Problem with SPICE calculation of the ripple in half-wave rectifier

[Gio47818732]

TL;DR

spice does show a different ripple voltage than the one calculated using spice I load or Vdc load and Req and C
spice shows a ripple voltage of about 31mV, but calculating the ripple with the formula i get 38mV wich is really weird.

spice does show a different ripple voltage than the one calculated using spice I load or Vdc load and Req and C.
spice shows a ripple voltage of about 31mV, but calculating the ripple with the formula i get 38mV wich is really weird.

as you can see in the photo i get a dc output voltage of around 3.511 wich divided by (Req*C*f) gives me 38mV. Spice tho shows me in output a ripple voltage measured of around 31mV wich makes no sense since formula is ripple voltage=Vdc/(Req*C*f) since capacitance equivalent impedence is supposed to be 0. also if that even counted that should increase the ripple voltage instead of giving me a ripple voltage smaller by 7mV than the calculated one.

thank you for the answers.

 

[Baluncore]

Welcome to PF.

I know it is a simple schematic file.asc, but there is critical embedded detail that does not show in screenshots. Please attach your LTspice file.asc to your next post so I can run the simulation.

You will need to change the extension from file.asc to file.txt, or make it file.asc.txt so it can be attached to your post.

 

[Gio47818732]

here the schematic, sorry i didn't add it before

 

[Baluncore]

You have a high series resistance 50R0 in the power supply. The capacitor charge time, (or phase angle), is therefore longer than expected, so the discharge time is proportionally less. I measure the discharge time, dt, to be 823.5 us, NOT the 1000 us you may assume.
C = Q / v ; The definition of capacitance.
C = i * t / v ;
C = i * dt / dv ;
dv = i * dt / C = ripple voltage.

Your LTspice .meas directive results in:
vripple: PP(v(n002))=0.031065 FROM 0 TO 0.05
vdc_out_load: AVG(v(n002))=3.51168 FROM 0 TO 0.05

C = 47.8 uF
Requ = 1.923 k
Average i discharge, is AVG(Vout) / Requ
dv = i * dt / C= ( 3.51168 / 1k923 ) * 823u5 / 47u8 = 0.03146 V
The 0.4 mV difference is to do with the switch on and off times of the diode, and my ability to put a cursor on a pixel.

You need to revise your approach to the computation of the ripple voltage.

Label your nodes, so the automatic numbering does not change your .MEAS directive node when the circuit changes.

Do not include the units in spice, only the multiplier.
Replace the decimal point with the multiplier on schematics.
20 ms = 20.0 m = 20m0 .
A one farad capacitor, 1F , is only one femtofarad in SPICE.

 

[Gio47818732]

thanks for the headups, they made me rethink my approach to the problem and made me realize things i didn't before when using spice; also going back to my theorical calculations of the ripple it seems they were also off due to me using the approximated to the first order formula for the ripple , the complete formula not including the generator resistance would have been (V-Von)*(1-exp(-(T-T{it takes to charge the condensator})/(R{equivalent}*C))).
where V-Von is the dc current i measure in out, also if we consider the equivalent series resistance of the condensator in case it matters we would need to add R_{of the capacitance}*(V-Von)/(R{equivalent}.
Correct me if i am wrong,i just wrote it in case someone has my same problem and forgets they are using the approximate formula instead of the complete one

 

[Baluncore]

The formula you used, works well for a conservative design, in that under most circumstances, it included a safety margin.

Attached is a cleaned-up schematic, that runs faster and sufficiently accurately. It measures ripple over two cycles at the end of the transient analysis.