Hi,
This is my first post on diyAudio (apart from the introduction page). I’ve been into audio for around 25years but only recently have I gotten more into the technical side of diy by learning about electronics and sound engineering. I’m currently learning about unregulated linear power supplies. Eventually I hope to apply my knowledge to building / modifying a PSU for a subwoofer power amp.
My questions:
I’m trying to understand the relationship between power, SPL and signal voltages. I think I have things by the wrong end of the pineapple, but I’ll only learn by asking, so here goes…
If I have a power supply for a power amp that has a transformer with twin secondaries capable of delivering 75VAC RMS at 12.8A each (so 960W RMS) what happens if I replace the transformer with one that supplies 60VAC RMS at 16A (also 960W RMS)?
From my limited understanding, I think I would be effectively reducing the voltage rails, which could lead to clipping (i.e. I would essentially have created a voltage sag for the amp, even though the power that the PSU can deliver is the same).
For example, in the first case I would have voltage rails of about 106VDC and in the second I would have rails of ~85VDC. As I understand it, an amplifier takes in a small signal (say 1vpk-pk) and can amplify it up to the value of the voltage rails. So in the first case I could have an output of 106VDCpk-pk and in the second case only 85VDCpk-pk. Because the only thing that would have changed between the two cases is the transformer, the load that the PSU sees is exactly the same in both cases. So if I had a 20 ohm load, in case one I could have amplified my signal to 106VDC pk –pk and drawn 5.3A (~560W) before clipping, but in case two I can only amplify my signal to 85VDCpk-pk and draw 4.25A (~360W) before clipping. Therefore I would have effectively created a less powerful amp for a given load.
Have I got this correct? Or can a reduction in voltage be made up for with an increase in current into the same load some how?
Any help is greatly appreciated.
Cheers!
This is my first post on diyAudio (apart from the introduction page). I’ve been into audio for around 25years but only recently have I gotten more into the technical side of diy by learning about electronics and sound engineering. I’m currently learning about unregulated linear power supplies. Eventually I hope to apply my knowledge to building / modifying a PSU for a subwoofer power amp.
My questions:
I’m trying to understand the relationship between power, SPL and signal voltages. I think I have things by the wrong end of the pineapple, but I’ll only learn by asking, so here goes…
If I have a power supply for a power amp that has a transformer with twin secondaries capable of delivering 75VAC RMS at 12.8A each (so 960W RMS) what happens if I replace the transformer with one that supplies 60VAC RMS at 16A (also 960W RMS)?
From my limited understanding, I think I would be effectively reducing the voltage rails, which could lead to clipping (i.e. I would essentially have created a voltage sag for the amp, even though the power that the PSU can deliver is the same).
For example, in the first case I would have voltage rails of about 106VDC and in the second I would have rails of ~85VDC. As I understand it, an amplifier takes in a small signal (say 1vpk-pk) and can amplify it up to the value of the voltage rails. So in the first case I could have an output of 106VDCpk-pk and in the second case only 85VDCpk-pk. Because the only thing that would have changed between the two cases is the transformer, the load that the PSU sees is exactly the same in both cases. So if I had a 20 ohm load, in case one I could have amplified my signal to 106VDC pk –pk and drawn 5.3A (~560W) before clipping, but in case two I can only amplify my signal to 85VDCpk-pk and draw 4.25A (~360W) before clipping. Therefore I would have effectively created a less powerful amp for a given load.
Have I got this correct? Or can a reduction in voltage be made up for with an increase in current into the same load some how?
Any help is greatly appreciated.
Cheers!
Yes, virtually all power amps amplify the input voltage, and if they are properly designed,
will supply the current to support that voltage into a load. The higher the voltage output,
the higher the power into the load, if the circuitry can supply the output current necessary.
the supply rails to the amplifier will determine the maximum output voltage from that amplifier.
If the amplifier load (speaker) draws significant current then the output voltage available can be very much less than the supply rail voltage.
eg.
a 35Vac transformer will give ~ 50Vdc supply rails.
The amplifier feeding a very high impedance load might be able to reach an unclipped 33Vac
But attach a 8ohms load and you will find that only good amplifiers can exceed 29Vac before clipping.
Now attach a 4ohms load and many amplifiers will not even reach 24Vac without clipping.
note the maximum output has changed from 33Vac to 24Vac as the load changed.
If the amplifier load (speaker) draws significant current then the output voltage available can be very much less than the supply rail voltage.
eg.
a 35Vac transformer will give ~ 50Vdc supply rails.
The amplifier feeding a very high impedance load might be able to reach an unclipped 33Vac
But attach a 8ohms load and you will find that only good amplifiers can exceed 29Vac before clipping.
Now attach a 4ohms load and many amplifiers will not even reach 24Vac without clipping.
note the maximum output has changed from 33Vac to 24Vac as the load changed.
Hi AndrewT,
Are you referring to voltage sag due to the current of the transformer being exceeded? Or are there other things within either the PSU or amp itself that contribute to the effect you are describing?
