Inductive Rectifier with internal DC load

In summary, the conversation discusses finding an expression for both i_in(wt) and average current in an inductive rectifier circuit. The circuit is being used as a choke input ripple filter and the resulting pulses contain a DC component, fundamental frequency, and harmonics. The peak ripple voltage and current can be found using Ohm's Law and the average current is 0.577 times the peak. One issue with the question is that the rectifier produces a DC voltage component and there is zero resistance in the circuit, leading to infinite direct current flow.
  • #1
Dan_D93
1
0
Hi,

I'm new here so I'm not sure how things work but I'd like some help on a question.

How do I go about finding an expression for both i_in(wt) and the average current?
This is the inductive rectifier circuit I'm working with.
http://i.imgur.com/HaWzFfQ.png
http://i.imgur.com/kefMiOd.png

I've tried using the inductive voltage equation V_L = L* (di/dt) and then integrating to get some kind of expression but haven't been able to come up with anything yet.

Any help would be greatly appreciated. Thanks.

Dan.
 
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  • #2
This question seems to relate to a choke input ripple filter. The battery causes the rectifier to base line clip the half sine waves coming from the rectifier. We deduct the battery voltage from the generator voltage. The resulting narrow pulses, if viewed in the frequency domain, consist of a DC component, a fundamental at power line frequency and all the harmonics. It is usual for engineering purposes to ignore the harmonics. We can take the height of the remaining pulse as our peak-to-peak ripple at fundamental frequency. Then the peak ripple voltage, Vp, is half the peak-to-peak. The peak ripple current, Ip, is then found from Vp and the inductive reactance, using Ohms Law. The ripple current at any instant is Ip sin wt. The average current is 0.577 times the peak.
One problem with the question is that the rectifier produces a DC voltage component, and the resistance in the circuit is zero. From Ohm's Law, this leads to infinite direct current flowing.
 

What is an inductive rectifier with internal DC load?

An inductive rectifier with internal DC load is a device that converts AC (alternating current) to DC (direct current) using an inductor and a diode. The inductor helps to smooth out the output voltage, while the diode ensures that the current flows in one direction.

What is the purpose of an inductive rectifier with internal DC load?

The purpose of an inductive rectifier with internal DC load is to convert AC power to DC power for use in electronic devices. It is commonly used in power supplies, battery chargers, and other applications where a steady source of DC power is needed.

How does an inductive rectifier with internal DC load work?

An inductive rectifier with internal DC load works by using an inductor to smooth out the AC voltage and a diode to allow current to flow in one direction. When the AC voltage is positive, the diode is forward-biased and current flows through the load. When the AC voltage is negative, the diode is reverse-biased and no current flows through the load.

What are the advantages of using an inductive rectifier with internal DC load?

There are several advantages to using an inductive rectifier with internal DC load. Firstly, it is a simple and cost-effective way to convert AC to DC. Additionally, it provides a steady output voltage and can handle high currents. It also has a low output ripple, which is desirable for many electronic devices.

Are there any limitations to using an inductive rectifier with internal DC load?

Yes, there are a few limitations to using an inductive rectifier with internal DC load. The main one is that it can only be used for low to medium power applications. It is also not efficient for converting high voltages to low voltages. Additionally, the output voltage may vary with changes in the input voltage, which can affect the performance of some electronic devices.

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