So, I'm designing a power supply with a 24v 1A transformer going through a bridge rectifier and now I'm trying to figure out what the value should be for the smoothing capacitor. I've done a little research and found the formula for the cap.
C = I*t/?V
Now, this is easy to figure out but I need to figure out the value for delta V. So, with the lm317, I'm trying to figure out what kind of ripple the lm317 can tolerate without the output voltage being affected.
Should be in the datasheets. Attached is a snippet from two datasheets (TI and OnSemi). Determine acceptable output ripple, divide by minimum ripple rejection, and you have the maximum ripple you should have at the input of the regulator.
Should be in the datasheets. Attached is a snippet from two datasheets (TI and OnSemi). Determine acceptable output ripple, divide by minimum ripple rejection, and you have the maximum ripple you should have at the input of the regulator.
Yes but this is in dB. How do I convert this to voltage?
Should be in the datasheets. Attached is a snippet from two datasheets (TI and OnSemi). Determine acceptable output ripple, divide by minimum ripple rejection, and you have the maximum ripple you should have at the input of the regulator.
Yes but this is in dB. How do I convert this to voltage?
maybe like so:
http://www.ti.com/lit/an/slyt202/slyt202.pdfripple_out = ripple_in / 10^(PSRR(dB)/20)
so a PSRR of 60 dB means three orders of magnitude reduction from ripple_in to ripple_out (e.g. 1 mV to 1 uV).
The amount of ripple is very hard to calculate, and usually dictated by the load being driven.
Charging a battery, nobody cares about ripple but power a microphone preamplifier and 50mV of ripple can be audible.
You can instead simulate using LTSpice but your transformer can have 30% regulation and is a great unknown. In other works, its voltage moves around and there is sine-wave distortion, so the math does poorly.
My rule of thumb 1,000-2,000uF per A for filter capacitors. Bigger is better, if it fits.
10uF capacitor on the ADJ pin increases LM317's ripple rejection, per the datasheet.
A 35V filter cap might be too low, as I said transformer will put out more than 24VAC under light loads. Consider a 50V part and the associated size.
Forget the ripple rejection (in dB) for this calculation. It is good enough on the LM317 to avoid worrying. Filtering the output can improve it even more.
You should look at minimum input-output differential instead. As the desired output voltage is a secret, it's anybody's guess.
Also, another question about design, I was planning on using a 1N4007 (thats what I have in my parts box) for the bridge rectifier, but that only goes to 1A, and since I want my power supply to have a max of 1A, It seems that I'm cutting it a bit close. Should I grab something like a 1N5401 just to be safe?
Take notice that ripple rejection is specified at a frequency. This number tells you that if you put a small 120Hz sine wave on the output, then look at how much of it made it to the output. The dB are the ratio between the two as others have explained, its also heavily application dependent.
Those 50dB are decent for 120Hz, but with these old linear regulators only work at these low speeds. Once you get above a few KHz they go "Ripple rejection? Whats that?" and pass all of that ripple straight trough itself.
So some mains harmonics might have a easier time getting trough, any high frequency noise makes it trough and you also have to remember that regulators also have internal noise that they create due to there operation.
So if you are worried about how clean your output is then a LM317 might not be the place to look. They are good enough for most applications, but if low noise is a important factor (Such as powering sensitive high gain audio circuits or sensitive RF stuff) then there are more modern higher spec regulator chips out there.
... I was planning on using a 1N4007 (thats what I have in my parts box) for the bridge rectifier, but that only goes to 1A, and since I want my power supply to have a max of 1A, It seems that I'm cutting it a bit close.
Remember that each of the diodes in the bridge will only be conducting for half of the time. So their average current will be 0.5A (at the supply's 1A maximum output).
If the transformer is rated for 24 V / 1 A AC, then drawing 1 A DC would be overloading the transformer. Something like 0.6 A DC might be a reasonable maximum with full bridge rectification (four diodes).