Why add a transformer, and what does it cost?

A transformer in a switching converter gives electrical isolation and a turns ratio, and because it runs at the switching frequency it can be small. But the core has to be magnetized and demagnetized every cycle, and each way of doing that gives a different converter. In this module you will watch three of them work, break them on purpose, and then use them to make design decisions.

By the end you can

1. Explain what the transformer does in a flyback, forward, and push-pull converter, and where the energy is stored or passed through.
2. Trace the current path in every switch state and derive each output equation from volt-second balance.
3. Decide from L, f, R, and D whether the key inductor current is continuous (CCM) or discontinuous (DCM), and say what DCM does to the output.
4. State the core-reset requirement and use it to limit the duty ratio and the switch voltage in a forward converter.
5. Predict what happens when the load, turns ratio, duty ratio, inductance, or frequency changes, then check the prediction.
6. Choose a topology and the main component values for an isolated supply, and defend the choice.

Why add a transformer?

In the buck, boost and buck-boost converters the input and output share a connection: if the source is grounded, so is the output. A transformer breaks that connection. Putting a 50 or 60 Hz transformer ahead of the converter would work, but it needs a large core, so it is big, heavy, and expensive. Putting the transformer inside the switching scheme lets it run at tens or hundreds of kilohertz, where it can be small.

A transformer also adds a design variable, the turns ratio, so the gain no longer depends only on the duty ratio. And extra windings give extra outputs. The price is that the core flux has to return to its starting value every cycle. Otherwise it creeps upward until the core saturates, and a saturated core cannot support voltage.

Three isolated converters

Comparison of flyback, forward, and push-pull converters
Converter	Isolated cousin of	How the energy moves	CCM output	Open-switch voltage	Typical power
Flyback	Buck-boost	Stored in the magnetizing inductance while the switch is closed, delivered after it opens	V_o = V_s {{D\|1 − D}} {{N₂\|N₁}}	V_s + V_o{{N₁\|N₂}}	to about 150 W, most often 10 W or less
Forward	Buck	Passes through the transformer while the switch is closed; a third winding returns the magnetizing energy	V_o = D V_s {{N₂\|N₁}}	V_s(1 + {{N₁\|N₃}})	to about 500 W
Push-pull	Buck, running at twice the frequency	Passes through, alternating between two primary halves so the core is driven both ways	V_o = 2D V_s {{N_s\|N_p}}	2V_s	to about 1000 W

Flyback: one switch, the primary carries current only while the switch is on, and the secondary only while it is off.

Forward: one switch, a buck-style output filter, and a third winding with D3 to reset the core.

Push-pull: two switches on a center-tapped primary, a center-tapped secondary with two diodes, and a buck-style filter.

Assumptions. Ideal switches and diodes, a large output capacitor for the equations (the labs also integrate the real ripple, including ESR), and steady state. Transformers use the model with a magnetizing inductance L_m (Hart Fig. 7-1d). Leakage inductance and losses are ignored, so real circuits need clamps or snubbers to handle switching spikes, and efficiency figures are not computed.

How to use this module

Each converter has its own tab with an animated circuit, the equations, a live lab, and activities. Move any slider and every number and graph updates. Activities with a number to find have a Load these settings button so everyone starts from the same circuit, and a Check button that compares your answer with the model. Open-response prompts are kept for you to download on the last tab. Nothing is sent anywhere; your answers stay in this page until you close it.

Suggested path: the transformer and core reset, then flyback, forward, push-pull, then What happens when… where you predict first and check second, then the converter comparison, the worked examples, and the design challenges.

The transformer, and the flux that must come back

Before the converters, two facts about transformers explain almost everything that follows: how voltages and currents scale, and why the magnetic flux in the core must return to where it started every switching cycle.

The ideal transformer

An ideal two-winding transformer scales voltage by the turns ratio and current by its inverse, so the power in equals the power out:

{{v₁|v₂}} = {{N₁|N₂}} {{i₁|i₂}} = {{N₂|N₁}}

The dots show polarity: if the voltage at the dotted terminal of one winding is positive, the voltage at the dotted terminal of the other winding is positive too. Current entering the dotted terminal of one winding corresponds to current leaving the dotted terminal of the other.

