@AndrewT with all do respect, nothing you have sayd there has answeared my question, it is just unnecessary complicated stuff, to make it simple from 35Vcc you can only obtain an amplified AC/audio signal of about 24V maximum, now 24V on 4 Ohm load ( witch i think could be the case here ) is about 140W of power, this not taking in consideration the amplifier losses witch as far as i know in class D are much smaller, now if let say you require about 150W power from the supply then 150/70=2,2A, or let say the worst case scenario, about 30% losses as it is with class B, 140/0,7=200W, that would be the power needed from the supply, and with 70V rai to rail that is about 2,9A, and something like 6A for 2 Ohms load but as i say i doubt that is the case, well from 6 to 20A is qouite a difference, so as i sayd 20A for just 35v is too large of a current.
@jlind54 i think that first you need to be sure what audio power you need for the speakers and what is theyr load, i mean are 2 Ohms or 4 Ohms, then a more correct specs could be drawn for the supply.
@jlind54 i think that first you need to be sure what audio power you need for the speakers and what is theyr load, i mean are 2 Ohms or 4 Ohms, then a more correct specs could be drawn for the supply.
Subwoofer is an 18" 4 Ohm. I'm looking to drive it with near 70 volts which is how I decided +-35 and at 4 ohms that equates to an approximate 17.5 amps correct? I figure being 20 amps allows some wiggle room with speaker impedance change at given freq./box combinations. Am I thinking incorrectly here. Perhaps the voltage should be bumped up a bit like say shot for 80 which would give +-40? or does my supply need to truely be +- 70 instead of 35 as I've outlined above?
@jlind54 i ask again: how much power do you need for that sub? if you mean you want a 70v signall for the load then it would be about 1,2Kw of power, now with 35v per rail you cannot get anywhere near that much power, 70vac signal would mean at least 98Vcc per rail ( that is +/-98Vcc ) and that is not taking into account the power devices losses, so maybe the power supply unit would need something like +/-100Vcc, that is much more than you have come to.
I think the reason for your mistake could be a poor understanding of what's going on on a power amplifier so that is why i think you have much to studdy before you can start working at a SMPS with that high voltage outputs.
I think the reason for your mistake could be a poor understanding of what's going on on a power amplifier so that is why i think you have much to studdy before you can start working at a SMPS with that high voltage outputs.
that appears to be your question.
I answered it rather comprehensively.
20Apk is not a large current, irrespective of whether the supply rails hold up at only +-35Vdc, or are +-100Vdc. I gave a detailed derivation of typical worst case transient currents for a 2ohms speaker driven from regulated +-35Vdc supply.
I disagree on this conclusion.
I would expect at least 30Vpk from an amplifier sitting on regulated (SMPS) +-35Vdc supply rails, if ordinary followers were used. If common emitter/common source is used I'd expect at least 33Vpk from +-35Vdc.
I think it was ampguru that did a design based on common source (vert mosFET) and there are others. Dr Cherry recommends this topology over the much more common follower topology (common collector=emitter follower).
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probably not as a gut reaction to an actively driven bass only speaker.
Maybe 1.5times to 2times, but this is only my guess, since I cannot reproduce the original testing for real music into actual speakers.
A well resourced builder/designer should be able to determine a current factor for a reduced bandwidth speaker without a passive crossover between it and the driving amplifier. I have not seen this style of test reported in this Forum, but my adopted 3times is based on the Forum report that I have referred to a few times. I do wish some members would get together and expand that original report to other speaker types that we typically use.
I guess (again) that many have not realised the importance of meeting speaker current demand and thus little priority is given to finding design data that would help with selecting amplifier and PSU topologies.
@AndrewT you cannot speak of Vpk when an AC sinus signal ( like audio signal ) is involved, on the load you speak of Vef, and from an DC current that can be considered Vpk you can only get and AC signal of Vcc/1,41, that is a well known fact, on the other hand when calculating the VA ratings of a PSU for an amplifier, you do not take into account the transients cus you can never exactly predict them, you think about the continuous current needed, and that is the current you design the PSU to deliver, it is cind of ridicoulus to make a PSU of 20 amps for just 35v just cus of the transients, what about for +/-100Vcc then? how much transients is involved here and how much more current you need to supply? with only 20A as it is as you say for 35v it would be a 4Kw PSU, for only 1,2-1,5Kw max audio power amplifier? that is beyound real, so again with all do respect you are wrong, and you need to better learn the difference between DC and AC, specially for sine wave ac.