Cheers.
Hi,
I received an email telling me that a reply had been made to this thread, but it hasn't shown up here. The contents of the message were as follows:
wenze has just replied to a thread you have subscribed to entitled - Is PSU power rating the whole story? - in the Power Supplies forum of diyAudio. This thread is located at: http://www.diyaudio.com/forums/power-supplies/304303-psu-power-rating-whole-story-new-post.html Here is the message that has just been posted: *************** Not just transformers; everything has a max output voltage and max output current and sometimes a max output power spec. Hence resulting in the rectangle graph that we're familiar with. Below is that of a solar panel but the concept is the same. Image: http://rimstar.org/renewnrg/sizing_...rid/solar_panel_maximum_power_point_graph.jpg It's the user that decides the operating point / line on this graph. In your case the second PSU is indeed less powerful with a 20ohm load. But subwoofers are rarely 20ohm. What if it is for example 4ohm? You're more likely to hit current limit, in which case the higher current transformer becomes better despite less voltage. And then you have to consider the limitations of the rest of the circuit too.
Firstly, my thanks to wenze for responding and secondly it raises another question:
What load does the PSU of a Power Amp actually see? To my logic, The PSU is driving impedance / resistive loads posed by the amplifier stage, xover componentry, speaker cabling and the speaker itself. So if I have a 4ohm speaker, the load seen by the PSU will be more than the 4ohms (4ohms + amplifier stage + xover + speaker cabling). Is this correct? If so, how is it measured / determined from a PSU designers perspective?
Thanks.
I received an email telling me that a reply had been made to this thread, but it hasn't shown up here. The contents of the message were as follows:
wenze has just replied to a thread you have subscribed to entitled - Is PSU power rating the whole story? - in the Power Supplies forum of diyAudio. This thread is located at: http://www.diyaudio.com/forums/power-supplies/304303-psu-power-rating-whole-story-new-post.html Here is the message that has just been posted: *************** Not just transformers; everything has a max output voltage and max output current and sometimes a max output power spec. Hence resulting in the rectangle graph that we're familiar with. Below is that of a solar panel but the concept is the same. Image: http://rimstar.org/renewnrg/sizing_...rid/solar_panel_maximum_power_point_graph.jpg It's the user that decides the operating point / line on this graph. In your case the second PSU is indeed less powerful with a 20ohm load. But subwoofers are rarely 20ohm. What if it is for example 4ohm? You're more likely to hit current limit, in which case the higher current transformer becomes better despite less voltage. And then you have to consider the limitations of the rest of the circuit too.
Firstly, my thanks to wenze for responding and secondly it raises another question:
What load does the PSU of a Power Amp actually see? To my logic, The PSU is driving impedance / resistive loads posed by the amplifier stage, xover componentry, speaker cabling and the speaker itself. So if I have a 4ohm speaker, the load seen by the PSU will be more than the 4ohms (4ohms + amplifier stage + xover + speaker cabling). Is this correct? If so, how is it measured / determined from a PSU designers perspective?
Thanks.
> if I have a 4ohm speaker, the load seen by the PSU will be more than the 4ohms
Each half of a push-pull power amp conducts alternately. Also we traditionally measure with Sines.
When driving 8 Ohms to full sine power, a totem-pole output such as you seem to be planning will "act like" about 50 Ohms. 6.28*Rl.
When idling, it will act-like whatever the idle current is. For say 50V 50mA it is more like 1,000 Ohms.
Class A idles with all the power it could ever use. Because of alternate action and Sine wave-factor, 4*Rl.
Output transformers allow different leverages.
These are for "Clean Sine". Motor-shakers and guitar amplifiers may approach Square Wave operation, which oddly is exactly equal to Class A idle condition. Hi-Fi amps are traditionally tested on a Full Power Sine, but spend 99% of the time below 1/10th of full power.
So to your question. With the Sine assumption, your amplifier "acts like" a 50 Ohms resistor. If you apply 50V, it sucks 1Amp, eats 50 Watts. If you apply 100V, it sucks 2A, eats 200 Watts. If you apply 50V from a source that "can" deliver 2A or 10A, it still only takes 1A and 50 Watts. So your lower-V higher-A PT may be the same size but will loaf; you over-bought current that you can not use.
This all ignores "sag". Yes, a higher-Amps winding will sag less. But it is uneconomic to begin to try to "eliminate sag". For the sizes you seem to plan, 20% sag is reasonable and a very small audible difference.
Each half of a push-pull power amp conducts alternately. Also we traditionally measure with Sines.
When driving 8 Ohms to full sine power, a totem-pole output such as you seem to be planning will "act like" about 50 Ohms. 6.28*Rl.
When idling, it will act-like whatever the idle current is. For say 50V 50mA it is more like 1,000 Ohms.
Class A idles with all the power it could ever use. Because of alternate action and Sine wave-factor, 4*Rl.
Output transformers allow different leverages.
