In-Rush Current for power supply

[chimera_786]

hello! Recently I decided to build a power supply. The one I have right now is big and really just gives fixed out put voltages.

Sure, I can add an LM317s or LT3080 to give me the variable voltages, but the output range is limited and then there is matter of the heat sinks...eh just a heap of mess to deal with if I want to modify my current power supply.

I have read up sufficiently on power supplies

I have a transformer with 50VA @ 24V secondary. I am using the 1n5408 diodes for the rectification (full wave) and 30,000uF filter caps. The output power of my intended power supply is 40W (20V @ 2A).
There are more elements present but for the sake of simplicity, I haven't included them here.

So this is what i need help with:

1- How do I calcuate the inrush current?

2- Do I really need to be worried of inrush current for a 40W power supply?

3- What are my options of implementing a in rush currrent limiting circuit---i.e soft start or an NTC thermistor.

Any suggestions?

[NiHaoMike]

Inrush is probably not a problem for that supply. If it is, the most usual solution is a NTC, although a high efficiency design might also have a relay to bypass it after startup.

[IanB]

Which inrush current are you concerned about--the inrush from the mains to the transformer, or the inrush from the secondary to the filter caps?

I doubt you have anything to worry about with such a small transformer on the primary side, and your diodes can handle a 200 A surge for several milliseconds so I doubt you have anything to worry about on the secondary side either.

Unless you have some special requirements, there seems nothing to be alarmed about.

[T4P]

There's a reason NOT to be afraid : Only be afraid if it's beyond 300W .

But , always check the startup current with a scope .

[chimera_786]

There's a reason NOT to be afraid : Only be afraid if it's beyond 300W .

But , always check the startup current with a scope .

thanks Dave for confirming by doubts, but can you please tell me how to find the start up current with a scope.. I have a rigol DS1502E

Which inrush current are you concerned about--the inrush from the mains to the transformer, or the inrush from the secondary to the filter caps?

I doubt you have anything to worry about with such a small transformer on the primary side, and your diodes can handle a 200 A surge for several milliseconds so I doubt you have anything to worry about on the secondary side either.

Unless you have some special requirements, there seems nothing to be alarmed about.

Hello IanB, so u think the diode internal resistance should limit the in rush current its self? The caps I am using are rated at 3.63 A Rms and an ESR of 0.0523 ohms each. U think they'll hold?

Inrush is probably not a problem for that supply. If it is, the most usual solution is a NTC, although a high efficiency design might also have a relay to bypass it after startup.

Hello NiHaoMike, I have looked at some literature that goes over using a relay to by pass the NTC, but I really dont have a visual idea of how to go about it. Can you recommend some material I can read up on and get an idea?

Thanks to every one who decided to help me out with my problem!!

[IanB]

thanks Dave for confirming by doubts, but can you please tell me how to find the start up current with a scope.. I have a rigol DS1502E

I don't think Dave's advice was very sensible. I'd pass that by if I were you.

Hello IanB, so u think the diode internal resistance should limit the in rush current its self? The caps I am using are rated at 3.63 A Rms and an ESR of 0.0523 ohms each. U think they'll hold?

Sure they'll hold. But your capacitance of 30,000 uF seems way more than necessary. You can reduce any concerns by lowering that to something more reasonable like 10,000 uF.

Hello NiHaoMike, I have looked at some literature that goes over using a relay to by pass the NTC, but I really dont have a visual idea of how to go about it. Can you recommend some material I can read up on and get an idea?

Thanks to every one who decided to help me out with my problem!!

I think you are worrying unnecessarily. You don't have a problem! (But what makes you think might have a problem anyway? Why did you start worrying?)

[T4P]

thanks Dave for confirming by doubts, but can you please tell me how to find the start up current with a scope.. I have a rigol DS1502E

I don't think Dave's advice was very sensible. I'd pass that by if I were you.

You just reminded me that i saw what i said 1 year ago on DIYaudio .

EDIT : There is a genuine reason to be afraid when it comes to anything other then audio amps .