@jlind54 it seams you do not know verry well the difference between AC and DC, so for example mains Ac current ( like audio signall as well ) is a sine wave with 2 ratings: Vpk and Vef, at AC state you measure the Vef witch is Vpk/1,41. Transforming this into DC is done with rectifiers and filter caps, the rectifiers allow the current to flow in only one direction so it makes the DC, and the filter caps raise rthe measured voltage at the Vpk. On the other hand the power amplifier ( it's final stage ) puts one side of the load alternatively at + than at - rails, never do both of them are at the same time on the load, so that means the load can only receive signall from one rail at a time, so if 35Vcc per rail that is the DC that will be delivered as Ac to the load, and as i stated above you can think of the DC as the Vpk, and you get the Vef that will always be the measured voltage on the speaker by Vpk/1,41 and 35vcc/1,41=24,8vac, in reality it is a bit smaller cus a few V drops on the power devices.
Now knowing all that, how do you calculate how much Vcc per rail you need for a wanted signall at the load? simple, for 70 as you stated, it is 70*1,41=98,7Vcc, you make sure you add some aditional volts needed for the power devices so you come up with at least 100V per rail, and as there are 2 rail ( positive and negative ) the supply needs the outputs to be +/-100Vcc, that is 200V rail to rail.
@jlind54 it seams you do not know verry well the difference between AC and DC, so for example mains Ac current ( like audio signall as well ) is a sine wave with 2 ratings: Vpk and Vef, at AC state you measure the Vef witch is Vpk/1,41. Transforming this into DC is done with rectifiers and filter caps, the rectifiers allow the current to flow in only one direction so it makes the DC, and the filter caps raise rthe measured voltage at the Vpk. On the other hand the power amplifier ( it's final stage ) puts one side of the load alternatively at + than at - rails, never do both of them are at the same time on the load, so that means the load can only receive signall from one rail at a time, so if 35Vcc per rail that is the DC that will be delivered as Ac to the load, and as i stated above you can think of the DC as the Vpk, and you get the Vef that will always be the measured voltage on the speaker by Vpk/1,41 and 35vcc/1,41=24,8vac, in reality it is a bit smaller cus a few V drops on the power devices.
Now knowing all that, how do you calculate how much Vcc per rail you need for a wanted signall at the load? simple, for 70 as you stated, it is 70*1,41=98,7Vcc, you make sure you add some aditional volts needed for the power devices so you come up with at least 100V per rail, and as there are 2 rail ( positive and negative ) the supply needs the outputs to be +/-100Vcc, that is 200V rail to rail.
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Yes, I can. I can see Vpk on a scope, it exists. I can measure Vaverage with a DMM, yes, it exists. I can meausre Vrms with a better DMM. Yes, it exists. I can use any of these to describe a voltage
Agreed.
No. Just because we cannot predict a transient is no reason to ignore the existance of current transients.
A linear PSU with enormous capacitance (by smps standards) is easily able to meet enormous current demands put out by the connected speaker. That is one of the advantages of using a linear supply. It CAN meet very high transient demands that a speaker makes on the PSU.
I think the reason you are not accepting (and we are having this extended argument), that transient demand exists and transient capability is required is because the SMPS has by comparison a relatively small output capacitance. That topology requires that the transient demand must be met by other means.
It is not sufficient as a design exercise to say that the transient demand does not exist, just because the SMPS will have to be designed with different priorties if one were to admit that transient current capability is a necessary PSU requirement.
I off to the dancing, not back till tomorrow.
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MarianB yes I somehow forgot the fact I would be creating an ac signal and there for taking what you said makes perfect sense now. So yes in general I would be shooting for the +-100 volt range. As for the amperage there would be no need to really increase to achieve 1200 watts of sinewave output correct?
