Let's say I've built a power supply with a transformer that converts 120v to 25v, and then a bridge rectifier, and some filter capacitors. Now I've got 25VDC to connect to my equipment. What determines the number of amps my equipment can draw? Is it the total capacitance of the filter caps? Then I guess there's the transformer, and then the wall.
To put it another way: If I double the capacitance of my power supply, did I just double it's maximum amperage?
To put it another way: If I double the capacitance of my power supply, did I just double it's maximum amperage?
Nor necessarily. It's everything together, as a system.
The AC line has current limits. The transformer has limits. Even the diodes.
The capacitance stores a certain amount of charge per rectification cycle, which sets
the voltage ripple, and has ESR, which causes internal heating and limits the useful current.
All these factors must be balanced to minimize the total cost for the needed performance.
The AC line has current limits. The transformer has limits. Even the diodes.
The capacitance stores a certain amount of charge per rectification cycle, which sets
the voltage ripple, and has ESR, which causes internal heating and limits the useful current.
All these factors must be balanced to minimize the total cost for the needed performance.
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So it’s not that we need to “make” the amps, the amps will come when called for, we just need to get out of the way?
We have to "attract" the current by supplying a favorable load and potential (often ground).
The current is "stored" in the form of charge on the plates of the power supply capacitor(s).
This storage is in the form of excess electrons on one plate, and an equal deficiency of electrons on the other plate.
The stored charge becomes current when the electrons flow from the capacitor plates to the load.
As current flows out of the capacitors into a resistive load, the voltage on the capacitor(s) decays exponentially.
Normally the capacitor(s) are recharged from the rectifier/transformer before the voltage decays below
a certain point (ripple voltage).
Current is the rate of electronic charge flow, or how many Coulombs per second.
A Coulomb is about 6.2 x 10^18 times the magnitude of an electron charge.
That's a lot of them little buggers, but current flow is actually in discrete parts, not continuous.
Don't worry about the current actually being in reverse (e is negative), it doesn't matter.
The current is "stored" in the form of charge on the plates of the power supply capacitor(s).
This storage is in the form of excess electrons on one plate, and an equal deficiency of electrons on the other plate.
The stored charge becomes current when the electrons flow from the capacitor plates to the load.
As current flows out of the capacitors into a resistive load, the voltage on the capacitor(s) decays exponentially.
Normally the capacitor(s) are recharged from the rectifier/transformer before the voltage decays below
a certain point (ripple voltage).
Current is the rate of electronic charge flow, or how many Coulombs per second.
A Coulomb is about 6.2 x 10^18 times the magnitude of an electron charge.
That's a lot of them little buggers, but current flow is actually in discrete parts, not continuous.
Don't worry about the current actually being in reverse (e is negative), it doesn't matter.
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Well, I've been known to expound a bit.
Water in a city water tower is a reasonable analogy to charge storage in a capacitor (not literally, though).
Water flows out if there is an appropriate (lower) gravitational potential presented to the tower outlet.
But if the gravitational potential there is higher, water will flow into the tower, instead.
To replenish the capacitor, once or twice per utility cycle a quick spurt of charge is sent into the capacitor,
so the process can repeat in a steady state. This must happen much more quickly than the charge is depleted.
So the peak capacitor input current can be much larger (though much briefer) than the actual load current.
It's the integral of the current (charge) that is constant over a cycle with a fixed resistive load.
The voltage at an electronic circuit node is a measure of the energy per charge at that node.
The absolute voltage (like absolute energy) does not matter, only the difference between two of the circuit nodes.
See Kirchhoff's laws. So a battery powered circuit still works, even with nothing being connected to "ground".
It's all relative.
Water in a city water tower is a reasonable analogy to charge storage in a capacitor (not literally, though).
Water flows out if there is an appropriate (lower) gravitational potential presented to the tower outlet.
But if the gravitational potential there is higher, water will flow into the tower, instead.
To replenish the capacitor, once or twice per utility cycle a quick spurt of charge is sent into the capacitor,
so the process can repeat in a steady state. This must happen much more quickly than the charge is depleted.
So the peak capacitor input current can be much larger (though much briefer) than the actual load current.
It's the integral of the current (charge) that is constant over a cycle with a fixed resistive load.
The voltage at an electronic circuit node is a measure of the energy per charge at that node.
The absolute voltage (like absolute energy) does not matter, only the difference between two of the circuit nodes.
See Kirchhoff's laws. So a battery powered circuit still works, even with nothing being connected to "ground".
It's all relative.
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Simplified follows Ohms law.