[chimera_786]

well.. about the whole deal involving measuring current using an o-scope, I dont know how to do it and prolly wont end up doing it..

about the cap, I figured that i can calculate the value by 1=c dv/dt, where dv= ripple voltage desired and dt = 8.333 ms for 60Hz fully rectified..so since my requirement for the output power is 20V @ 2A---> this means that a 10% ripple @ 20V is 2V.

I i can get away with 10,000uf!!---however... a 20,000uF will get me 5% ripple and 30,000uF will get me a 2.5% ripple..... at least in theory... however, the cap size isant really an issue.. the issue is the in rush current..

The reason why I began to worry about the in rush current is because I dont understand it and anything that I dont understand, I dont trust and can't be too sure about. I can believe the words of experienced individuals such as ur self.. but if I was calculate it for instance, I would be more assured that an in rush current wont be an issue..

I just dont want to fry anything...i.e over load the transformer, or have the massive 10,000uF blow in my face.. (i have been wearing safety goggles ever since I started to mess around with power stuff )

BTW.. i wish an engineering degree had prepared me for this..it really didnt..

[IanB]

about the cap, I figured that i can calculate the value by 1=c dv/dt, where dv= ripple voltage desired and dt = 8.333 ms for 60Hz fully rectified..so since my requirement for the output power is 20V @ 2A---> this means that a 10% ripple @ 20V is 2V

Lets try it out using your numbers:

I = C * dV/dt

I = 2 A

dV/dt = 2 V / 0.00833 s = 240 V/s

Hence,

C = I / (dV/dt) = 2 / 240 = 0.00833 F = 8300 uF

So 10,000 uF should be fine.

Inrush current is only going to be a problem if it blows a fuse (not an issue in your case), or if it blows up a diode. Since your 1N5408 diodes have a good surge rating you are golden. Nothing to worry about.

[chimera_786]

yeah...10,000uF works for a ripple of 2V.

The diodes are fine, but is there no way to calculate a surge current? or is it way too complicated for such a simple low wattage power supply?'

I tried finding a paper or an article on it and I came up with a pdf. Please look at page 11 and you will see what I am concerned about. I have attached it to this post. I do not take credit for the pdf and the original author is responsible for its content.

[T4P]

But , in any case .
10,000uf is rather huge already ,

Still not a problem .

[cybergibbons]

Transformers and big electrolytics are hard to model - so inrush current will be hard to predict. You can make some guesses, but there probably isn't really any need.

A low value resistor in the current path would be used with the scope to measure inrush current.

[alm]

Either a shunt resistor or ideally a current probe, since the resistor may limit the surge current. I wouldn't suggest buying a current probe for just this measurement, however. A DMM with fast min/max or peak hold may also be able to give you an estimate, although it won't tell you about the duration.

[SeanB]

A cheap and simple current probe is easy to make. Take an old PC power supply and remove the output filter toroid from it. Strip off all the windings, leaving a bare toroid. Take some thin insulated wire ( strip some network cable to get a metre or so of one of the cores) and wind 10 turns evenly around the core. Tape down the ends and insulate with a single turn of electrical tape. Twist the 2 free ends together to make a twisted pair and connect a 1ohm 2W non wirewound resistor across them. Place your scope probe across the resistor. Run one of the power leads through the core ( that is why you insulated it, paranoia is good with mains connections) and turn it on with the scope in single shot mode to get the single event of a high input current that decays away to a low level in a few cycles.

You can calibrate it by realising that you have made essentially a 10: current transformer, so every volt on the secondary is equal to 10A on the primary wire. Will cook the resistor with long high current input, but perfect to do short measurements. Will be quite linear with low current inputs, but eventually the core will saturate. It will still work, but with low outputs and nasty spikes as the core goes in and out of saturation during each input cycle.

[IanB]

yeah...10,000uF works for a ripple of 2V.

The diodes are fine, but is there no way to calculate a surge current? or is it way too complicated for such a simple low wattage power supply?'

