Duty Ratio and Volt-Second Balance

Period and duty ratio. The duty ratio is a ratio of times, and it is dimensionless:

The average is the area divided by the period. The waveform is 200 V for a fraction \(D\) of each cycle and zero otherwise:

The RMS is the root of the mean square, and squaring happens before averaging — which is why the duty ratio enters under a root:

Note that \(V_{rms} > V_{avg}\) by a factor of \(1/\sqrt{D}\). Using 80 V to compute the load heating would understate it by a factor of 2.5.

Power delivered. For a resistive load it is the RMS value that does the work, never the average:

Equivalently \(P_o = D V_{in}^2/R = (0.4)(200^2)/10 = 1600\) W — the duty ratio scaling the power the load would draw if connected continuously.

Check from the source. Current is drawn from the 200 V supply only while the switch is closed, at \(200/10 = 20\) A, so the average source current is \(D(20) = 8\) A:

Input and output agree exactly because the switch is ideal. Every watt of discrepancy in a real converter is a loss you can name — Problems 4, 5 and 6.

Form factor and ripple factor, the two figures used to describe how far a chopped waveform is from DC:

A ripple factor above 1 means the AC content exceeds the DC content — the raw chopped waveform is a poor DC supply. Every converter in Part 3 exists to fix that, and it fixes it with an inductor.

Duty ratio from the ideal conversion ratio. For the buck in CCM the output is the input scaled by \(D\):

The inductor voltage in each subinterval. With the switch closed the inductor sits between the input and the output; with the switch open it sits between the diode (at 0 V) and the output:

Verify the balance. The areas must cancel over one period:

Written symbolically, that identical statement is the conversion ratio itself: \((V_{in}-V_o)D = V_o(1-D)\) rearranges directly to \(V_o = D V_{in}\). The formula is not a separate fact to memorise; it is volt-second balance with the algebra already done.

The current ripple, from a constant voltage acting on the inductor for a known time:

The off-interval must give the same number, and does: \(\Delta I_L = (9/100\,\mu\text{H})(6.25\ \mu\text{s}) = 0.5625\) A. If the two disagree, the balance was written wrongly — this is the cheapest self-check in the subject.

Maximum and minimum currents. The average inductor current in a buck equals the load current, because the capacitor carries no DC:

Both the switch and the diode must be rated for \(I_{max}\), not for the 3 A average.

The edge of continuous conduction is where \(I_{min}\) reaches zero, i.e. where the load current has fallen to half the ripple. Note that \(\Delta I_L\) does not depend on the load at all:

Above 32 Ω the converter enters DCM and \(V_o = DV_{in}\) ceases to hold — the output would rise above 9 V at the same duty ratio. Problem 7 works that case out in full.

Duty ratio. For the boost in CCM the output always exceeds the input, and the ratio is set by the off-time fraction:

Verify by volt-second balance. With the switch closed the whole input is across the inductor; with it open the inductor is clamped between the input and the output:

The inductor current is the input current here, not the output current — the single most important structural difference from the buck:

Consistent with \(I_L = I_o/(1-D) = 2/0.4 = 5\) A. The inductor carries 2.5 times the load current, so it — and the switch — are sized for 5 A, not 2 A.

Ripple and peak current:

Output ripple, from charge balance. While the switch is closed the diode is reverse biased and the capacitor alone supplies the load, losing charge at a constant \(I_o\) for the whole on-time:

This is the dual of the volt-second argument: the capacitor's charge must also balance over a period, so the charge lost during \(DT_s\) is exactly the charge returned during \((1-D)T_s\). Note it depends on \(D\) and \(I_o\) only — the inductor plays no part.

Device stresses. When the switch is open it stands off the output, not the input:

A converter that raises voltage raises the switch's blocking requirement with it. At \(D \to 1\) the ideal ratio diverges, but the real one collapses — the inductor's resistance eventually dominates, which Problem 6 illustrates on a buck.

The linear regulator, in one line. A series pass device carries the full load current while standing off the difference in voltage, so:

Note the general result: for a linear regulator \(\eta = V_o/V_{in}\) exactly, regardless of load. The device is being used as a resistor, and a resistor with 12 V across it and 5 A through it dissipates 60 W — there is nowhere else for the energy to go.

The buck converter: itemise the losses. With \(D = V_o/V_{in} = 0.5\) and ripple neglected, the MOSFET carries 5 A for half the period and the diode for the other half:

The inductor conducts for the whole period, so its loss carries no duty-ratio factor — a common slip.

Switching loss, using the triangular-overlap approximation for a clamped inductive load:

This is the only term that grows with frequency, and it is why the choice of \(f_s\) is a trade against the size of \(L\) and \(C\) rather than a free parameter.