3 constants Voltage ,Current, Resistance
Long as you know 2 constants
Can use formula to solve
missing value.
You question was how many amps the equipment or load will
draw.
You stated the voltage 25 volts
If you knew you load was 8 ohms resistance.
Then current drawn is 3.3 amps or 78 watts
Again simplified the transformer would need to be large enough
and be able to provide 3.3 amps or 78 watts.
Very simplified but ohms law will solve the answer.
Of course in the real world there is always losses depending on the
efficiency of the circuit. So very generalized if you know the efficiency
and how much is lost. Power supply is usually larger to make up
for losses.
Again very simplified because every application has its details
and further formulas
Transformer have rated voltage and Volt Amps.
Volt Amps = Watts
100va is 100 watts
You can also solve watts to current
100va = 100watts = 3.54 amps
Lot more details of course to understanding transformer behavior.
More to it since your usually converting AC to DC.
Generalized if load efficiency is good around 80%
You determined a 78 watt load so a 100va transformer
or 100 watt transformer would be suitable.
If load efficiency was poor or low around 50 to 65%
then your 78 watt load could require closer to a 200va
transformer or 200 watts.
Most hobbyist usually buy " off the shelf" transformer.
30va 80va 100va 180va 250va pretty common values.
Of course can get much larger than that.
3 constants Voltage ,Current, Resistance
Long as you know 2 constants
Can use formula to solve
missing value.
You question was how many amps the equipment or load will
draw.
You stated the voltage 25 volts
If you knew you load was 8 ohms resistance.
Then current drawn is 3.3 amps or 78 watts
Again simplified the transformer would need to be large enough
and be able to provide 3.3 amps or 78 watts.
Very simplified but ohms law will solve the answer.
Of course in the real world there is always losses depending on the
efficiency of the circuit. So very generalized if you know the efficiency
and how much is lost. Power supply is usually larger to make up
for losses.
Again very simplified because every application has its details
and further formulas
Transformer have rated voltage and Volt Amps.
Volt Amps = Watts
100va is 100 watts
You can also solve watts to current
100va = 100watts = 3.54 amps
Lot more details of course to understanding transformer behavior.
More to it since your usually converting AC to DC.
Generalized if load efficiency is good around 80%
You determined a 78 watt load so a 100va transformer
or 100 watt transformer would be suitable.
If load efficiency was poor or low around 50 to 65%
then your 78 watt load could require closer to a 200va
transformer or 200 watts.
Most hobbyist usually buy " off the shelf" transformer.
30va 80va 100va 180va 250va pretty common values.
Of course can get much larger than that.
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If the capacitor in our analogy is in the power supply of an amplifier, is the fluctuation of current drawn by the speakers equivalent to raising and lowing the output faucet? Where “level with the tower” is clipping?
Your last comment was very helpful. What is a “utility cycle”?
Your last comment was very helpful. What is a “utility cycle”?
Fluctuation of load current would be like changing the gravitational potential at the tower outlet up and down.
Clipping would sorta be like the potential at the outlet and load being equal, so no water flows.
The utility cycle is 60Hz around here. It can be half rectified (still 60Hz) or full wave rectified (twice, or 120Hz).
Clipping would sorta be like the potential at the outlet and load being equal, so no water flows.
The utility cycle is 60Hz around here. It can be half rectified (still 60Hz) or full wave rectified (twice, or 120Hz).
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Ahh! Gotcha, thank you!
Does the power supply in an amp see a constant voltage potential, or does it fluctuate with signal level?
The reason that supply current flows is that the load is at a different potential than the power supply,
due to the signal applied to the output devices. An amplifier can be viewed as a regulated power supply,
with the reference voltage being the input signal, instead of a fixed reference, so the output varies with the signal.
The power supply is an approximate voltage source. Ideally, the supply remains at a perfectly constant voltage,
but without an electronic regulator, the voltage will fluctuate significantly with the load and signal level.
If the electrical potential (voltage) at the load were equal to that of the power supply, no current would flow.
due to the signal applied to the output devices. An amplifier can be viewed as a regulated power supply,
with the reference voltage being the input signal, instead of a fixed reference, so the output varies with the signal.
The power supply is an approximate voltage source. Ideally, the supply remains at a perfectly constant voltage,
but without an electronic regulator, the voltage will fluctuate significantly with the load and signal level.
If the electrical potential (voltage) at the load were equal to that of the power supply, no current would flow.
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On the other hand, a regulated power supply can be seen as an amplifier with a fixed, constant (DC) input signal.
It also includes warts like stability considerations, etc. In this view, it would seem that the regulated supply "amplifier"
should be several times better in performance than the actual audio amplifier circuit, to avoid restricting its performance.