I tried finding a paper or an article on it and I came up with a pdf. Please look at page 11 and you will see what I am concerned about. I have attached it to this post. I do not take credit for the pdf and the original author is responsible for its content.

The essence of engineering is only to calculate what you need to calculate.

According to the paper you linked, the main danger from surge current is damaging the rectifying diodes between the transformer secondary and the filter capacitor. On looking at the 1N5408 data sheet, it says: "Non-repetitive Peak Forward Surge Current (8.3 ms Single Half-Sine-Wave) : 200 A"

So you can ask yourself, what would it take to get a surge current of 200 A?

If the transformer has a 24 V secondary, it may produce (say) 35 V peak when unloaded. Using Ohm's law, it would need a resistance less than 0.175 ohms (35/200) to produce 200 A on switch on. This is conservative, since the total circuit impedance will be higher than the DC resistance. So what is the likelihood of your circuit having a resistance or impedance less than 0.175 ohms? Very unlikely, in fact. Your transistor secondary is likely to be in the range of 0.5 to 1 ohm, and then you have the diodes, and then you have the capacitor ESR.

Since a simple estimate indicates there is no danger, there is no need to do a more detailed calculation.

[chimera_786]

The essence of engineering is only to calculate what you need to calculate.

According to the paper you linked, the main danger from surge current is damaging the rectifying diodes between the transformer secondary and the filter capacitor. On looking at the 1N5408 data sheet, it says: "Non-repetitive Peak Forward Surge Current (8.3 ms Single Half-Sine-Wave) : 200 A"

So you can ask yourself, what would it take to get a surge current of 200 A?

If the transformer has a 24 V secondary, it may produce (say) 35 V peak when unloaded. Using Ohm's law, it would need a resistance less than 0.175 ohms (35/200) to produce 200 A on switch on. This is conservative, since the total circuit impedance will be higher than the DC resistance. So what is the likelihood of your circuit having a resistance or impedance less than 0.175 ohms? Very unlikely, in fact. Your transistor secondary is likely to be in the range of 0.5 to 1 ohm, and then you have the diodes, and then you have the capacitor ESR.

Since a simple estimate indicates there is no danger, there is no need to do a more detailed calculation.

From my understanding, all of the inrush current is caused becaz the caps, initillay, they are not charged to any voltage level. The only thing stopping the in rush is what u mentioned ( ESR and series coil resistance and the resistance from the diodes)


The capacitor I have has an ESR of 0.030 ohms and a Arms of 5.75A @ 120Hz (caps data sheet). Now I am planning on using two of these 10,000uF caps in parallel, so the effective ESR drops to have half of this.

I think this Arms rating of the capacitors is what is throwing me off regarding the entire in rush current.

I really am not trying to argue a mute point here, but I dont want anything to go bad which I could have prevented, had I take a more detailed approach to this power supply.

[IanB]

Maybe you are mixing up inrush current and operating current?

Inrush current is what happens one time only when you switch on the power supply. It is a single, instantaneous event. As you observe the capacitors initially are not charged to any voltage, and so they appear like a short circuit across the transformer/rectifier section. This may produce a very large current during the first AC cycle. That current will not harm the capacitors, but it may let the magic smoke out of the rectifier.

Operating current is what is flowing through the capacitors during normal operation. If this is too large it may cause the capacitors to overheat and fail prematurely. However, operating current is rarely a problem with mains frequency linear supplies. It is more of a problem in high frequency switching supplies. In your case you may consider that with an output current of 2 A the smoothing capacitors are not going to handle a current much different from this. If they are good high quality capacitors you have nothing to worry about.

[chimera_786]

Maybe you are mixing up inrush current and operating current?

Inrush current is what happens one time only when you switch on the power supply. It is a single, instantaneous event. As you observe the capacitors initially are not charged to any voltage, and so they appear like a short circuit across the transformer/rectifier section. This may produce a very large current during the first AC cycle. That current will not harm the capacitors, but it may let the magic smoke out of the rectifier.