Ideal transformer calculator

A better model includes the magnetizing inductance

A real transformer needs a magnetizing current to set up flux in its core. It is modeled as an inductance L_m in parallel with an ideal transformer. Leakage inductances and winding resistances exist too, but they matter mostly for switching spikes and losses, so the models in this module use only the ideal transformer plus L_m.

This one inductance explains how the three converters differ. In the flyback, L_m is the energy-storage inductor. In the forward and push-pull converters it is an unwanted parasitic: it draws a magnetizing current that does no useful work and must be dealt with.

The flux must return

The voltage across L_m sets how fast its current (and the core flux) changes: v = L_m·di/dt. In steady state the current must end each cycle where it began, so the positive and negative volt-seconds applied to the core must cancel. If they do not, the flux ratchets up each cycle until the core saturates and the current can no longer be controlled.

Here is the idea in its simplest form. A voltage is applied for a fraction D of the period (the switch is on). Then a reset voltage, r times as large, is applied in the opposite direction until the current reaches zero, as it is in the forward converter, where a diode stops it from going below zero. Try it.

Does the core reset?

Three answers to the same problem

Flyback. The energy stored in L_m is delivered to the output every cycle. The magnetizing current is the working current, and the output diode takes care of the reset: when the switch opens, the secondary takes over and discharges the core.

Forward. The magnetizing current is a side effect, so a third winding and a diode send it back to the source after the switch opens. The reset voltage must be large enough, or the off-time long enough, to finish the job, which limits the duty ratio.

Push-pull. The two primary halves drive the core in opposite directions, so the volt-seconds cancel by symmetry. This works only as long as the two halves are equal.

Activities

Flyback: store the energy, then deliver it

The flyback is the isolated buck-boost converter. While the switch is on, the source charges the transformer’s magnetizing inductance. When the switch opens, that energy “flies back” through the secondary and the diode to the output. Its simplicity (one switch, one diode, no output inductor) makes it the favorite for low-power supplies.

How it works

Choose a state below and follow the current. Notice that the primary and the secondary never conduct at the same time. Notice also where the dots are: the secondary is wound so that its diode conducts only when the primary voltage reverses.

Show the discontinuous version (a third state appears)

Where the equations come from

Switch closed. The source is across the magnetizing inductance, so its current rises linearly. The diode is reverse-biased by V_o + V_sN₂/N₁, so the secondary carries nothing.

V_s = L_m {{di_Lm|dt}} ⇒ (Δi_Lm)_closed = {{V_sDT|L_m}}

Switch open. The magnetizing current cannot jump, so it transfers to the secondary through the ideal transformer. The output voltage is reflected to the primary and appears across L_m with the opposite sign:

v₁ = −V_o{{N₁|N₂}} ⇒ (Δi_Lm)_open = −{{V_o(1 − D)T|L_m}}{{N₁|N₂}}

Volt-second balance on L_m (the two changes cancel in steady state) gives the gain. It matches the buck-boost converter, with a turns-ratio factor:

V_o = V_s {{D|1 − D}} {{N₂|N₁}}

The open switch carries the source plus the reflected output voltage, and the diode current is the magnetizing current times N₁/N₂:

v_sw = V_s + V_o{{N₁|N₂}} = {{V_s|1 − D}} i_D = i_Lm{{N₁|N₂}}

Average magnetizing current follows from power balance (V_s·I_LmD = V_o²/R):

I_Lm = {{V_o²|V_sDR}} Δi_Lm = {{V_sD|L_mf}}

Continuous or discontinuous?

The magnetizing current stays positive if

L_m ≥ L_m,min = {{(1 − D)²R|2f}}{{N₁²|N₂²}}

Below that, the current falls to zero before the next cycle and the converter is in DCM. Power balance then gives an output that no longer follows the CCM formula:

V_o = V_sD [[{{R|2L_mf}}]] (DCM; no turns ratio)

Ripple with an ideal capacitor is ΔV_o/V_o = D/(RCf). The capacitor current is the diode current minus the load current, a tall pulse, so the ESR often dominates: ΔV_o,ESR = I_Lm,max(N₁/N₂)·r_C.