This seldom happens.
It also includes warts like stability considerations, etc. In this view, it would seem that the regulated supply "amplifier"
should be several times better in performance than the actual audio amplifier circuit, to avoid restricting its performance.
This seldom happens.
Can one throw a butt-ton of capacitance at low-power amp and say it can handle super-high peaks? What becomes the limiting factor if you keep adding capacitance?
Ideally the voltage rails of the amplifier is constant.
Audio signal voltage varies and impedance curve or resistance of the load
varies. So current draw also varies.
If load exceeds the current rating of transformer.
It can still provide the current but voltage will drop.
And will basically get hotter and hotter if you exceed
the rating.
Given the chance could melt itself to death if shorted
But we have fuses and mains line breakers.
Audio signal voltage varies and impedance curve or resistance of the load
varies. So current draw also varies.
If load exceeds the current rating of transformer.
It can still provide the current but voltage will drop.
And will basically get hotter and hotter if you exceed
the rating.
Given the chance could melt itself to death if shorted
But we have fuses and mains line breakers.
No, the signal peak is the voltage, which is relatively fixed. It could handle larger current peaks, if they are brief enough.
More supply capacitance will increase the audible bass range and quality, since the supply caps are in series with the speaker.
Yes, even with feedback.
More supply capacitance will increase the audible bass range and quality, since the supply caps are in series with the speaker.
Yes, even with feedback.
Converting AC to DC
So depends on the overall acceaptable ripple on the DC.
Can be calculated depending on frequency of AC and expected
load current.
On startup or power up a large capacitor bank causes big current surge
so excessively large banks will stress the transformer and rectifiers over
life time. So if you start getting obsessed with unneeded capacitance
Your looking at soft starts and additional circuitry to not stress the rectifier
and transformer.
With large transformers and needed banks could be considered mandatory.
Opinions vary
On realistic level 30 to 100 watt class AB
10,000uf to 12,000uf more than enough.
Doesn't meet your ripple expectations then go regulated.
Large enough capacitance to consider simple soft start
if transformer around 180 to 250va
Once up to 330va or 500va and higher
Be a more elaborate soft start and consider it mandatory.
Dont want the fuse too large if a fault occurs should blow.
Slo Blow is common because rated fuses can still blow at startup
from the current surge caused by capacitor bank.
So excessively large banks are ridiculous and cause additional problems or cost.
If unregulated supply doesnt meet your accepted ripple. Go regulated.
More cost and design consideration and new can of worms to open.
So depends on the overall acceaptable ripple on the DC.
Can be calculated depending on frequency of AC and expected
load current.
On startup or power up a large capacitor bank causes big current surge
so excessively large banks will stress the transformer and rectifiers over
life time. So if you start getting obsessed with unneeded capacitance
Your looking at soft starts and additional circuitry to not stress the rectifier
and transformer.
With large transformers and needed banks could be considered mandatory.
Opinions vary
On realistic level 30 to 100 watt class AB
10,000uf to 12,000uf more than enough.
Doesn't meet your ripple expectations then go regulated.
Large enough capacitance to consider simple soft start
if transformer around 180 to 250va
Once up to 330va or 500va and higher
Be a more elaborate soft start and consider it mandatory.
Dont want the fuse too large if a fault occurs should blow.
Slo Blow is common because rated fuses can still blow at startup
from the current surge caused by capacitor bank.
So excessively large banks are ridiculous and cause additional problems or cost.
If unregulated supply doesnt meet your accepted ripple. Go regulated.
More cost and design consideration and new can of worms to open.
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The amount of current you can draw from a traditional transformer+rectifier+capacitor power supply is determined by how much current the transformer can tolerate without overheating, how much current the diodes can handle, and how much ripple voltage you can tolerate.
The power rating of the power transformer is typically specified in VA - volt-ampere. It's simply the product of the voltage and current the transformer is rated for. For a resistive load 1 VA = 1 W, but for a reactive load such as a reservoir capacitor, 1 VA is not the same as 1 W because the current and voltage are out of phase. But basically figure that the VA rating is the amount of power you can draw from the transformer before it overheats.
Transformers tend to be pretty forgiving of short-term overload, which is probably why they're still favoured in the audio world.
Rectifier diodes are specified by their average current handling and their peak repetitive current handling abilities. If you exceed the average current the diode will overheat and blow. Same for the peak repetitive current, but that's often overlooked. The peak current is a function of the conduction angle of the diode, so basically the amount of time out of each mains cycle the diode is conducting. If the diode is only conducting for a very short amount of time but needs to supply a lot of charge to the reservoir capacitor during that time the peak current gets to be enormous. Recall, I = Q/t, so a large Q or a small t will give you a high I.