Total and efficiency:

Where the linear regulator becomes acceptable. Setting \(\eta = V_o/V_{in} = 0.9\):

That leaves only 1.3 V of headroom — at or below the dropout voltage of most series regulators, and with no margin for a bus that sags. High efficiency and useful regulating range are mutually exclusive in a linear regulator, which is the whole argument for switching.

Why there is any loss at all. An ideal switch has \(v = 0\) when it conducts and \(i = 0\) when it blocks, so the product \(vi\) is zero at every instant. A real device cannot change state instantaneously, and during the transition it has substantial voltage and substantial current together. The energy lost is the area under that product:

Turn-on, first subinterval. With a clamped inductive load the freewheeling diode holds the device voltage at 400 V until the device has taken over the full load current. So \(v = 400\) V while \(i\) ramps \(0\to20\) A, giving an average product of \((400)(10) = 4000\) W:

Turn-on, second subinterval. Only now can the voltage collapse, and it does so at full current:

Turn-off is the same construction reversed, and slower — which is typical, because the IGBT's current tail is the slowest event in the cycle:

Total loss at 10 kHz. Switching loss is energy per cycle times cycles per second; conduction loss is not:

The frequency ceiling. Conduction loss is fixed by the operating point, so the whole of the remaining budget belongs to switching:

This is why IGBTs are used at a few kilohertz to a few tens of kilohertz while MOSFETs run at hundreds. Raise \(f_s\) beyond this and the answer is not a better filter but a destroyed device — unless the overlap itself is removed, which is what the resonant converters of Chapter 20 do.

Write the inductor voltage with every drop included. During the on-time the current passes through the MOSFET and the winding before reaching the output; during the off-time it passes through the diode and the winding:

Balance the volt-seconds and solve for \(D\):

The ideal answer would have been \(12/48 = 0.25\). The converter must run 6% longer on each cycle to make up for the drops — and a controller with insufficient duty-ratio headroom is precisely how a supply fails to regulate at low line.

Itemise the conduction losses at the duty ratio just found:

The diode alone accounts for 70% of it, because a fixed 0.7 V drop carrying 10 A for three-quarters of every cycle is simply expensive.

Efficiency, and an independent check. The output is \(P_o = (12)(10) = 120\) W, so:

Now verify from the input, which draws current only during the on-time:

The two routes agree, which they can only do if the duty ratio from step 2 was right. Whenever a loss budget and an input-power calculation disagree, the error is almost always a duty ratio taken as ideal.

The synchronous buck. Replacing the diode with a MOSFET replaces a fixed 0.7 V drop with \(I_o R_{DS(on)} = 0.25\) V, so the off-time inductor voltage changes and the duty ratio moves with it:

The new loss budget:

Loss falls by 45% and the heat to be removed falls from 7.3 W to 4.0 W. The low-side MOSFET conducts for the longer interval, so it is the one whose \(R_{DS(on)}\) matters most — the opposite of the intuition that the "main" switch dominates.

Test the conduction mode before assuming a formula. The standard test compares the dimensionless parameter \(K\) with its critical value:

Since \(K < K_{crit}\) the converter is in discontinuous conduction. Writing \(V_o = DV_{in} = 6\) V here would be wrong by a factor of more than two.

Volt-second balance over the intervals that actually carry current. The third interval contributes nothing, because with \(i_L = 0\) the inductor voltage is zero too:

One equation, two unknowns — which is why DCM needs a second relation. That relation is the load itself.

Close the system with the average current. The inductor current is a triangle of peak \(I_{pk}\) and base \((D_1+D_2)T_s\), and its average must equal \(V_o/R\):

Eliminating \(D_2\) and \(I_{pk}\) gives the standard DCM buck result in terms of \(M = V_o/V_{in}\):

Substitute and solve:

Against 6 V from the CCM formula. A lightly loaded buck does not gently drift off its characteristic — it rises towards the input voltage, and an unloaded one reaches it.

All three intervals. With \(T_s = 20\ \mu\text{s}\):

The inductor is doing nothing for 58% of every cycle, and during that interval the output node floats free of both the input and ground — which is exactly why the output voltage is no longer pinned to \(DV_{in}\).

Peak current, and the check that closes the problem:

The two agree, confirming both \(V_o\) and \(D_2\). In DCM this check is not optional — it is the only thing that distinguishes a correct solution from a plausible one.

The boundary load. Setting \(K = K_{crit}\):

Confirm it from the CCM side, where the boundary is where the load current equals half the ripple:

The 100 Ω load of this problem draws about a sixth of that, so the converter is deep in DCM — consistent with \(K = 0.05\) against \(K_{crit} = 0.7\).