# Synchronous Buck Converter - PWM modulation

Let's say that our Vin = 9V, Vout = 5V. Hence our duty cycle D = 5/9.

Let's say that our switching frequency is 100kHz.

Consider T_on as the time the HSFET is on. So our T_on in this case is:

T_on = (1/100kHz)*(5/9) = 5.5us

If I_Load increases, we will need more time that HS FET is kept on. So that will mean that T_on increases. So this 5.5us will increase.

My question : Does the equation I used to calculate 5.5us only apply for no load conditions? I assume I need to have an extra scaling factor which repersents the PWM reacting to increased load and hence inceasing Ton?

• Yes, the longer T_on compensates the losses. You have virtually a higher Vout
– Jens
Jun 13 at 18:11
• Just a side note, the "PWM Drivers" as shown may be asking for trouble. There must be a dead-time, meaning "time when neither are on." Otherwise both could be on at the same time due to delays, causing large current spikes and destroy the FETs. Jun 13 at 18:42

My question : Does the equation I used to calculate 5.5us only apply for no load conditions?

When you have a synchronous buck regulator (as per your schematic), the duty cycle multiplied by the input voltage equals the output voltage irrespective of loading conditions (within reason of course). In other words, the volt drop in both MOSFETs will eventually eat into the above equality but that will be on very heavy loads.

For light loads and medium loads the equality holds. For normal heavy loads it nearly holds but on really heavy loads you see several percent discrepancies.

Maybe you are thinking of a regular buck converter where the bottom MOSFET is replaced by a diode and, you are not operating in continuous conduction mode?

For a synchronous buck converter, the circuit is operating like CCM or in full CCM; it can never operate in DCM (discontinuous mode).

I assume I need to have an extra scaling factor which represents the PWM reacting to increased load and hence increasing Ton?

No, not when you have a push-pull MOSFET output circuit aka a synchronous buck converter.

• I don't understand. Let's say my buck is happily regulating at 5V under 0A condition with a 5/9 duty cycle at 500kHz. If I now change my load to 3A, to my understanding, the loop should increase the on-time (= duty cycle increase) since it is a PWM control and that's all it can do it. 3A will deplete the output cap much faster right, so PWM needs to comepnsate with higher duty cycle?? You can ignore FET resistance. Jun 13 at 20:37
• Or is that duty cycle increase just a transient behaviour and then it will go back to the equality? Jun 13 at 20:39
• @alayoiskgfbfqhxjiw think of a pure hard square wave of duty cycle 5/9 (55.555%) and amplitude 9 volts falling back to 0 volts then back to 9 volts again.... Nothing on earth alters the fact that its average level is 5 volts so.... follow that with a low pass 2nd order lossless filter and the average level of the output is still 5 volts. Then choose that filter to have a low pass 3 dB point that is much lower than the switching frequency and.... you end up with 5 volts plus a little ripple. Jun 13 at 22:12
• Ah, that makes sense now. Jun 13 at 22:46

The equation applies for no-resistance condition and instantaneous switching. Include Rds(on), inductor ESR, and capacitor or supply impedances, and you will see it needs slightly more under load.

Since most of these impedances are also losses, we tend to choose components with relatively little of them, so the ideal duty case still gets us pretty close to the real thing.

Put another way, the dashed box can be converted to a Thevenin source, with those impedances converted to its resistance, and this gives the output regulation (dV_out/dI_Load) for fixed duty. Which looks like any other resistance, such as in the inductor, or wiring to the load, so we merely compensate for that with feedback.