Because of the very high peak current in the rectifier diodes, the RMS current (which is what the transformer is sized for) is generally assumed to be 1.4 to 1.6 times the average load current of the power supply. So when you have a transformer custom made, you either specify VA = V*1.6*I_average unless the transformer manufacturer asks for the average current directly (and does the math for you).
The capacitance of the reservoir capacitor impacts the ripple voltage. Basically figure that the transformer supplies power to the load through the rectifier diodes during the small amount of time where the diodes are conducting and the capacitor supplies power during the rest of the mains half-cycle (assuming a full-wave rectifier). C = Q/V and Q = I*t, so you can estimate the ripple voltage as follows: C = I*t/V <-> V = I*t/C. Thus, a large C will give you a low ripple voltage (V). Similarly, a small I (current draw) will give you a small ripple voltage. And for completeness, a higher mains frequency will give you smaller ripple voltage as t = 1/f.
How much ripple voltage is acceptable will depend on the circuit attached to the power supply. For an audio amp with high power supply rejection ratio you may be fine with a few volt of ripple. But in a low-noise amp you might want much, much less.
Simple question (and a good one). But not so simple answer. 🙂
Tom
The power rating of the power transformer is typically specified in VA - volt-ampere. It's simply the product of the voltage and current the transformer is rated for. For a resistive load 1 VA = 1 W, but for a reactive load such as a reservoir capacitor, 1 VA is not the same as 1 W because the current and voltage are out of phase. But basically figure that the VA rating is the amount of power you can draw from the transformer before it overheats.
Transformers tend to be pretty forgiving of short-term overload, which is probably why they're still favoured in the audio world.
Rectifier diodes are specified by their average current handling and their peak repetitive current handling abilities. If you exceed the average current the diode will overheat and blow. Same for the peak repetitive current, but that's often overlooked. The peak current is a function of the conduction angle of the diode, so basically the amount of time out of each mains cycle the diode is conducting. If the diode is only conducting for a very short amount of time but needs to supply a lot of charge to the reservoir capacitor during that time the peak current gets to be enormous. Recall, I = Q/t, so a large Q or a small t will give you a high I.
Because of the very high peak current in the rectifier diodes, the RMS current (which is what the transformer is sized for) is generally assumed to be 1.4 to 1.6 times the average load current of the power supply. So when you have a transformer custom made, you either specify VA = V*1.6*I_average unless the transformer manufacturer asks for the average current directly (and does the math for you).
The capacitance of the reservoir capacitor impacts the ripple voltage. Basically figure that the transformer supplies power to the load through the rectifier diodes during the small amount of time where the diodes are conducting and the capacitor supplies power during the rest of the mains half-cycle (assuming a full-wave rectifier). C = Q/V and Q = I*t, so you can estimate the ripple voltage as follows: C = I*t/V <-> V = I*t/C. Thus, a large C will give you a low ripple voltage (V). Similarly, a small I (current draw) will give you a small ripple voltage. And for completeness, a higher mains frequency will give you smaller ripple voltage as t = 1/f.
How much ripple voltage is acceptable will depend on the circuit attached to the power supply. For an audio amp with high power supply rejection ratio you may be fine with a few volt of ripple. But in a low-noise amp you might want much, much less.
Simple question (and a good one). But not so simple answer. 🙂
Tom
You know.... entire reams... heck entire years of college courses describe this stuff. Some of the stuff you are describing applies only to a linear power supply.
In general ( forget Stasis ) the power supply provides a static voltage and the amount of current it drives into the load (amplifier section in our case) is dependent on the gain of the amplifier and the load. Some amplifiers have variable gain ( preamplifiers mostly ) and others fixed gain ( amplifiers mostly ). And, of course, we must not forget the load proper which presents some impedance ( resistance ) to the amplifier, which then throttles the amount of current.
Perhaps a more important term should be the POWER being delivered, not just the current. Because power is what makes the work and it a better measure of the system. After all, it is the Electro-Motive-Force (EMF) that defines the potential (in volts) of those amperes (current) to do work.
And the truly evil part, naturally, is that the flow of charge (electrons) is actually backwards to the flow of measured current.
In general ( forget Stasis ) the power supply provides a static voltage and the amount of current it drives into the load (amplifier section in our case) is dependent on the gain of the amplifier and the load. Some amplifiers have variable gain ( preamplifiers mostly ) and others fixed gain ( amplifiers mostly ). And, of course, we must not forget the load proper which presents some impedance ( resistance ) to the amplifier, which then throttles the amount of current.
