Can someone verify my understanding of a boost converter?
For a basic boost (step-up) converter, once the inductor has been charged and a magnetic field has formed, the switch to ground is opened and the charge gets transferred to the capacitor through the inductor.
The inductor voltage spikes because the current cannot change by a finite amount instantaneously. Once the voltage spike reaches to 0.7 V (assuming a silicon diode) above the capacitor voltage, the diode forward biases and transfers the charge to the capacitor.
Thus, the capacitor is only charged each cycle by a voltage slightly above the diode turn on voltage.
Specifically then besides verifying my above observation, is my idea about the inductor and diode also valid?
Nope. It's driven by a current... _almost_ current-source-like.
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
Sorry, yes, the capacitor is charged by the charge that gets placed on it from the inductor. But only after the voltage generated by the inductor forward biases the diode - correct?
Well let me put it this way. The complete, utterly horrific detailed story is:
When the switch opens, there is some current I in the inductor (due to V = L
dI/dt (inductor equation) integrated over the switch-ON period, minus series resistance, which causes a slight RL exponential curve). As the switch turns off (over perhaps 10 to 1000 nanoseconds), voltage rises and current is transferred to the parasitic capacitance (of the inductor, switch, diode and anything else, incidental or intentional) according to dV/dt = I/C (capacitor equation). The capacitance limits rise time to the shape of the leading edge of a decaying resonant pulse. For slow turn-off times and fast (i.e., low capacitance) parts, the turn-off curve can control the rise time instead.
Note that, when using solid state switches (transistor, diode), capacitance is often a function of voltage; for large capacitances, the rise can start out slow, speed up in the middle, then slow down again, because capacitance is highest near one side or the other. I have something of an example here:
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This is an inductor switched with IGBTs. Rise/fall time is very quick (comparable to the listed data of the transistors themselves), but still, there is a noticable change in slope at the toe and shoulder of the edges.
When voltage reaches the diode turn-on voltage (in perhaps another 10 to
1000 ns), current is transferred from the capacitances to the diode. The change in current in the diode causes an overshoot in voltage (which can be seen in the picture as well, as that circuit uses diodes to restrain the inductive flyback), in proportion to the loop inductance (from capacitor to diode to inductor; note that two inductances are in series, so in a manner of speaking, the overshoot is partly error from measuring it from some given point along that loop!). Voltage remains here until the inductor discharges (again, dt = L*dI/V). When current reaches zero, the parasitic capacitance (and also charge stored in the diode, depending on reverse recovery time) holds voltage for a short time, causing a small negative current in the inductor. This negative current causes the voltage to alternately swing down and up, according to the resonant frequency of the inductor vs. capacitances. If the downward swing is sufficient, voltage may be clipped by a diode connected in reverse across the switch (usually intrinsic, since the switch is usually a MOSFET), clipping the peak before it continues to resonate.
At this point, inductor current is essentially zero, and the switch can turn on again, consuming a short peak current (depending on current risetime) due to capacitance, and a longer current ramp due to the inductor charging.
Tim
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Yes but . . . the collapsing magnetic field is busy producing a high voltage spike and that spike is allowed to charge the capacitor to a value higher than the supply (the diode merely subtracts its forward voltage drop from the total the inductor produces)
The overshoot caused by the collapsing field is high, but only present for a short period of time - so you run the converter at a high frequency and use diodes that switch quickly.
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Yup, pretty much true. Note that, since the inductor-diode current is at a low duty cycle, the peak diode current must be a good deal higher than the average output current.
There are lots of numerical subtleties, of course: inductor saturation, copper loss, distributed coil+diode capacitances, stuff like that. Semiconductor manufacturers' appnotes have lots of examples that actually work.
Pretty much. If you want an analogy, think of a pendulum which pushes against a spring and ratchet on each stroke. It slowly compresses the spring, but it has to go further each time to add a little more to the compression.
The pendulum is the inductor, the ratchet is the diode and the spring is the capacitor.
You answered your own question. Study it. (And don't top post ;-)
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
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