Electronics > Beginners
Understanding boost converter
mtimmermans:
In derivations people use to get to Vo/Vi = 1/(1-D), they assume that the output voltage settles to some value e.g. http://www.simonbramble.co.uk/dc_dc_converter_design/boost_converter/boost_converter_design.htm or http://www.ti.com/lit/an/slva061/slva061.pdf. I would however think that the output voltage of a boost converter should keep on increasing when in continuous current mode.
As I understand the boost converter, it works as follows. First you charge the inductor, which means that current flows through the inductor. Then you flip the switch. Since the current must keep flowing, the voltage across the voltage must increase in order to forward bias. This allows current to flow through the diode and thereby charging the capacitor. This means that each time you flip the switch, some current will flow through D1 and thus the capacitor will keep charging forever (theoretically). Why does this not happen though? Furthermore, why is the output voltage incorrect in the simulation? The duty cycle is set to 50%, so according to the formula the output voltage should be 20V?
Thanks for your help!
capt bullshot:
Welcome to the difference in theory and real world ;)
The formula you've quoted is correct for either a synchronous rectified converter, or a converter like your circuit with a minimum load.
With no load, the voltage would increase to infinity in theory. In practice, there are losses in L1, Q1 and D1 that stop the voltage rising at some point. In each cycle the inductor gets charged and must discharge into the capacitor, that would cause the capacitor voltage to increase endlessly. But in each cycle there are also losses of all kind in the components, that rise with the output voltage, and at some point it reaches equilibrium and all the energy that you put into the circuit gets wasted as heat, causing the output voltage to settle.
These losses are difficult to model, so a simulation would settle at a different voltage than the real circuit.
Zero999:
1) As soon as the switch opens, the inductor behaves as a constant current source, i.e. it passes the same current irrespective of the load across its terminals.
2) When there's no load, the duty cycle makes no difference. The inductor will just have a lower current flowing through it, but the theoretical open circuit voltage will be the same: infinity. Of course in reality, it will not be infinite, but very high.
mtimmermans:
Thanks for the quick responses!
I am however still slightly confused. Assuming we are in continuous current mode, then the output voltage is fixed and given by Vo/Vi = 1/(1-D). Also the current that flows each cycle through the diode is fixed. If you don't want the voltage to build up in the steady state, then you need a resistance of R = Vo / Id where Id is the average current that flows through the diode. If you pick a larger resistance, then current will flow to the capacitor, which causes the output voltage to increase. If you decrease the resistance, then you will draw current from the capacitor, causing the output voltage to drop. Where did my reasoning go wrong?
Zero999:
--- Quote from: mtimmermans on August 23, 2018, 03:05:06 pm ---Thanks for the quick responses!
I am however still slightly confused. Assuming we are in continuous current mode, then the output voltage is fixed and given by Vo/Vi = 1/(1-D). Also the current that flows each cycle through the diode is fixed. If you don't want the voltage to build up in the steady state, then you need a resistance of R = Vo / Id where Id is the average current that flows through the diode. If you pick a larger resistance, then current will flow to the capacitor, which causes the output voltage to increase. If you decrease the resistance, then you will draw current from the capacitor, causing the output voltage to drop. Where did my reasoning go wrong?
--- End quote ---
When your circuit has no load, the steady state voltage will be theoretically infinite. It will carry on attempting to supply the same current into the load, determined by the current flowing through the inductor, its inductance and the repetition rate. The higher the duty cycle, the higher the average current going into the output capacitor will be, but the theoretical steady state voltage will be the same, i.e infinity, as long as the duty cycle is greater than zero and less than one.
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