Operating current is what is flowing through the capacitors during normal operation. If this is too large it may cause the capacitors to overheat and fail prematurely. However, operating current is rarely a problem with mains frequency linear supplies. It is more of a problem in high frequency switching supplies. In your case you may consider that with an output current of 2A the smoothing capacitors are not going to handle a current much different from this. If they are good high quality capacitors you have nothing to worry about.

Nope, I am not confusing the continuous operating current and In rush current. The magic smoke (lol) is not an issue to the ratings of the rectifier.

Okay, there is a catch: I am going to be using a switching device--LM2576. It'll regulate the output voltage to a manageable 16V from the higher out voltage of the rectified voltage. So its not exactly a linear power supply.

However, the switching should not be a problem. The datasheet of LM2576 has an application note of using an unregulated power supply to provide a input source and yield a regulated output voltage upto 3A. The switching frequency of the device is 20Khz.

Yeah the caps are good quality. They are made by CORNELL DUBILIER and buying them from Element 14.

[codeboy2k]

Okay, there is a catch: I am going to be using a switching device--LM2576. It'll regulate the output voltage to a manageable 16V from the higher out voltage of the rectified voltage. So its not exactly a linear power supply.

However, the switching should not be a problem. The datasheet of LM2576 has an application note of using an unregulated power supply to provide a input source and yield a regulated output voltage upto 3A. The switching frequency of the device is 20Khz.

Yeah the caps are good quality. They are made by CORNELL DUBILIER and buying them from Element 14.

Ian has done a fantastic job explaining why you don't need to worry about in-rush current with your 30mF setup.

But now that you revealed you are using a post-regulator, then it's a whole different ballgame.  That switcher has excellent line regulation, changing only 1% in output voltage for an input swing of 10-50VDC.  With smaller input ripple, it changes output even less than 1%.  There is absolutely no reason for you to pre-filter  the input voltage as much as you are. You don't need the input to be that smooth, and you certainly don't need to spend money on 30,000 uF , and knowing that you are going to use an LM2576 after the transformer, you don't even need the original calculation of 8300 uF any more either. 4700uF on the input, before the regulator, will be fine. That will give you roughly 20% input ripple, but the regulator can handle that just peachy.  The output capacitance will depend on what kind of load step regulation you are looking for.

 

[chimera_786]

Ian has done a fantastic job explaining why you don't need to worry about in-rush current with your 30mF setup.

But now that you revealed you are using a post-regulator, then it's a whole different ballgame.  That switcher has excellent line regulation, changing only 1% in output voltage for an input swing of 10-50VDC.  With smaller input ripple, it changes output even less than 1%.  There is absolutely no reason for you to pre-filter  the input voltage as much as you are. You don't need the input to be that smooth, and you certainly don't need to spend money on 30,000 uF , and knowing that you are going to use an LM2576 after the transformer, you don't even need the original calculation of 8300 uF any more either. 4700uF on the input, before the regulator, will be fine. That will give you roughly 20% input ripple, but the regulator can handle that just peachy.  The output capacitance will depend on what kind of load step regulation you are looking for.

I Absolutely agree that Ian has been very patient with my questions plus lack of experience.

The only reason why I wanted to add that 30mF capacitor was bcaz i figured I need to feed a relatively 'ripple free' input to the LM2576 to promote good line regulation characteristics. 

In the application note ( I have attached the pdf for this device) on page 20, fig15.. you'll see that the input is unregulated. The term unregulated in my mind means that the voltage is free to swing based on the load; so it was no brainer (again, in my mind) to have the input be clean.

Okay so this what im going to do:

Rectify the voltage, smooth it out with a 8600 uF cap (an over kill), pass the voltage to the LM2576, let it do its thing and have a nice stable output voltage.

But before I do this, please, can u explain to  me one thing:

On page 20, fig 15.. the input cap is 100uF. But I am going to be using a 8300uF. Do you think thats an issue? The reason why I ask that is for instance, when I turn of the power supply and disconnect the load, all the charge stored in the input cap will be dumped into the LM2576. Would it not damage the device?

