Discussion in 'General Electronics Chat' started by anhnha, Feb 16, 2016.

1. ### anhnha Thread Starter Active Member

Apr 19, 2012
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47
Here is a boost converter operating in discontinuous conduction mode (DCM).
The load current Io is equal to the average diode current (ID).
Can anybody explain it?

2. ### AnalogKid Distinguished Member

Aug 1, 2013
4,534
1,251
100% of the load current comes through the diode. That current comes through in pulses, not continuously. Those pulses are averaged (integrated) by the output capacitor C. The load and the diode form a series circuit, so the average power in the load must come through the diode.

ak

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
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But what you do not understand?. When the switch is closed Id = 0A and the cap provide all the current. But when switch is open the coil will provide Io and in the same time the coil will charge the cap.

4. ### anhnha Thread Starter Active Member

Apr 19, 2012
773
47
Thanks but why load and diode are in series?
The load current will be affected by the capacitor and diode current. Why capacitor doesn't have influence in load current?

5. ### ErnieM AAC Fanatic!

Apr 24, 2011
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1,605
Because the net current into and out of the cap is zero.

Any current sourced by the cap to load must come from thru the inductor, and later be returned thru the diode.

(nitpickers may note I am actually referring to charge not strictly current)

6. ### AnalogKid Distinguished Member

Aug 1, 2013
4,534
1,251
The capacitor does not pass any current at DC, so it has no affect on the long-term average output current. It affects the shape of the load current, but not its average value. The diode and the load are in series because that is what is on your schematic. If you are asking why they are that way on the schematic, it is because of Faraday's Law:

ALL switching power supplies function by turning the input DC into some form of AC so it can use a transformer or inductor to modify the energy, then rectifying and filtering the output to turn the power stream back into DC.

ak

7. ### crutschow Expert

Mar 14, 2008
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3,232
When S1 is closed Vi causes the current to increase in L, storing energy equal to 1/2 LI^2.
At this point the diode keeps the voltage the capacitor from being discharged to ground
Also the load is being supplied current from the charge stored on the capacitor.

When S1 opens, the stored energy in L's inductance keeps the current moving through D to recharge the capacitor and provide load current.
This continues until all the inductor energy is transferred to the capacitor.

The charging and discharging of the capacitor creates a ripple voltage on the output DC as determined by the C value, the load current, the input voltage, the inductor size, and the switching frequency.

Last edited: Feb 16, 2016
8. ### anhnha Thread Starter Active Member

Apr 19, 2012
773
47
Thanks for these responses.
I think I've got it now. The said load current is the average not the instantaneous.