Understanding boost converter - Page 1
Forum section: electronics / beginners

author: mtimmermans
author_rank: Newbie
timestamp: August 23, 2018, 11:34:53 am
content: 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.htmorhttp://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!

author: capt bullshot
author_rank: Super Contributor
timestamp: August 23, 2018, 12:17:05 pm
content: Welcome to the difference in theory and real worldThe 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.

author: Zero999
author_rank: Super Contributor
timestamp: August 23, 2018, 12:20:05 pm
content: 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.

author: mtimmermans
author_rank: Newbie
timestamp: August 23, 2018, 03:05:06 pm
content: 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?

author: Zero999
author_rank: Super Contributor
timestamp: August 23, 2018, 04:48:58 pm
content: Quote from: mtimmermans on August 23, 2018, 03:05:06 pmWhen 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.

author: capt bullshot
author_rank: Super Contributor
timestamp: August 23, 2018, 06:29:16 pm
content: If the converter runs in continuous mode, the formula applies for the output voltage. The converter enters continuous mode when its output current is large enough that the inductor current "touches" zero just before the switch turns on again. If the output current now increases, the inductor DC current increases, but AC (pp) current stays the same, determined by the switch duty cycle. The inductor doesn't get fully discharged anymore, and thus the output voltage doesn't rise.As long as the converter runs in discontinous mode (inductor current reaches zero before the switch turns on again), it fully discharges into the capacitor and causes the voltage to rise. Think of some minimum energy quanta that must be transferred each cycle. As soon as continuous mode is reached, more than the minimum quanta is transferred per cycle, implicit determined by the output current. The minimum quanta is determined by inductor value and its AC (p-p) current, not the DC current.

author: David Hess
author_rank: Super Contributor
timestamp: August 23, 2018, 07:03:21 pm
content: The output voltage only increases beyond Vo/Vi = 1/(1-D) when the output current is not sufficient to maintain continuous current mode.

author: mtimmermans
author_rank: Newbie
timestamp: August 23, 2018, 07:08:54 pm
content: Yes thanks, that makes a lot of sense! I forgot about the DC current, and thought that only the AC current could power the load.