[alm]

On page 20, fig 15.. the input cap is 100uF. But I am going to be using a 8300uF. Do you think thats an issue? The reason why I ask that is for instance, when I turn of the power supply and disconnect the load, all the charge stored in the input cap will be dumped into the LM2576. Would it not damage the device?

Why would that cap suddenly dump its load into the regulator? Lets consider the regulator as a resistor (which is a major simplification, but it should suffice for this question). You have a voltage source, a 8300 uF cap, a series resistor (the regulator), another cap, and a load resistor. The current through the resistor is determined by the voltage across it. What happens to the voltages across both caps and the current through the resistor if you disconnect the voltage source? Note that the time it takes to discharge a capacitor is proportional to 1/RC. What happens if you disconnect the load? You can simulate this in LTspice if you have problems intuitively analyzing an RC circuit.

Something that can cause issues is if the cap on the output of the regulator is much larger than that at the input, although I don't believe the LM2576 is susceptible to it. What happens to the voltage across the resistor, and the direction of the current, if you disconnect the voltage source with a small input cap and a large output cap?

[chimera_786]

On page 20, fig 15.. the input cap is 100uF. But I am going to be using a 8300uF. Do you think thats an issue? The reason why I ask that is for instance, when I turn of the power supply and disconnect the load, all the charge stored in the input cap will be dumped into the LM2576. Would it not damage the device?

Why would that cap suddenly dump its load into the regulator? Lets consider the regulator as a resistor (which is a major simplification, but it should suffice for this question). You have a voltage source, a 8300 uF cap, a series resistor (the regulator), another cap, and a load resistor. The current through the resistor is determined by the voltage across it. What happens to the voltages across both caps and the current through the resistor if you disconnect the voltage source? Note that the time it takes to discharge a capacitor is proportional to 1/RC. What happens if you disconnect the load? You can simulate this in LTspice if you have problems intuitively analyzing an RC circuit.

Something that can cause issues is if the cap on the output of the regulator is much larger than that at the input, although I don't believe the LM2576 is susceptible to it. What happens to the voltage across the resistor, and the direction of the current, if you disconnect the voltage source with a small input cap and a large output cap?

For your first scenario, when the voltage source is disconnected, the voltage on both the caps starts to drop as there is no voltage source present; what ever voltage the input cap had at the time of the disconnection would be used up by the LM2575 to regulate the output. The output, in the mean time, trying to keep the output voltage stable would also experience a diminishing voltage, since the voltage source (which is now the input cap) no longer has the any energy for the LM2576.


For your second scenario, when the voltage source is disconnected (assuming the load is also disconnected), the enegery stored in the output cap will be dumped back into LM2576. But I am not to sure of this scenario. Can you please explain  what whill happen when a source is disconnected as well as the load, and now the only thing completing the circuit is a small input cap, LM2576 and a big output cap.

btw: I am learning for this discussion .. lets keep it up

[codeboy2k]

For your second scenario, when the voltage source is disconnected (assuming the load is also disconnected), the enegery stored in the output cap will be dumped back into LM2576. But I am not to sure of this scenario. Can you please explain  what whill happen when a source is disconnected as well as the load, and now the only thing completing the circuit is a small input cap, LM2576 and a big output cap.

Think about where the reverse current will flow.  As it is, the way the LM2576 is designed, there is no low resistance DC path from the output of the LM2576 to ground.
So only leakage current will flow back through the device. This will not be very much, and will not harm the device.
The feedback circuit on the output will provide the least resistance to ground, and the output cap will slowly drain through the feedback divider.

Although large capacitors are capable of storing massive amounts of energy, it's not going to suddenly dump all that energy unless there is a low-resistance path to ground for
the reverse current to flow through iit. The LM2576 doesn't have this problem and thus also doesn't need any reversed biased diode from output to input as you'll often see on
the ubiquitous 78xx series linear regulators with large output caps.

