Vdc= 1.654Vm where Vm is the peak phase voltage
Idc= 1.654Vm/R where R is the load resistance, 26ohms
Hello, my best guess is that you only partly understand what you are currently doing. This applies to me too.
So I'd suggest to first use a scope with a current clamp to monitor one of the AC side currents. (The others will look the same - just phase shifted.)
Then you might be able to refine the question above.
Dear Muhammed,
I would like that you see the reference in the Link: www.springer.com/cda/content/.../cda.../9780387293103-c1.pdf?...0...
you will find complete analysis of the three phase 6 diodes rectifier.
The dc output power of your rectifier= Vdc Idc= 260x 10 =2600 w,
The input ac power Pac= Pdc assuming the diodes are ideal, which is more or less valid for sake of estimating the input parameters of the rectifier.
Pac= 3 Vp Ip cos phi, where vp is the rms phase voltage and Ip is the rms phase current and cos phi is the power factor.
Vp=Vm/1.41,= Vdc/ 1.654x1.41= 111.48 V,
Given that cos phi= 0.958 according to the given reference above, table 2-2, we can calculate the the phase current Ip
Ip= 2600/ (3x111.48 x 0.98)= 8.11 A,
If this power is supplied from the utility power supply with A phase voltage of 220V,
Then one needs a step down transformer from 220 to 110 V. Thus the primary current will be half its secondary current. That is Ip primary= 4.055 A.
Wish you success.
Dear Abdelhalim Zekry ,
Thank you so much for ur detailed answer. But I am getting different result from my simulation. So when i googled i got a paper. Ive attached that file. in which it has given a formula for RMS Secondary current which gives 20A! Please see page 41 of the attached file and please explain me whats the difference between ur method and above method. And the saddest part is im neither getting 20A in my simulation! Waiting for ur reply
Dear Mohammed, tell me please what is value you obtained from the simulation.
I made the calculation according to the formulas given in the attached reference and i got the same Vm and the same Vp. For the current , i calculated the DC current in every in every diode Id to be the DC load current divided by 3= 3.333,
Then i used the relation between the dc diode current and the maximum value so that Im= Id/.318= 10.47 A,which gives 7.41 which is slightly deviated from the above answer. So, the estimations are not deviated from each other as you report.
wish you success
The basic problem: if you want to get a "real" DC voltage/current (with low ripple) you need an output capacitor. The solution greatly depends on the value of this capacitor as the diodes are only conducting for a short time (when line voltage is higher than the output voltage). This leads to a peak diode current much higher than the average current - resulting in the diode voltage drop no longer being negligible.
I am involved with a similar situation: I am using a 3-phase diode rectifier, the rectified voltage being then inverted to feed a 3-phase induction motor.
Harmonic analysis of the input current of the rectifier shows a predominant DC component in the current of each phase (the sum of these 3 DC components is seen in the DC component of the neutral conductor current).
How can these DC components be justified? Can it be due to motor saturation?
Hi ,
Can anyone help me understand why my DC-link voltage is as shown in the attached picture for 3-phase diode rectifier connected to 1-phase inverter.
PLECS file attached.
Your answers would be very helpful.
Kind regards,
Silpa
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