Based on the description in the attached figure, you can switch between buck and boost operation by controlling the duty cycle (D) of the PWM.devonsc said:I've read about how does the flow of current in this circuit, the Buck-Boost Converter. But still don't quite understand. Do you guys mind explaining how do we control the circuit to work as a Buck and on the otherhand work as a Boost? Is there away of controlling? As in, having it switch to perform a buck converter and having it switch to perform as a boost whenever it is needed.
a) Inductor Current Rating for a Buck-Boost Converter at **broken link removed**devonsc said:a.) Regarding the component selection of the buck-boost converter, can I implement the same method I used for a boost DC/DC converter?
d.) Is it true that the output voltage of the buck-boost converter will be in negative? I'm trying to develop this circuit for charging sealed lead acid battery purposes, does it mean that I can apply the output negative voltage of the converter to the negative terminal of the battery? Can charging be made in this manner?
The negative pin of the ouput capacitor is more 'negative' compared to the positive pin of the output capacitor. In order to push the charging current through the battery, you should connect as follows:devonsc said:... Can I charge the battery by using a negative output voltage from the converter being apply to the negative terminal of the battery?
I think the 5V of function generator is not 'powerful' enough to drive the mosfet.
The driving signal should be able to supply enough charging gate current. Also, it's advised that you put a Rg in series with the gate of the power mosfet.
Besides you have to remember that in order to turn on the power mosfet, the Vgs (not Vg) should be > than the gate threshold voltage (Vgs(th)).
Function generator usually have two output terminals. How did you connect both output terminals to your circuit or the gate of the power mosfet?
By the way, what was the VDS (pls refer to the datasheet of the power mosfet) when you tested the circuit?
I remember something about signal generator. The one in my lab has output impedence of 600ohm. This output impedance will result in voltage drop and therefore the voltage at the gate might not be 5V as you expected.
1. the entire energy required for one switching cycle is stored in the inductor. In the case of the boost, during the energy transfer phase, part of the energy comes directly from the input power, since the inductor is effectively in series with the input voltage: E=Vin*Iout*Tsw, so the inductor only needs to store:
Eind=(Vout-Vin)*Iout*Tsw. For the buck-boost you need to store:
Eind=Vout*Iout*Tsw
3. the transistor is now on the "high side", it has either the source or the drain connected to the input power, so you will need a "high-side driver".
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