0
\$\begingroup\$

I'm trying to make a boost converter, from a 9V battery, but I'm not sure if understand the schematic entirely:

Valid XHML.

The load appears to be a resistor, leading me to my question: how does the voltage even get stepped up? As you begin to charge the capacitor, would it not discharge, through the resistor, immediately?

\$\endgroup\$
  • 1
    \$\begingroup\$ It would improve chances of a meaningful answer if you gave real component values, rather than leaving the schematic editor's default values. \$\endgroup\$ – Peter Bennett Mar 26 '17 at 6:12
  • \$\begingroup\$ Related: How to make a boost converter circuit \$\endgroup\$ – Roger Rowland Mar 26 '17 at 8:01
  • \$\begingroup\$ The schematic is good for understanding the basic concept of a boost regulator. The tricky part is turning the switch on and off not only really fast, but with a duty cycle that results in correct output voltage. The switch would typically be an N-channel MOSFET. Some kind of control system (switching regulator) would control the MOSFET. \$\endgroup\$ – mkeith Mar 26 '17 at 8:23
1
\$\begingroup\$

Capacitor would discharge, but not immediately. If your frequency is high enough and duty cycle is right, it will remain charged.

The operation is very simple and is based on two basic properties: inductor's current can't change immediately and capacitor voltage can't change immediately. So you charge your inductor by closing the switch, the open it and current flows to capacitor and load. Since the frequency is high, voltage on the capacitor doesn't change.

With you values you will have to go to quite high frequency...

\$\endgroup\$
0
\$\begingroup\$

When thinking about "boost convertors" / "switching regulators" you have to think about current not voltage.

A good linear regulator, regulates voltage and is current limited. That is, it adjusts the voltage and prevents the output current from going over a set limit.

A switching regulator does the inverse. It adjusts the current and makes sure the output voltage is within a set range.

In your circuit, the "switch" is used to establish an average current in the inductor that is enough to supply the load AND keep the capacitor charged to a target voltage.

When first starting, the capacitor voltage is low so the pulse width ratio is made high to pump the maximum current through the inductor. This current is higher than the load current so the excess quickly charges the capacitor.

When the voltage on the capacitor is OVER the target voltage the ratio is reduced considerably, sometimes even stops completely. At that point there is no longer enough average current in the inductor to supply the load, and the capacitor will begin to discharge.

This all repeats at a high frequency and the regulator, if controlled correctly is a stable, self balancing system.

Of course the above is an over simplification. The Error Voltage = Target Voltage - Actual Output Voltage is used to slew the pulse width according to some slope or formula in a predefined manner that effectively attempts to quickly maintain the average coil current at the same level as whatever the load current is at any given time. All be it within an acceptable lag time.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.