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 .
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.
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.
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
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?
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!!
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.
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
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.
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.
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.
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?
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 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.
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.
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.
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.