our ubiquitous 50W push-pull amp model with these specs:
output transformer: 4kaa
100pk requires 316Vpk at 316mApk
tube saturation voltage is 60V
minimum plate voltage at full load: 316Vpk + 60V = 376V, call it 380Vdc
The effective load is then 380V / 316mA = 1k2
Assuming the plate winding on the power transformer has the usual 20% regulation, unloaded Va is then about 1.25 x 380V = 475Vdc. This would be derived from 330Vac. At full load, the winding voltage has fallen by 20% to about 270Vac and its 380Vpk is the relevant value to use.
We always use full-wave rectification for ALL windings, so the ripple frequency is twice the mains frequency, making the ripple 100Hz or 120Hz depending on where we live.
The approximate ripple voltage from a rectifier can be calculated using the ripple frequency, the peak voltage from the rectifier, and the filter cap value. Say we use 100Hz ripple for 50Hz mains and a 100uF cap to begin with.
Vripple = (Vpk x Tau) / (C x R-load)
where Tau is the reciprocal of frequency >> 1 / 100Hz = 0.01s, or 10ms,
then
Vripple = (380Vpk x 10ms) / (100uF x 1k2)
= 31V7ac
Truly, this is going to be marginal with respect to not having the output signal being modulated by AC ripple.
If we double the filter cap value the ripple will be half - 16Vac - a straight linear relationship BUT with diminishing returns if we go too large. 1mf (1,000uF) would cut the load ripple to just over 3V and this is what hifi guys do in their tube amps, often followed by a massive choke into another 1mF cap Great if you do not want to add solid-state hum filters.
In the past, high-voltage capacitors of this value range were not available and series-parallel banks would be needed. The cost would be prohibitive and the bulk of components needed to be managed. Now that snap-mount caps are affordable and of tremendous quality, there is no excuse for under-filtered power supplies EXCEPT for equipment that is never expected to be driven to its maximum capability. A music amp used to be such a device, although nowadays there is so much compression used in recording that peak-versus-average levels have narrowed. In a guitar amp, under-filtering is historic and thus copied widely. This means that any amp driven to clipping is also likely to exhibit a lot of hum - witness Komet amps, Tranwrecks, some plexis, and all their clones.
A very reasonable observation is to use a PT with better regulation, or at the very least use one rated for a bit higher voltage so that the sagged voltage will not modulate the output even at clipping. In TUT5's Stentorian chapter, you may recall that polypropylene caps were used as supply filter. The caps were 47uF each and that with three caps the clipped signal output was hum-modulated but at four caps it was clean - 141uF changed to 188uF. This result would have been the same were the caps electrolytic.
When you go to lower voltages as found in solid-state amps, the cap values are necessarily much higher.
Say we have a 100W amp at 8-ohms, which requires a peak signal of 40V at 5A. The saturation voltage of BJTs and mosfets is quite low and we may only need a few volts. Let's say, this is 5V. We now have an effective load of 45V / 5A = 9-ohms. We will use the same 50Hz mains >> 100Hz ripple >> Tau = 10ms.
We are being thrifty and try 1,000uF first (1mF):
Vripple = (45V x 10ms) / (1mF x 9)
= 450mV / 0.009
= 50Vac ???
This is more ripple than AC voltage coming in to the rectifier, so this value must be ridiculous or we have a wonky equation? If we go to a more common value of 10mF, at least the ripple is more reasonable at 5Vac.
Since the solid-state amplifier is likely to use symmetric supplies, it is a common assumption that half the power comes from either side of the PT center-tap - indeed it does over the complete signal cycle, but each peak is carried fully by just one side at a time. The "averaging" assumption would have the effect of making the load seen by each side twice as high in value and potentially reduces the ripple voltage to 2V5. This assumption is often used in assessing the ripple current rating required for the filter caps.
In both the tube and solid-state examples, it is clear that it is highly beneficial to have an idle DC voltage quite a bit higher than the expected loaded value, not just to accommodate supply and mains regulation, but to accommodate insufficient ripple reduction.
In any case, the first filter capacitor is the main element in fighting ripple through the supply line, so skimping on its value is never a good idea.