Dear All,
I have a query about the inductor design. The simplified application schematic is attached here with. I explain you the application and my query as concise as possible here.
Application and aim:
The application is three phase diode bride rectifier which is loaded with an SMPS operating at 100kHz (See 1.JPG). The idea behind putting an inductor on DC side is to limit the inrush current.
Work done till now:
I did the simulation in MATLAB for the circuit. However, I connected only the resistive load and not the SMPS. In that case the inductor current contained 300Hz component (See 2.JPG).
Query:
In above case,
1. Will the inductor current reflect 100kHz frequency?
2. WIll the inductor be designed at 100kHz or 300Hz in this case?
Regards
Aalok
6xfsource. By fsource=50Hz - 300 Hz. 300 Hz is first harmonic frequency.
Hi,
the approximated value of peak current is given by a following formula:
I_max = E*sqrt(C/L)
where: E is the mean value of input voltage (in Europe about 560VDC).
To limit the peak current during capacitor charging u should choose a proper combination of inductor and capacitor.
yes, only for the six pulse of the ac sources connected to the rectifier.
@Lukasz Niewiara:
The AC input voltage is 85Vrms. Therefore, developed average DC voltage would be 85*1.35=115Vdc. In this case, Can you please tell me how did you get this equation? Would be great if you can share me the reference for this.
If you have a big capacitor after the bridge, you should use a precharge circuit to limit the inrush current . Depending on the power, this can be done with a PTC (small power) or a resistor and a relay which shorts the resistor after the capacitor is fully charged. The inductor only helps to shape the current from on the ac side or to block conducted EMI from the dc/dc converter.
If you use relay and resistor, keep the dc/dc off to let the capacitor charge.
In this case the input rectifier circuit can be considered as an DC circuit. Some assumptions should be made:
1) the voltage behind the rectifier is constant
2) no load during capacitor charging
3) initial voltages and currents are zero.
Fullfiling the assumptions, the circuit can be considered as an DC circuit during switching on. Therefore, a equivalent circuit can be obtained (see LCE_circuit.wmf). II Kirchhoffs law gives us:
E = vC(t) + vL(t), where E - equivalent input voltage, vC(t) - capacitor voltage, vL(t) - inductor voltage
A following differential equation can be formulated:
E = L*di(t)/dt + vC(t)
assuming i(t) = C * dvC(t)/dt gives us a second order differential equation:
E = L*C*d^2vC(t)/dt^2 + vC(t)
the solution of this equation is:
vC(t) = E * ( 1 - cos( 1/sqrt(LC) * t ) )
i(t) = C dvC(t)/dt = E * sqrt(C/L)*sin(1/sqrt(LC) * t)
the maximum value of the current is obtained when the sinus function is equal to 1, so it gives:
i_max = E*sqrt(C/L)
In your case u should use for calculations the RMS value of the voltage. It gives E = 115 VDC.
Btw folks, I am now using the thermistor in this application with adequate capacity.
Good question.. Please share me the best answer might you trust...
Regards…
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