Secondary is 36V rms
So it’s about 50.9 peak
Diode drops of 1.4V rectified so about 49.5V into the filter

R surge = 10
C= 100 u
RL = 3.3k

I’m using Vr(pp) =[1/(f x RL x C)] x Vpk

I calculate 1.25V ripple and my LTSpice gives me over 8v. I have a feeling it has to do with the RC filter and Rsurge but not sure what I’m missing in my calculation or is it spice?

 

I’m very new with LTSpice and I believe it’s that. I must be doing something wrong. I wasted hours on this but at least I won’t forget this formula. It simulated fine using another program.

 

There must be something wrong with your simulation (you probably place the ground in the wrong place).
Can you upload your simulation file? Because my simulations show Vr(pp) around 1V (see my file).

 

Awesome, I will check again when I get home. I’m glad that you’ve confirmed my calculations. As I said I have the correct results from a different simulator. I am completely new to LTSpice and I’m sure I did something wrong.

 

Hi,

Just to note, the resistor you call Rsurge would actually make the output smoother because it lowers the current a little bit. It may not be noticeable with a load of 3.3k though. It also drops the DC average a little bit, and again might not be noticeable unless you increase it to maybe 100 Ohms.

Like many things in nature the complete solution to a rectifier circuit comes about with a partial differential equation because there are two independent variables to be solved for. However it isnt something most people want to get involved in so i wont post it here again. It's nice to know that somewhere, someplace, there is an exact solution to any rectifier circuit full wave or half wave or just about anything else.

 

The Vp dc formula does not take into account the RC time constant of the charging through Rsurge. I've noticed the waveform is smoother and peak and ripple is smaller. I have been puzzling over how to calculate for this, it won't be on the exam so it's for personal understanding. I believe I have to calculate for peak DC again but start at the bottom of the ripple to account for it... but again, not important in this case. It makes the saw tooth looking wave look more like a filtered DC and makes a difference in the scope.

 

Without using spice an approximate value can be found just by considering the filter capacitor and the load.

The change in voltage across a capacitor is dv = (i * t) / c

i is 50v/3.3k -> about 15 mA (15e-3), t is just the time the power line is off between peaks. so 1/120. -> 8.3e-3 seconds

so dv = (15e-3 * 8.3e-3) / 100e-6 -> 1.24 V drop from the peak voltage...

 

That's another way of looking at it... thanks! It's the same as Vr(pp) =[1/(f x RL x C)] x Vpk

it still does not account for the filter charging through R surge.

 

the rectifier introduces all half-cycles as the positive 1-s to your (GND) (~V) (L.ii)(R.L.ii) (ΔV) (R.su)((C.F)||(R.L))(GND)
-- the C.filter does not follow the Sine averaging product " X.C = 1/(2πfC)" you need to define 1 for ... (and the diodes' clipping (while charge gathers to C.F) makes it even more "interesting") ...
(goes to that dif. eq. thing) somewhat fuzzy examples for sine → http://cns-web.bu.edu/~eric/EC500/attachments/ON(2d)LINE(20)READINGS/LabManual_Chpt3.pdf
perhaps : https://www.google.com/search?tbs=ic:gray&tbm=isch&sa=1&q=formulas+for+rectified+clipped+sine
returns also ... (Fig.5) and (a correction to your above formula)

 

Hi,

The LTS solution is in the time domain and it takes into account the diode model.
More importantly, the diode does not conduct for the entire half cycle and that is what makes it hard to determine what the ripple with be without actually solving for the diode "on" time. The duty cycle is small compared to the half cycle sometimes so it could be just 1/8 of the half cycle for example or 1/4 of the half cycle or 1/2 of the half cycle. There may be some min Rsurge resistance that keeps the diode duty cycle near some limit though so we could look into that.

The big difference between this kind of circuit and many other circuits is that the diode does not conduct all the time so the output of the bridge sometimes goes to high impedance and that makes it harder to solve without allowing two independent variables instead of just the usual just one.

What you could try is to simulate a bunch of rectifier circuits and see if you can find a simpler relationship.
If you use an input inductor it helps keep the diode conducting for longer and is easier to solve but yes that is not always an option.

One advantage to using Rsurge though is that the resistance will swamp the series resistance of the diode and thus help make the specific diode model needed less important. If i get a chance today i'll look into this.

 

I spoke with the instructor last night and he mentioned that this is ignored. The impedance of the transformer has a significant effect. I stated that the RC charge time is visible and he agreed. hmm it's bugging me... not sure how important it is.

Rsurge has the effect of reducing the peak voltage as the filter capacitor is charging. The load resistor has more of the current reduction through the diodes. In any case, very interesting to think about.

