LTspice - problem simulating simple rectifier

[Pinkevin]

This is a simple question, but at this hour I can't figure out what I'm doing wrong.

I'm using LTspice to simulate a simple rectifier circuit using a sine wave generator (24 V, 50 Hz), a full wave diode bridge, a reservoir capacitor (2200 µF) and a load resistor (1k).
I was expecting to see approximately 24 * sqrt(2) - 2 * 0.65 (diode drop) = 32.5 V, across the load resistor (plus some ripple).
Instead, I'm seeing 23.3 V.

What am I doing/thinking wrong?
Screenshot attached.

Regards,
Alexander

[jcx]

24 V is the amplitude of the sine source - ie the peak values are +/-24 V

you have to multiply your argument in the sine source parameter list by sqrt(2) if you want "rms"

[gootee]

Yoiu should enter the PEAK value for the sine's AMPLITUDE field, in the AC voltage source, not the RMS value.

[janneman]

Originally Posted by gootee

Yoiu should enter the PEAK value for the sine's AMPLITUDE field, in the AC voltage source, not the RMS value.

Yes. The entered value shows 24V which is the peak value. For 24VRMS he should enter about 34V peak (or 24V RMS; not sure you can spec RMS in LTspice).

jd

[Pinkevin]

Well, of course. Thank you!

Ever since I started playing/working with electronics (which isn't many years, but still), I've had the intuitive view of the RMS voltage of an AC signal, as a rectangle which occupies the same area under the graph as a sine wave does over the same range.

Mathematically speaking; integrate abs(sin(x)) (a rectified sine wave) over the range of 0 to a (a = 2*pi*k), and divide the sum by a to get the height/amplitude of the rectangle that you've created.

Example:
Integrate abs(sin(x)) over 0 to 2*pi = 4.
Divide 4 by 2*pi = 2/pi ≈ 0.637.

0.637 V would then be the RMS voltage of an AC signal with a peak voltage of 1 V.

I did some calculations and found this terribly wrong.
The real RMS voltage of an AC signal with a Vpk of 1 V is 1/sqrt(2) ≈ 0.707 V.

What am I missing?

[AndrewT]

Why is LTspice reporting one diode Vdrop?
24Vpk and 23.3Vdc equates to a 700mV Vdrop?

[runebivrin]

Originally Posted by AndrewT

Why is LTspice reporting one diode Vdrop?
24Vpk and 23.3Vdc equates to a 700mV Vdrop?

Probably because those are Schottky diodes, and because the current is quite low. Forward voltage drop is a function of the forward current.

[AndrewT]

so that means the simulator is reporting 350mVdrop for each diode passing 23.3mA.
Correct?

[runebivrin]

I'm sure Spice aficionados can chime in here, but I'd prefer to say that "the model used for the diodes result in a 350mV drop at 23.3mA". LTSpice isn't really reporting the voltage drop across the diodes. It uses whatever the model tells it to.

[AndrewT]

Originally Posted by runebivrin

I'm sure Spice aficionados can chime in here, but I'd prefer to say that "the model used for the diodes result in a 350mV drop at 23.3mA". LTSpice isn't really reporting the voltage drop across the diodes. It uses whatever the model tells it to.

yes, we agree.