Hero999 said:Here's the plans for a power supply I'm intending to build. By my claculations it should provide me with +-15.7V@1.5A. The mains transformer I'm using is a 15V-0-15V torroid rated to 2.67A which is slightly over-rated which is perfect as the regulators might not current limit untill 2.2A.
Anyway, I just thought I'd ask you lot if you can think of any improvements before I etch the PCB and build this.
230V.cadstarsucks said:Briefly if you are on 115V:
220V to 250V.the mains normal operating point is 100-130V
I don't know what the regulation is but it's certainly better than that, it's an 80VA toroidal transformer and these give pretty good regulation, about twice as good as an E-core transformer so let's say 7.5%. Anyway isn't the voltage of a transformer specified at the full load?the transformer is probably 15%
About 311V to 354V but what difference it makes is beyond me, I normally use the output voltage to calculate ripple.the mains peak voltage is RMSx1.414 or 141-184V
It's rated for 2.67A, but I disagree, I want to be able to draw at least 1.5A so the transformer needs to be rated for at least [latex]1.5 \times sqrt{2} = 2.21A[/latex].The safe transformer current is rating/1.8 or 1.48A
You're mistaken, providing it's not too hot or not too cold the LM217 has a dropout voltage of 2.25V@1.5A so minimum DC voltage needs to be 17.25V,therefore a 15V transformer will do, let's calculate the ripple assuming a 4700µF capacitor.The regulators need 3V to operate - so you need 18 volts at the valley of the filtered waveform, or preferably an 18V transformer
They only conduct for half the cycle so they only need to be rated for 1.11A, either way I've used a 3A 400V bridge from an old PC power supply so it doesn't matter.The diodes need to be rated for 2A or better
Why?and the caps should have a 4.5A RMS rating...
That is what is supposed to be done. For a complete discussion see the end of this catalog and cut corners as you like:
**broken link removed**
I prefer switchers too, but they can be too noisy and troublesome to build.I prefer offline switchers myself but most people do not have the resources that I do.
In your DC voltage and ripple calculations, you need to keep in mind that a diode bridge drops about 1.5V.The minimum voltage in-between the ripple is 21.2 - 3.2 - 0.7 = 17.3V
I ran the simulation, and got 3.06V p-p ripple. Your equation fails to account for the fact that the cap is only discharging for about 6.4ms. For the remaining ~2ms, it's charging.A 2000µF capacitor will give 4.167V of ripple @60Hz when given a 1A constant current load. The formula used in my calculation above if pretty accurate for constant current loads, try simulating it in a SPICE simulator.
That's true when measured from +15V to -15V but when measured from +-15V to 0V it's half that value, 0.75V.Ron H said:In your DC voltage and ripple calculations, you need to keep in mind that a diode bridge drops about 1.5V.
You sure you used a constant current load?I ran the simulation, and got 3.06V p-p ripple. Your equation fails to account for the fact that the cap is only discharging for about 6.4ms. For the remaining ~2ms, it's charging.
I'm saying that your peak DC voltage will be about 19.7V. There are always 2 diodes in series with the transformer between GND and the rectified output.Hero999 said:That's true when measured from +15V to -15V but when measured from +-15V to 0V it's half that value, 0.75V.
Yep, I used a constant current load. Furthermore, a with resistive load and 3V ripple, the current only changes about 15% for a 20V supply, so the load could be approximated by a constant current without too much error.You sure you used a constant current load?
I'll have another go I suppose but if what you're saying is correct then I haven't undersized my filter capacitors so I haven't cut any after all.
I still disagree.Ron H said:I'm saying that your peak DC voltage will be about 19.7V. There are always 2 diodes in series with the transformer between GND and the rectified output.
Which is how a linear regulator should behave, providing its load is resistive the current will be the same regardless of the input voltage because the load current it always the same.Yep, I used a constant current load.
I agree, I've just simulated it and you're right.Furthermore, a with resistive load and 3V ripple, the current only changes about 15% for a 20V supply, so the load could be approximated by a constant current without too much error.
As long as your valley voltage stays above the drop out sure...Hero999 said:It's rated for 2.67A, but I disagree, I want to be able to draw at least 1.5A so the transformer needs to be rated for at least [latex]1.5 \times sqrt{2} = 2.21A[/latex].
You're mistaken, providing it's not too hot or not too cold the LM217 has a dropout voltage of 2.25V@1.5A so minimum DC voltage needs to be 17.25V,therefore a 15V transformer will do, let's calculate the ripple assuming a 4700µF capacitor.
[latex]V_{RIPPLE} = \frac{I_{LOAD}}{2FC}= \frac{1.5A}{2 \times 50 \times 4700 \times 10^{-6}} = 3.2V[/latex]
The peak voltage is [latex]15 \times \sqrt{2}= 21.2V[/latex]
The minimum voltage in-between the ripple is [latex]21.2 - 3.2 - 0.7 = 17.3V[/latex]
I agree on the diode rating, but that as you pointed out about some of the transformer mfgs equations, is based on a resistive load. Given that your transformer only represents 1They only conduct for half the cycle so they only need to be rated for 1.11A, either way I've used a 3A 400V bridge from an old PC power supply so it doesn't matter.
Why?
Sure, the peak current might be 4.5A but the RMS current is probably nearer to I/√2.
I dissagree with some of the formulae and figures.
For a start the formula for the capacitor is incorrect if you use a linear regulator, it says "2000µF/amp for 3V p-p ripple", this assumes a resistive load rather than a constant current load. A 2000µF capacitor will give 4.167V of ripple @60Hz when given a 1A constant current load. The formula used in my calculation above if pretty accurate for constant current loads, try simulating it in a SPICE simulator.
The claim that rectifiers can't handle surges and need to be rated to the full output current is rubbish, a 1N4001 rectifier can withstand a non repetitive peak surge of 30A and the WOO5 can take 50A for a 8.3ms half sine wave (this is just a ballpark figure I didn't use 1N4001s or a WOO5 in my design). I might see the point if this was a huge power supply but not in my little power supply where the internal impedance of the transformer will limit the surge current to a safe level.
The only corner I have cut is undersizing the filter capacitors slightly so I might see some ripple on the output if the mains voltage is a little too low, the regulator is too hot/cold, the capacitors are in the lowest tolerance band, or the voltage is whacked up to 15.7V (15V is the design maximum but due to component tolerances it can go 0.7V higher). I know I should've used 6800µF capacitors but I didn't have any handy, for a start as the transformer is slightly over-sized the output voltage will be higher than expected at full load, 1A is good enough for most applications and I'm not doing this professionally, besides there's nothing stopping me piggy backing a couple of 2200µF capacitors on there.
I prefer switchers too, but they can be too noisy and troublesome to build.
A fast op-amp would not help things as far as transient response is concerned, the LM337 takes care of that.Optikon said:FWIW, transient load recovery can be improved by using something better than that aweful 741 opamp and you can do away with the bias current error correction resistor (5k1) as well by using a FET input type.
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