I'm looking at the designs for a simple buck converter which has the P mosfet, inductor, capacitor and reverse based diode. Instead of PWM, I'm trying out to use an op-amp to switch the P mosfet but one thing that I notice is that if the load changes from low resistance to high resistance then the inductor spikes the voltage on the load to double what it is until it can re-stablise. (This is all in simulation)

Is this happening because I'm using the op-amp? How does using PWM affect the inductor in such a way that if the load would change from say 10 ohms to 100 ohms that the inductor won't spike the voltage? Perhaps PWM / fast switching keeps the inductor with such a low charge that when the load changes, the inductor doesn't have nearly enough voltage to do anything?


  • 3
    \$\begingroup\$ Using an op-amp instead of a PWM doesn't convey enough information. Can you give a schematic of your control circuit? \$\endgroup\$ Sep 24, 2013 at 14:32
  • \$\begingroup\$ Yes, we really need a schematic of what you are trying to do. \$\endgroup\$
    – Joe Hass
    Sep 24, 2013 at 15:31

2 Answers 2


For any given load, a switcher will transfer a given amount of energy thousands of times per second. This is how the buck regulator works.

Let's say your op-amp is switching at 10kHz (because it's a slow sort of device and will have slew rate problems compared to other devices). Let's also say you are aiming to deliver 5V across a 10 ohm resistor. Resistor power is 25/10 watts = 2.5 watts.

To calculate energy per switching cycle divide this power by frequency because power = joules per second. At 10kHz, the energy you transfer per switch cycle is 250\$\mu J\$.

This energy powers your load resistor but, if you removed your load resistor, this energy gets dumped into the output capacitor and its voltage rises a little (or a lot) higher than normal.

Let's say your output capacitor is 10uF - if suddenly it was imbibed with 250\$\mu J\$, how much would it rise in voltage?

We know that capacitor energy is \$\dfrac{C V^2}{2}\$ therefore we can calculate the voltage rise and this is: -

\$\sqrt{\dfrac{250\times 10^{-6} \times 2}{10\times 10^{-6}}}\$ = 7.07V.

It's a little bit subtler than this - in the above I assumed the capacitor was being charged with energy from a zero voltage state. In fact it already has 5V across it and this means that the previously stored energy + influx energy (from the inductor) is 125\$\mu J\$ + 250\$\mu J\$ = 375\$\mu J\$.

If you do the reverse math, the peak voltage on the capacitor becomes 8.66V i.e. 3.66 volts higher than the 5V rail.

You could put an argument together to consider the losses in the diode also - this may trim half a volt of the absolute peak voltage.

So, you either need to increase the capacitance a lot or, decrease the transfer energy by increasing the operating frequency. Modern switchers regularly operate at 500kHz and this means the energy per cycle reduces from 250\$\mu J\$ to 5\$\mu J\$ in this example.

Should this be the case (500kHz operation), the rogue energy from the inductor would make the capacitor's stored energy 130\$\mu J\$ and this means a peak voltage of 5.1 volts - probably quite acceptable for load dumping on a switcher.

Operating at higher frequencies requires faster silicon but, the ability to control load variations (and their repercussions), on a cyclic basis, means much tighter control of the output voltage.

This is just an example to see where you might be going wrong.


An OpAmp will not be able to directly control the state of a MOSFET. The OpAmp will be too limited in output impedance, slew rate, and output voltage swing. You are going to want a device that can switch quickly while driving the gate capacitance of a FET.

There are different ways to control the on and off time of the FET. PWM is the most used, but there is also constant on time and constant off time. Constant on time and constant off time can be used in a hysteretic regulation approach, using a comparator to sense the output voltage relative to a reference. Whatever control technique you try, you will need to bound the on or off time of the switch somehow to have any reasonable control of output voltage.

As to the overshoot you see when the load changes:

Did you put resistance in series with the inductor and capacitor in the simulation? Real parts have loss. If you have no loss in the buck inductor and filter capacitor, filter Q will be very high. High Q LC filters will ring to 2 times Vin on application of a step.


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