These are for "Clean Sine". Motor-shakers and guitar amplifiers may approach Square Wave operation, which oddly is exactly equal to Class A idle condition. Hi-Fi amps are traditionally tested on a Full Power Sine, but spend 99% of the time below 1/10th of full power.
So to your question. With the Sine assumption, your amplifier "acts like" a 50 Ohms resistor. If you apply 50V, it sucks 1Amp, eats 50 Watts. If you apply 100V, it sucks 2A, eats 200 Watts. If you apply 50V from a source that "can" deliver 2A or 10A, it still only takes 1A and 50 Watts. So your lower-V higher-A PT may be the same size but will loaf; you over-bought current that you can not use.
This all ignores "sag". Yes, a higher-Amps winding will sag less. But it is uneconomic to begin to try to "eliminate sag". For the sizes you seem to plan, 20% sag is reasonable and a very small audible difference.
Attachments
voltage sag of the supply rails is normal.
Think about a PSU that is passing zero current.
The capacitors are charged up to max and the ripple is zero. The voltage output is Vmax.
Increase the current to 100mA (quiescent bias) and the Vmax stays roughly the same, but the ripple is now 100mVpp.
The average output voltage is now Vmax-(50% of Vripple). The voltage at a low output current is lower than Vmax.
Increase the output current to 1Apk. Vripple will be very much higher and the average voltage at the amplifier will have sagged
Increase the output current to 3Apk and the average voltage will have sagged even more.
This has NOTHING to do with overloading the transformer. It is simply the increase in Vripple that reduces the average voltage.
The Vmin at the amplifier is Vmax- Vripple this could be 2 to 3 Volts below the quiescent voltage you measured. Vmin sets the onset of clipping of the output signal.
Now take account of the transformer.
Vmax is affected by the source impedance before the smoothing capacitors. A smaller transformer (higher regulation) will have a higher source impedance. As current increases, the Rs does reduce Vmax. This effect is in addition to Vripple.
The combined effects usually exceed 4V of Vsag at the amplifier when maximum power is being delivered and can reach 7V. A very poor PSU could approach 15Vsag, (AJT reported ~ this figure some years ago, claiming it as normal for solid state PSUs).
Then you have the losses through the amplifier. These too can be around 5V
A typical good current delivery 100W into 8ohms power amplifier operating on 50Vdc supply rails often has a maximum output of around 40Vpk to 41Vpk at the speaker terminals. That's a loss from quiescent to output of 9V to 10V.
PRR said
10V sag from 50V supply is 20% But I would claim that very much better Vsag can be achieved.
5Vsag is only 10% and 7Vsag is 14%. These would be my suggested targets. If you got down to 8% for Vsag @ full power, then you have a very current capable PSU.
Last edited:
Hi AndrewT,
A few more questions: what aspects of a linear unregulated power supply would a designer look at to minimise sag? Does it boil down to choosing components with the lowest impedance and resistance that still meet other circuit objectives, as well as choosing a quality transformer? What can be done to minimise the Vsag due to changes in ripple?
Cheers.
I would not over-fret sag.
The low limit is: you must meet your Power Output Spec or the customer will complain.
The high limit is: parts to stand that voltage must be available/affordable.
In DIY audio, both goals are "soft". An exact power output is not required; really we want "infinite power" within finite budget. "Requires 20 Watts" speakers play on 15 Watt amps, just not as loud as the designer intended. I remember when 80V devices cost a lot more than 60V devices; today high-power audio devices start at 140V and go up with prices that look silly-cheap compared to early days. Maybe you pencil-budgeted $80 for output transistors, and discover it will cost $100.... you wait for next pay-day, it isn't like your company will go broke and your workers thrown out in the alley.
The low limit is: you must meet your Power Output Spec or the customer will complain.
The high limit is: parts to stand that voltage must be available/affordable.
In DIY audio, both goals are "soft". An exact power output is not required; really we want "infinite power" within finite budget. "Requires 20 Watts" speakers play on 15 Watt amps, just not as loud as the designer intended. I remember when 80V devices cost a lot more than 60V devices; today high-power audio devices start at 140V and go up with prices that look silly-cheap compared to early days. Maybe you pencil-budgeted $80 for output transistors, and discover it will cost $100.... you wait for next pay-day, it isn't like your company will go broke and your workers thrown out in the alley.
Most power amplifiers running on a dual 35Vac transformer will hit the 100W into 8r0 target.
Some might only get 95W and others might go as high as 110W.
It depends very much on the components chosen and on the mains voltage used during the test.
But the real test is when you try half the rated impedance. This really shows up a poor current capability PSU+AMP.
If the maximum output voltage into a 4r0 test load is more than 1dBv down from the 8r0 load then in my opinion the amp fails. A really good amplifier+PSU will drop around 0.4dBv to 0.6dBv changing from rated load to half rated load.
The next test is for one third rated load. The max will fall even more and the amp should not trigger the IV or current limiting. More than 3dBv drop from rated load is a fail. I aim for better than -1.5dBv into 2r67 for an 8ohms rated power amplifier.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.