 

Here is a workup of a pulse charging your network:

v0 = average voltage = (3300*D)/(3300*D+10)*E
voltage at peak = (1-e^(-(TH/2)/RPC))*(E-v0)+v0
voltage at valley = v0*e^(-(TL/2)/RLC)
where
E is the peak voltage out of the rectifier
D is duty cycle: TH/(TH+TL)
TH is time high
TL is time low
RPC is both R's in parallel times C
RLC is load R times C

so peak to peak ripple would be the peak voltage minus the valley voltage.

Just keep in mind this is for a rectangular pulse. You'd have to find a correlation between the diode conduction time and the pulse width duty cycle. I did not get that far yet.

 

the TF follows the magnetic circuit sat/de-sat hysteresis curve about Google . . . ( returns 1 2 3 4 5 . . . )

"charging the cap " : {∫ "clipped Sin from t0 t1 " ({V.pk·Sin(2·π·f·t) - (a·t+b)"simplified charge to cap through R.srg"}/R.srg - (a·t+b)/R.ld)·dt} / C.flt = ΔV =
"dis-charging the cap " : {∫ (A·t+B - Const.)/R.ld·dt} / C.flt -- where ΔV is added to Const. corresponds to t0 ,
Const. = a·t0 + b ,
B = a·t1 + b
A = {(Const. + ΔV) - B} / Δt ,
Δt ≈ (1/f)/2 - (t1-t0) ← i don't know if it defines (by that yet ...) . . . and how exact it would be * -- note : as the post quick Err.scan revealed as it's a "full wave rect." ←↑

basically you need to predict (trial out) Const. , t.0 & t.1 (a & b)

. . . since Sin(2·π·f·t) is not cool to Go! in a sense of integrating you can norm the time to t¹ = 2·π·f·t thus a·t = a·t¹/(2πf) etc.

 

Yeah you know what? We solved this on another site i forgot about that.
I'll look up the details.

The strategy is to find the point where the RC discharge leaves the sine it follows while the diode conducts.
Only thing missing then is the diode model, which will show a larger voltage drop than normal due to the peak current. Then find the point where it meets the sine again.
So it involves a sine and an exponential.

 

Yes that’s my approach exactly! Charging starts where Vdc peak - Vr pp. But it only affect the charge time of the filter capacitor so assuming that the transformer is able to supply the load and the charge, solve for tRC for the charge curve based on R surge and C.

 

Without using spice an approximate value can be found just by considering the filter capacitor and the load.

The change in voltage across a capacitor is dv = (i * t) / c

i is 50v/3.3k -> about 15 mA (15e-3), t is just the time the power line is off between peaks. so 1/120. -> 8.3e-3 seconds

so dv = (15e-3 * 8.3e-3) / 100e-6 -> 1.24 V drop from the peak voltage...

This method, based on the simple relation I = C ⋅ dV/dT, ignores a lot of factors needed to arrive at an exact value for the ripple voltage. But this is one situation where exactness is rarely needed; most often, what you're after is simply estimating how big a filter capacitor you'll need to avoid exceeding "xxx" volts ripple while supplying a load with "yyy" amount of current.

And for that, this simple method is almost always adequate.

 

Hello again,

Yes the solution for the first equality goes like this:

Vpk*cos(t*w)=Vc(t)+Rs*i(t)

and unfortunately Vc(t) comes out quite complicated, although still manageable:
(
((v0*w^2*C1^3*R1^3+v0*C1*R1)*R2^3+(2*v0-Vpk)*C1*R1^2*R2^2+(v0-Vpk)*C1*R1^3*R2)*
%e^(-(t*(R2+R1))/(C1*R1*R2)))/(C1*R1*R2*((w^2*C1^2*R1^2+1)*R2^2+2*R1*R2+R1^2))+
(cos(t*w)*(Vpk*R1*R2+Vpk*R1^2))/((w^2*C1^2*R1^2+1)*R2^2+2*R1*R2+R1^2)+(Vpk*w*sin(t*w)*C1*R1^2*R2)/((w^2*C1^2*R1^2+1)*R2^2+2*R1*R2+R1^2)

That looks too big and it can be reduced somewhat, but that makes the solution possible.
It ultimately ends up being an equation which is solved for 't' numerically, but we just use a software numerical solver to get the value.

I would like to find the solution we found several years ago first though. That's complete.
I think @The Electrician took part in that endeavor.
.

 

So that’s why it’s ignored. It’s a large equation for such a small effect.

 

So that’s why it’s ignored. It’s a large equation for such a small effect.

And that's just the first equality

Once it is all solved though it gets a bit simpler, but you do have to do a numerical solution.
If it was just the cap and that one load resistor it's 1000 times simpler.

 

Or my favorite go to, when it’s multiple iterations like that excel or simulate it.