Attached is a graph of the diode current with the first cycle being 5Amp approx. and the rest of 3 Amps peak. What diode rating would you choose, if the graph showed the max load current? Is there a rule of thumb?
It looks like a 1 amp diode would be adequate. However, if I was doing this I would determine the worst case power (worst case diode parameter) using SPICE, and make sure the diode junction temperature was not more than 100 degrees C at the highest ambient temperature.
I am working out an exercise where I am giving the schematic attached and I have to proof that the components fail on the long run. I managed to derive some values but I am not sure on what to argue the most. As you can see, 1 amp diodes are used 1N4001. The input from the secondary windings are 12V rms. The load attaches is the max load that will be used 24Ohms.
I used these formulas to work out the conduction angle, conduction period and peak current:
∆θ= √((2Vr)/Vpk )
∆T= ∆θ/2*pi*f
Ic=(CVr)/∆T
Io=Ic+Iload
Vr = ripple voltage
Are these formulas familiar ? If yes are they derived from some rules of thumb?
I found the Io (peak current to be 2.39A)
After I need to work out a safe current for which these components are not stressed.
I used these formulas to work out the conduction angle, conduction period and peak current:
∆θ= √((2Vr)/Vpk )
∆T= ∆θ/2*pi*f
Ic=(CVr)/∆T
Io=Ic+Iload
Vr = ripple voltage
Are these formulas familiar ? If yes are they derived from some rules of thumb?
I found the Io (peak current to be 2.39A)
After I need to work out a safe current for which these components are not stressed.
The equations you listed look somewhat familiar, especially the last 2. The first 2 are derived from the fact that the diodes conduct when the output from the diodes is higher than the capacitor voltage.
There is no (absolutely) obvious reason why any of the components should fail prematurely. However, one thing that could be a problem is ripple current in the filter caps. All caps have an ESR (equivlaent series resistance) and some electrolytics are pretty poor. The power loss in the cap is Irms^2*ESR, and this causes heating, and failure if the cap gets too hot. Find out what the ESR and ripple current rating of the cap is. Then add them to your simulation and find out what they are in this circuit.
There is no (absolutely) obvious reason why any of the components should fail prematurely. However, one thing that could be a problem is ripple current in the filter caps. All caps have an ESR (equivlaent series resistance) and some electrolytics are pretty poor. The power loss in the cap is Irms^2*ESR, and this causes heating, and failure if the cap gets too hot. Find out what the ESR and ripple current rating of the cap is. Then add them to your simulation and find out what they are in this circuit.
Lets assume the typical 470uF caps have an esr of 0.25Ohms.Therefore Parallel esr = (0.25*0.25)/(0.25+0.25)=0.125Ohms Irsm is 1 Amps approx so 1^2*0.125=0.125W. Is this a lot for the caps?
No, .125 watts would not be much for a 35 volt 470 uF cap, but .25 ohms is pretty small for this cap. Take a look at vendor data sheets on caps with these specs; I don't think you will find an ESR that low.
Giving a 1A diode more than 1A average is not a proof it will fail. The 1A is only a minimum guaranteed current spec, not a "we guarantee it will pop at 1.nnn A" (random value above 1.0A), spec. Certainly it's a bad design pushing any part to it's limits but you can't prove failure like this unless you have absolute maximum /failure specifications provided from the manufacturer too.
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@sawreyrw you are right, I got that value from a general table I found online just for the an example. I will get a more practical value.
@! Given a question like this;
A design of a mains plug in power supplies with a fixed output voltage of 12V and a rated current of 500mA. A student decided to dismantle the power supply and traced the schematic shown in Figure 1. Using calculations prove that under full load conditions the power supply will not perform as advertised and will actually cause components to fail over a long term. Output from secondary is 12V rms.
Hence calculate the load current that will not cause undue stress to the components.
Do you think this question is a little bit misleading?
@! Given a question like this;
A design of a mains plug in power supplies with a fixed output voltage of 12V and a rated current of 500mA. A student decided to dismantle the power supply and traced the schematic shown in Figure 1. Using calculations prove that under full load conditions the power supply will not perform as advertised and will actually cause components to fail over a long term. Output from secondary is 12V rms.
Hence calculate the load current that will not cause undue stress to the components.
Do you think this question is a little bit misleading?
Yes, completely misleading unless there is a sufficient sample size of the "components" to statistically analyze the probability of failure and that data is provided at which point a failure threshold would need be provided. If it were simply 50% or higher chance of failure then the "long term" period length would need be provided to compare against a failure rate curve to see where it met the 50% period length.
The fairer question would be one that avoids the word failure, OR that defines a failure threshold as beyond a specific excessive ripple value but even then merely exceeding a diode's rating won't tell you the effect until (similar to above) you have sufficient sample size and data collection to determine whether different specimens of the same (diode) component can vary the % chance vs failure rate. Thus the question is best asked given a component sufficiently undersized for the job that it can be assumed to fail and it is stated that it will fail and that this failure will damage the load so that aspect doesn't have to be proven.
However, there is a different real-world side to the question which is what type of power supply and what the load is. 12V/500mA power supplies are in some sense a commodity item to the extent that many products are designed to run from one when they may not need very close to 12.0V or 500mA (may use load-side regulation to arrive at lower voltage levels needed by ICs), and today's consumer electronics are more and more often using a switching power supply which should have a design that shuts down the output if it cannot stay within a built-in voltage range limit (vs sample rate).
Given a 500mA power supply, odds are fair the components it is comprised of are picked from those commonly available and cost effective based on an arbitrary human standard of rounding off to whole digits. IE - 1A diode, 2A diode, etc. rather than 0.91A diode or 1.37A diode.
Certainly I am reading too much into the "question", but still feel it is invalid as worded.
The fairer question would be one that avoids the word failure, OR that defines a failure threshold as beyond a specific excessive ripple value but even then merely exceeding a diode's rating won't tell you the effect until (similar to above) you have sufficient sample size and data collection to determine whether different specimens of the same (diode) component can vary the % chance vs failure rate. Thus the question is best asked given a component sufficiently undersized for the job that it can be assumed to fail and it is stated that it will fail and that this failure will damage the load so that aspect doesn't have to be proven.
However, there is a different real-world side to the question which is what type of power supply and what the load is. 12V/500mA power supplies are in some sense a commodity item to the extent that many products are designed to run from one when they may not need very close to 12.0V or 500mA (may use load-side regulation to arrive at lower voltage levels needed by ICs), and today's consumer electronics are more and more often using a switching power supply which should have a design that shuts down the output if it cannot stay within a built-in voltage range limit (vs sample rate).
Given a 500mA power supply, odds are fair the components it is comprised of are picked from those commonly available and cost effective based on an arbitrary human standard of rounding off to whole digits. IE - 1A diode, 2A diode, etc. rather than 0.91A diode or 1.37A diode.
Certainly I am reading too much into the "question", but still feel it is invalid as worded.
Did you quote Ripple Voltage as 5V from the normally quoted or did you work it by design blacknight with a frequency of 50Hz😕?
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