[chimera_786]

For your second scenario, when the voltage source is disconnected (assuming the load is also disconnected), the enegery stored in the output cap will be dumped back into LM2576. But I am not to sure of this scenario. Can you please explain  what whill happen when a source is disconnected as well as the load, and now the only thing completing the circuit is a small input cap, LM2576 and a big output cap.

Think about where the reverse current will flow.  As it is, the way the LM2576 is designed, there is no low resistance DC path from the output of the LM2576 to ground.
So only leakage current will flow back through the device. This will not be very much, and will not harm the device.
The feedback circuit on the output will provide the least resistance to ground, and the output cap will slowly drain through the feedback divider.

Although large capacitors are capable of storing massive amounts of energy, it's not going to suddenly dump all that energy unless there is a low-resistance path to ground for
the reverse current to flow through iit. The LM2576 doesn't have this problem and thus also doesn't need any reversed biased diode from output to input as you'll often see on
the ubiquitous 78xx series linear regulators with large output caps.

Thanks codeboy, that was a the basic idea in my head as well. I am glad I was on the right track. Well, I think the discussion has been very helpful and it definitely has given me a few things to consider while I go about designing the power supply.

I actually am going to document my progress and once I am done editing and compiling it in  a neat fashion ( a video/pdf), I'll post the file/link on this forum and then ppl can leave feedback

A sincere thanks to everyone who helped on  this thread.

[alm]

For your first scenario, when the voltage source is disconnected, the voltage on both the caps starts to drop as there is no voltage source present; what ever voltage the input cap had at the time of the disconnection would be used up by the LM2575 to regulate the output. The output, in the mean time, trying to keep the output voltage stable would also experience a diminishing voltage, since the voltage source (which is now the input cap) no longer has the any energy for the LM2576.

Exactly, so the current flowing through the regulator would not reach harmful levels, since the voltage between the input and output is fairly small. The voltage on the output will probably drop faster than the voltage on the input, but nothing worse than the regulator experiences during power-on. The LM2576 (which is not all that much like a resistor) may behave funny if the input voltage drops below a certain threshold, but this is not related to the amount of capacitance.

The LM2576 doesn't have this problem and thus also doesn't need any reversed biased diode from output to input as you'll often see on
the ubiquitous 78xx series linear regulators with large output caps.

This was what I was referring to. Regulators like the 78xx have a reverse-biased diode between the input and output. If the output cap is very large and the input cap is small or the input is shorted, the voltage on the output will be higher than the voltage on the input, resulting in a current through the (now forward-biased) diode. This diode can only handle a limited amount of current, so it's recommended to place a beefier diode in parallel if the output capacitance is large. Not something that's an issue with the LM2576, but something to be aware of when determining output capacitance.

[chimera_786]

well I have simulated the circuit. I will post the schematic and graphs here soon but for now, there is one thing thats alluding me constanly.

What is the 'average DC value' and why is it important to consider it when designing a power supply.

I have read other resources and this is what I come up wiith:

1- The average DC value is the value measured by a volt meter at a given load

2- The formula to calculate is: DC_avg = (2*Vp)/pi, where Vp = Vrms x1.414-------other in other words, DC_avg = 0.636Vp

Now, the above points contradict each other: becaz DC_avg formuale does not take the load into account.

Let me rephrase:

I have a 24V rms from my transfomer. This means that Vp = 24*1.414 = 33.94 volts. Now, once I substract 2 volts becaz I am using a bridge rectification (4 diodes), then I am left with Vp=31.94V. Now, this is at NO load

So, according to the forumala, my DC_AVG value is 20.311 volts @ NO Load.

What is if I draw 2A current to a load. What is  the change in DC_avg now ?

I am really not clear on the DC_avg part 

[alm]

If you refer to the LM2576, for a constant load, average DC current is equal to load current, and just the average load current (eg. Vp/2 for a pulse train with 50% duty cycle). RMS current may be lower or higher, depending on the signal (eg. sinusoidal, low duty-cycle square wave). The same applies to voltages. Not sure where that (2*Vp)/pi formula comes from. Vp = Vrms x sqrt(2) is only valid for sinusoidal signals.