@jlind54 you make the same mistake i've seen many people do... ok let us look at a verry simplistic schematic of audio power amplifier:
What you should look at are Q9 and Q10, they are the output devices, they take an voltage amplified signall from the other stages and delivers it to the load with current amplification ( it's sort of simplistic sayd but enough to get you started ), so how do they deliver the signal? well imagine a sine wave, as i hope you know it is a wave that goes up and down a center line, now immagine that center line to be the ground of the load/speaker and the upper part of the wave as Q9 and the lower part of the wave Q10, now as the wave goes up above the center line ( that is the ground here as we have established ) it develops positive voltage and it powers up Q9 witch is an N channel one and it conducts current to the load, as the wave is up Q10 witch is a P channel, cannot conduct any current cus it has no negative gate voltage to get it started, so only Q9 conducts the current to the load, this is the first part of the cicle, now as the wave is a sinus one it goes down too so that means it generatesc negative voltage this time, what that means is that Q9 cannot conduct cus it has no positive voltage any more at it's gate so it cuts of, Q10 on the other hand likes negative voltage on it's gate so it starts conducting current to the load, so when the wave goes down only Q10 conducts current to the load, that is the full cicle.
What did we learn from all of this?:
-The sine nature of audio signall means that only one rail at a time can power the load/speaker, so the voltage signall on the speaker can only be provided from one single rail Vcc at a time. The speaker has one side connected at the ground, the other side of the speaker is connected by the power devices alternatively at possitive and negative side, up and down, and so on again and again, at the signall frequency.
Hope you have better understanding now of what happenes in an audio power amplifier, and also understand why i have sayd that i do not recommend any tinkerring with ofline SMPS for now, you lack too much basic knowledge and ofline SMPS is extreemely dangerous for one's life when you do not know it well enough!
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What you should look at are Q9 and Q10, they are the output devices, they take an voltage amplified signall from the other stages and delivers it to the load with current amplification ( it's sort of simplistic sayd but enough to get you started ), so how do they deliver the signal? well imagine a sine wave, as i hope you know it is a wave that goes up and down a center line, now immagine that center line to be the ground of the load/speaker and the upper part of the wave as Q9 and the lower part of the wave Q10, now as the wave goes up above the center line ( that is the ground here as we have established ) it develops positive voltage and it powers up Q9 witch is an N channel one and it conducts current to the load, as the wave is up Q10 witch is a P channel, cannot conduct any current cus it has no negative gate voltage to get it started, so only Q9 conducts the current to the load, this is the first part of the cicle, now as the wave is a sinus one it goes down too so that means it generatesc negative voltage this time, what that means is that Q9 cannot conduct cus it has no positive voltage any more at it's gate so it cuts of, Q10 on the other hand likes negative voltage on it's gate so it starts conducting current to the load, so when the wave goes down only Q10 conducts current to the load, that is the full cicle.
What did we learn from all of this?:
-The sine nature of audio signall means that only one rail at a time can power the load/speaker, so the voltage signall on the speaker can only be provided from one single rail Vcc at a time. The speaker has one side connected at the ground, the other side of the speaker is connected by the power devices alternatively at possitive and negative side, up and down, and so on again and again, at the signall frequency.
Hope you have better understanding now of what happenes in an audio power amplifier, and also understand why i have sayd that i do not recommend any tinkerring with ofline SMPS for now, you lack too much basic knowledge and ofline SMPS is extreemely dangerous for one's life when you do not know it well enough!
Thanks for the above explination as it helps visualize and step through the workings of the amp. However, is it not possible to have two circuits like the above creating a situation were the speaker would be tied from the +35volt side to the -35volt side and not tie the speaker to a ground leg?
When I mentioned the +-35 I was speaking of Vref not the actual supply rails shown above in the example. As you mentioned previously they would need to be higer (near 50). So in theory a +-55 volt power supply with a bridged configuration would be capable of supplying 70 volts across the speaker and be capable of supplying 1.2kw of power correct?
For 1,2kW of power on 4 Ohm load you need about +/-75Vcc supply rails, but do you have any ideea how much more stress this config at this voltage rails puts on the output devices? i think you do not, in fact i think you do not have a grasp of most basics on the power amplifier, you really need to learn more, to read more, there are way too much stuff you do not understand, you cannot espect us to teach you all the basics, trust me there is much you need to know.
