Hello I have designed a current boost converter:



As you can see in the final equation if D<2/3 then the output current has a 180 degrees phase shift from the input current even though I am using DC.I dont understand why this is happening and for D=3/2 we get a 0 which means that the output current is always 0.Why are these weird stuff happening?

 

In the bottom circuit, are you shorting out the input supply? It looks like the switch is across the input.

 

Sure does.

 

That is not a proper boost converter circuit.
Try again.

 

In the bottom circuit, are you shorting out the input supply? It looks like the switch is across the input.

Yes im shorting out the input supply.But since a capacitor is a short for current sources current is divided exactly into 2.

 

That is not a proper boost converter circuit.
Try again.

It is a current boost converter not a voltage boost converter.We take for granted that the input is a voltage source but what if its a current source?Why is it not proper?Due to the inversion of the I/O?

 

Define a “current boost converter.”

 

Define a “current boost converter.”

it boostes current of a current source

 

it boostes current of a current source

I now see the input is a current source.
Where did you get this information? We have never seen this. Does not work.

 

I now see the input is a current source.
Where did you get this information? We have never seen this. Does not work.

I havent seen it anywhere.But why should we be limited to voltage sources?I just thought it would be a nice exercise to create a current source boost converter.Also many electronic circuits behave like current sources(MOSFET) because they have almost infinite input resistance.Also in nature there are current sources (the photovoltaic cell)so I was wondering if we could boost up its current somehow.

 

I havent seen it anywhere.But why should we be limited to voltage sources?I just thought it would be a nice exercise to create a current source boost converter.Also many electronic circuits behave like current sources(MOSFET) because they have almost infinite input resistance.Also in nature there are current sources (the photovoltaic cell)so I was wondering if we could boost up its current somehow.

So, if I understand you correctly, it should produce higher current at a lower output voltage. Can you explain your theory of operation?

 

So, if I understand you correctly, it should produce higher current at a lower output voltage. Can you explain your theory of operation?

Yes if we assume there arent any losses , due to conservation of energy if the output current doubles , the output voltage is halved.In the first picture the switch ( a PWM controlled MOSFET) is open->the diode conducts so the current through the capacitor is Ii-Io.By the equation for the IV relationship of the capacitor we find the dVc.In the second picture the switch is closed so the current through the capacitor is Ii/2.Ahhh here is the mistake guess need to work on it a bit

 

This is what a proper boost converter should look like.

 

I retract I am correct(I think).

In the first picture the switch ( a PWM controlled MOSFET) is open->the diode conducts so the current through the capacitor is Ii-Io.In the second picture the switched is closed so the current through the capacitor is -Io.From the IV equation of the capacitor and the duty cycle we arrive to the equations already posted.

 

This is what a proper boost converter should look like.

View attachment 317121

You are correct 100% if my source was a voltage source but my input is a current source.

 

How can you boost the current when the constant current is all the current you have?
Where do you think the extra current is going to come from?
Your equations don't prove the circuit works.

It's not the same as boosting the voltage with a reduced current.

 

How can you boost the current when the constant current is all the current you have?
Where do you think the extra current is going to come from?
Your equations don't prove the circuit works.

It's not the same as boosting the voltage with a reduced current.

By that reasoning how can we control the output voltage on a "normal" boost converter when in the input we have a fixed voltage reference???We do have fixed current but we dont have fixed voltage thats the trick I believe.

 

You are correct 100% if my source was a voltage source but my input is a current source.

IMHO that is a ridiculous proposition. Even a small current source into a small resistance will produce a very large voltage. I'm not sure your understanding of circuits runs very deep.

 

By that reasoning how can we control the output voltage on a "normal" boost converter when in the input we have a fixed voltage reference???We do have fixed current but we dont have fixed voltage thats the trick I believe.

The inductor provides the means to accomplish this from a fixed voltage by controlling the derivative of the current through the inductor.

 

This is my final push for today


When the switch is open the diode conducts so the current through the capacitor is Ii-Io and by the IV relationship for a capacitor we end up with a equation.

When the switch is closed the diode doesnt conduct so the current through the capacitor is -Io(the capacitor is discharging through the load) and by the IV relationship for a capacitor we end up with another equation.

We equate them and we end up with our relationship.Note that when D=0.5 the output current is infinite so the output voltage is 0(there is a pole at D=0.5)

Feel free to comment , I want to discuss this because I dont understand it very well either so lets do the first baby steps together.