[chimera_786]

If you refer to the LM2576, for a constant load, average DC current is equal to load current, and just the average load current (eg. Vp/2 for a pulse train with 50% duty cycle). RMS current may be lower or higher, depending on the signal (eg. sinusoidal, low duty-cycle square wave). The same applies to voltages. Not sure where that (2*Vp)/pi formula comes from. Vp = Vrms x sqrt(2) is only valid for sinusoidal signals.

No, I havent included the LM2576 as yet. I am just running a simulation. I have attached a schematic and an output graph. The graph contains, Vin (the input from the AC source at 24V rms=33.94 V) , Vout (load voltage) and Output Current.

If you look at the schematic, I have included a reading from a voltmeter--it reads 27.2 volts. Now, if you read the output voltage from the graph, its a saw tooth voltage (as expected) and goes from 28.1 to 27.1( the ripple being 1V peak to peak).

Now, my question is that the 27.2V that I read off the voltmeter-------is that the Average DC voltage? If I was to build it this circuit, will I read that same reading across my DC voltmeter?

So, as you can see, I want to know:

1- How to calculate the DC average voltage
2- Why is the average DC value important

I hope I am explaining my question(s) properly. Thanks for all your help though!!

[alm]

Average DC voltage is a somewhat strange term, since the definition of DC is that it's not changing with time. A time varying signal can be composed of a (constant) DC signal and a periodic AC signal, for example a sawtooth superimposed on a (constant) DC voltage. Since AC signals by definition have an average of zero, the DC offset is equal to the average value of the signal.

In theory a DC voltmeter only measures the DC component, although in real life the rejection of AC signal is limited, especially at lower frequencies, so superimposed AC signals may change the DC reading, especially with cheap meters.

I'm not convinced that average DC value is that important, did you find a document claiming that? The minimum and maximum values may be important, since they have to be in range of your regulator (eg. duty cycle limits, or dropout voltage in the case of linear regulators). A signal with an average of 12 V with 50 V peaks would probably be bad for your regulator. To calculate the total energy, RMS voltage (and current) are important. RMS is just another way of averaging.

[chimera_786]

yeah I found a few documents going a little over the average DC value but found nothing that hints on the importance of this quantity. I think its just value that a DMM reads when connected to varying DC signal. Since the output voltage is in saw tooth form, the meter will just average out around 27.5 volts at steady state.

The reason why I have been pain painstakingly going over every question/confusion is that I am trying to compile a paper of power supply construction. Sure, there are several available online and tonnes of schematic. But every single one lacks some kind of detail or doesn't cover a point.

I figured that if I am successful at compiling a paper, then maybe people later on can use it as a reference. Kind of like a one-stop place for 40W to 60W bench power supply.

Power side of electrical engineering has never been my strong point. I am really comfortable with analog electronics, DC to DC converters, DSPs etc. But I guess this is as far as I am willing to go regarding 'power'

Now I can finally start having fun with LM2576!!

Thanks for all your help. I will post back if I am having some problem with the LM2576.

Best

[IanB]

As alm asked, where did you read that the average DC value is important? If you have a source that says it is important, yet it doesn't tell you why it is important, then perhaps it isn't that important after all? 

When it comes to averages, average voltage is like average anything. You add up all the values over a certain time period and then divide by the number of samples. Or if you have the voltage as a graph you can add up the area under the graph and divide by the time interval.

Mathematically the average looks like this:



This is how you write "the area under the graph divided by the time interval" algebraically. A simple voltmeter will do something like this.

However, sometimes it is useful to consider the power delivered to a resistive load. There is a different average voltage, the RMS average, that would give the same power as the voltage being measured. To calculate it, we can see that the power into a load is given by the following formula:



W is the power, V is the voltage and R is the resistance.

Now if we look at the total power over a period of time, we get this equation:



This is now a different average voltage, one which will produce equal power to the true voltage. If we rearrange this equation to find the average voltage, and call the average voltage the RMS voltage, we get:



This is called the root-mean-square average, for reasons indicated by the equation.