1200W into 4r0 is equivalent to 98Vpk & 24.5Vpk from a "normal" single ended amplifier.
If you go bridged you need two amplifiers each capable of 600W into 2ohms.
Each of the pair delivering 600W into 2r0 is equivalent to 49Vpk and 24.5Apk.
The total "load" on the PSU becomes 49Apk since the two amplifiers draw current at the same time, but from opposite rails.
Allow an overhead through the amplifier and cabling of ~5V. Allow an unregulated PSU droop of 5V to 10V, depending on how you have designed it. If you use a regulated PSU, you will need to work out the droop on load.
If you go bridged you need two amplifiers each capable of 600W into 2ohms.
Each of the pair delivering 600W into 2r0 is equivalent to 49Vpk and 24.5Apk.
The total "load" on the PSU becomes 49Apk since the two amplifiers draw current at the same time, but from opposite rails.
Allow an overhead through the amplifier and cabling of ~5V. Allow an unregulated PSU droop of 5V to 10V, depending on how you have designed it. If you use a regulated PSU, you will need to work out the droop on load.
@AndrewT with all do respect, that is ridicoulus, you really should research more.
How do you calculate the power output for a given load and voltage signal? it is:
Now if the reverse is the case and you need to calculate the signal voltage on a given load and power you do the reverse math:
And in this case we have P witch is 1200 and R witch is 4, now 1200*4=4800, the square root of this is 69, so U=69Vac, that is the signall voltage needed to generate 1,2Kw of power on 4 Ohm load, now ther power rails must be 69*1,41=97,2Vcc minimum, not accounting for power device losses, and let say for ease of argument a PSU of +/-100Vcc. Now how much power does the PSU needs? simple, foe 1200W of audio power on a B class audio power amplifier for example, accepting an efficiency of about 70% that means that 30% of the power from the supply will be lost as heat on the cooling system, and we need to provide enough power from the supply to cover both the audio power and losses, so at 70% eff we need 1200W/0,7=1700W, that is the power needed from the PSU, now we have established earlier that the power rails voltage is +/-100Vcc so that means from tail to rail 200Vcc, the total power of the PSU is related to that entire voltage so 1700W/200Vcc=8,5Amps, that is the minimum current the PSu must supply the amp, but that is on a resistive load with a continous voltage on the load, but the musick is not like that and the speaker coil is reactive, so in reality the current is drawn more like in pulses and most of it from the lower spectrum of audio witch is anything but continous.
So again quit with that Apk current and transients, and step down to earth, audio is not SMPS, do more research ( much more ) test more amplifiers for current consuption and you will realise how wrong you are.
best of wishes.
Marian.
How do you calculate the power output for a given load and voltage signal? it is:
An externally hosted image should be here but it was not working when we last tested it.
Now if the reverse is the case and you need to calculate the signal voltage on a given load and power you do the reverse math:
An externally hosted image should be here but it was not working when we last tested it.
And in this case we have P witch is 1200 and R witch is 4, now 1200*4=4800, the square root of this is 69, so U=69Vac, that is the signall voltage needed to generate 1,2Kw of power on 4 Ohm load, now ther power rails must be 69*1,41=97,2Vcc minimum, not accounting for power device losses, and let say for ease of argument a PSU of +/-100Vcc. Now how much power does the PSU needs? simple, foe 1200W of audio power on a B class audio power amplifier for example, accepting an efficiency of about 70% that means that 30% of the power from the supply will be lost as heat on the cooling system, and we need to provide enough power from the supply to cover both the audio power and losses, so at 70% eff we need 1200W/0,7=1700W, that is the power needed from the PSU, now we have established earlier that the power rails voltage is +/-100Vcc so that means from tail to rail 200Vcc, the total power of the PSU is related to that entire voltage so 1700W/200Vcc=8,5Amps, that is the minimum current the PSu must supply the amp, but that is on a resistive load with a continous voltage on the load, but the musick is not like that and the speaker coil is reactive, so in reality the current is drawn more like in pulses and most of it from the lower spectrum of audio witch is anything but continous.
So again quit with that Apk current and transients, and step down to earth, audio is not SMPS, do more research ( much more ) test more amplifiers for current consuption and you will realise how wrong you are.
best of wishes.
Marian.
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