Perhaps a more important term should be the POWER being delivered, not just the current. Because power is what makes the work and it a better measure of the system. After all, it is the Electro-Motive-Force (EMF) that defines the potential (in volts) of those amperes (current) to do work.
And the truly evil part, naturally, is that the flow of charge (electrons) is actually backwards to the flow of measured current.
I only buy transformers that have an associated spec sheet. eg
Iron Loss Copper Loss Efficiency Temperature Rise Ambient Rated
1.49 Watts16.9 Watts Approx.92% typical56°C Approx.40°C
From a 225VA transformer available at several output voltages. The rest are in this spec sheet.
https://www.farnell.com/datasheets/2712881.pdf
So what happens if I load it at 225VA. The temperature will rise to ambient+56C. The output voltage will fall to the stated output voltage which is AC. Rectify that and add a capacitor and without load the DC voltage will be root 2 times the transformers output plus some more due to no copper losses. Some specs may give the open circuit voltage or the regulation. Say that was 10% the DC voltage open circuit would be 10% higher.
So the transformer is loaded at 225VA. It's temperature will increase by 56C. If the internal temperature of the case it's in reaches 40C whoops the insulators etc in it are reaching their max temperature. If it was loaded at 112.5VA the temperature rise would be 28C. All in all it's best to check what temp it does get under load but it will get hotter but they are designed to function at the temperatures they specify. It's not going to cause it any problems. These are industrial transformers.
So it's loaded at some amperage. There will be ripple on the DC that will reduce as the value of the filter capacitor are increased. There are calculators around on the web. If too much a regulator can be used to reduce it but those aren't perfect. They can reduce it significantly. The power output from an amp varies in use as does the load it represents to the supply. This influences the ripple.
OK so say this 225VA transformer is used to power a 200w RMS amp if it can remain in spec. In many situations that level of power is unlikely to be needed. We don't listen to sine waves close to clipping we often listen to music. Advertising blurb gives a clue - so called music power with will generally be twice the RMS rating. DIN power is another but more complicated. So your 200w amp would be fine for testing close to clipping. The important thing is to have enough volts to handle peaks in the music. Actual load on the transformer is likely to be 1/2 what the transformer is capable of giving in RMS terms.
The other question is how much power is needed in the listening situation. In the peak of the must have hifi era 100w rms was more of a badge of honour than what was actually needed in domestic situations. In fact peak power at that level is much the same. So the transformer loading goes down again in actual use.
Iron Loss Copper Loss Efficiency Temperature Rise Ambient Rated
1.49 Watts16.9 Watts Approx.92% typical56°C Approx.40°C
From a 225VA transformer available at several output voltages. The rest are in this spec sheet.
https://www.farnell.com/datasheets/2712881.pdf
So what happens if I load it at 225VA. The temperature will rise to ambient+56C. The output voltage will fall to the stated output voltage which is AC. Rectify that and add a capacitor and without load the DC voltage will be root 2 times the transformers output plus some more due to no copper losses. Some specs may give the open circuit voltage or the regulation. Say that was 10% the DC voltage open circuit would be 10% higher.
So the transformer is loaded at 225VA. It's temperature will increase by 56C. If the internal temperature of the case it's in reaches 40C whoops the insulators etc in it are reaching their max temperature. If it was loaded at 112.5VA the temperature rise would be 28C. All in all it's best to check what temp it does get under load but it will get hotter but they are designed to function at the temperatures they specify. It's not going to cause it any problems. These are industrial transformers.
So it's loaded at some amperage. There will be ripple on the DC that will reduce as the value of the filter capacitor are increased. There are calculators around on the web. If too much a regulator can be used to reduce it but those aren't perfect. They can reduce it significantly. The power output from an amp varies in use as does the load it represents to the supply. This influences the ripple.
OK so say this 225VA transformer is used to power a 200w RMS amp if it can remain in spec. In many situations that level of power is unlikely to be needed. We don't listen to sine waves close to clipping we often listen to music. Advertising blurb gives a clue - so called music power with will generally be twice the RMS rating. DIN power is another but more complicated. So your 200w amp would be fine for testing close to clipping. The important thing is to have enough volts to handle peaks in the music. Actual load on the transformer is likely to be 1/2 what the transformer is capable of giving in RMS terms.
The other question is how much power is needed in the listening situation. In the peak of the must have hifi era 100w rms was more of a badge of honour than what was actually needed in domestic situations. In fact peak power at that level is much the same. So the transformer loading goes down again in actual use.