LTspice is a simulator, not a circuit designer. It's up to you to calculate the capacitor value. Your simulation shows about 800 mV peak-to-peak with 1800 µF, so to get 300 mV peak-to-peak, you need 8/3 times 1800 = 4800 µF. A 1N4148 is not a very good choice; because the inrush current at switch-on would probably destroy it; a 1N4001 would be better,

Hello,

 

In a diode rectifier circuit, the rectified signal is then filtered by a filter capacitor connected in parallel. How to calculate the minimum value of the filter capacitor by LTSpice?
See my rectification-filter.asc file.

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Best Wishes
John Woodgate
Keep trying

LTspice does not "calculate" the value of the capacitor. It simply gives you the operating values with the parameters you have entered.

It's up to you to modify whatever value (capacitance, load resistance, voltage...) until the results reache a desired value.
As John wrote, 1N4148 would probably explode in real life.
LTspice has undestructible components. You can apply any voltage or current, it will not explode. Again it's up to you to check that the simulated values are within the capacity of the "real" components.

LTspice offers the possibility to test in a single simulation several values. Check the .step command.

On 12/16/24 4:17 AM, jacfev via groups.io wrote:

Hello,
In a diode rectifier circuit, the rectified signal is then filtered by a filter capacitor connected in parallel. How to calculate the minimum value of the filter capacitor by LTSpice?
See my rectification-filter.asc file.

Sounds like you need a primer on power supply design more than how to use simulation:

https://www.worldradiohistory.com/Archive-Poptronics/70s/1975/Poptronics-1975-06.pdf

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David Schultz

In a diode rectifier circuit, the rectified signal is then filtered by a filter capacitor connected in parallel. How to calculate the minimum value of the filter capacitor by LTSpice?
See my rectification-filter.asc file.

It is easy to calculate the (approximate) required capacitor value based on load current and input frequency:

C ≅ Iload/(2*F*Vpp)

where:

Iload is  DC load current
F is the sinusoidal input frequency
Vpp is the target peak to peak ripple voltage

The formula isn't exact as it assumes Vpp/Vrect → ∞, i.e. the ripple is linear sawtooth rather than exponential, but it's close enough given the tolerance of capacitors.

You can parametrise this to put the value straight into schematic components, e.g.:

C1 LOAD 0 {Cval} Rser=0
.
.
.param Freq 50
.param Vpp 100
.param Iload 1
.param Vrip 100m
.param Cval Iload/(2*Freq*Vrip)

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Regards,
Tony

Yes Tony, but I don't know how to integrate these formulas into my diagram.

And what is the LOAD command, I can't find any help in LTSpice.

-> rectification-filter_m3.asc

I  uploaded a fixed version of your file to show you how it's done: rectification-filter_m3_ATC.

I hadn't realised that your circuit was not a bridge rectifier, so the formula has to be reworked a bit to calculate the approximate capacitor value required.

I also added two .MEAS directives, so you can inspect the measured ripple voltage and the calculated capacitance. Just press Ctrl-L to view the Error Log.

Unfortunately, the "Load" parameter of the current source element isn't mentioned in the Help. It's to avoid any unexpected behaviour when V(load)=0, in some circumstances.

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Regards,
Tony

On 16/12/2024 15:06, jacfev via groups.io wrote:

Yes Tony, but I don't know how to integrate these formulas into my diagram.

And what is the LOAD command, I can't find any help in LTSpice.

-> rectification-filter_m3.asc

I  uploaded a fixed version of your file to show you how it's done: rectification-filter_m3_ATC.

I hadn't realised that your circuit was not a bridge rectifier, so the formula has to be reworked a bit to calculate the approximate capacitor value required.

I also added two .MEAS directives, so you can inspect the measured ripple voltage and the calculated capacitance. Just press Ctrl-L to view the Error Log.

Unfortunately, the "Load" parameter of the current source element isn't mentioned in the Help. It's to avoid any unexpected behaviour when V(load)=0, in some circumstances.

Also, as you used real diodes instead of ideal ones, the formula is less accurate, because the diode voltage isn't negligible compared to the peak input voltage. You can compensate, if necessary, for this in the formula, but it's hardly worthwhile given that the capacitors usually have quite a wide tolerance.

It's definitely still useful as a first step in the design process.

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