In-Rush Current for power supply

[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.