Why is the secondary current zero while the switch is closed? The polarity of the windings is such that the diode is reverse-biased. All the energy is stored in the magnetic field (L_m), not in the ideal transformer. This is why the transformer in a flyback acts partly as an inductor, and why its core must be larger than a forward converter’s core at the same power.

Flyback lab

The starting point is Example 7-1: 24 V in, 5 V out, N₁/N₂ = 3. Look for three things: the mode badge, the peak voltage across the switch, and what happens to V_o in the last graph as R grows.

Activities

Forward: pass the energy through, then reset the core

The forward converter is the isolated buck converter. When the switch is closed, energy flows straight through the transformer into an output inductor and capacitor. The transformer is not an energy store here, so its magnetizing current has to be sent back to the source, which is the job of a third winding and a diode.

How it works

There are three states. In the first, the switch is on and the secondary delivers power. In the second, the switch is open and the third winding resets the core. In the third (if there is time), the core is already reset, and only the output filter is active.

Where the equations come from

Switch closed. The secondary voltage is V_sN₂/N₁. D1 conducts, D2 and D3 are off, and the filter inductor current rises:

(Δi_Lx)_closed = {{[V_s(N₂/N₁) − V_o]DT|L_x}} Δi_Lm = {{V_sDT|L_m}}

Switch open. The magnetizing current continues through N₃ and D3 into the source, so the primary voltage becomes −V_sN₁/N₃ and the switch blocks V_s(1 + N₁/N₃). D1 turns off and D2 carries the filter current, which falls at V_o/L_x:

(Δi_Lx)_open = −{{V_o(1 − D)T|L_x}} v_sw = V_s(1 + {{N₁|N₃}})

Volt-second balance on L_x gives the buck-converter result with a turns-ratio factor:

V_o = D V_s {{N₂|N₁}}

Core reset

The magnetizing current peaks at V_sDT/L_m and falls at V_s(N₁/N₃)/L_m, so it needs ΔT_x = DT·N₃/N₁ to reach zero. That must finish before the next cycle starts:

D(1 + {{N₃|N₁}}) ≤ 1 ⇒ D_max = {{1|1 + N₃/N₁}}

With N₁/N₃ = 1, a common choice, D must stay below 0.5. A smaller N₃ (larger N₁/N₃) resets faster and allows more duty ratio, but the open switch then blocks more voltage. This is the central trade-off of the forward converter.

Filter current and ripple

The output stage is a buck converter fed by V_sN₂/N₁, so it is continuous when L_x ≥ (1 − D)R/2f, and the ripple is ΔV_o/V_o = (1 − D)/(8L_xCf²) for an ideal capacitor. The ESR contribution is Δi_Lx·r_C, and it often dominates. The primary current at the switch is the reflected load current plus the magnetizing current: I_Lx,maxN₂/N₁ + V_sDT/L_m.

Double-ended (two-switch) forward. Two switches turn on and off together, and two diodes return the magnetizing current to the source. The primary voltage is then −V_s during reset, so each open switch blocks only V_s rather than V_s(1 + N₁/N₃). The core resets for any D below 0.5. The price is a second switch, one of whose drive circuits must float.

Forward lab

The starting point is Example 7-4: 48 V in, N₁/N₂ = 1.5, D = 0.4. Move D above 0.5 and watch the notice and the last graph. Then cut L_x and watch the mode change.

Activities

Push-pull: drive the core both ways

The push-pull converter uses two switches on a center-tapped primary. They conduct in turn, so the core is magnetized in one direction and then the other, which makes the volt-seconds cancel without a reset winding. The output stage is a buck-style filter that sees two pulses per switching period.

How it works

There are three kinds of interval in each period: Sw1 closed, Sw2 closed, and both open (twice, once after each pulse). When both switches are open, no primary current flows, and the filter inductor current divides equally between the two secondary halves. Press Play the cycle to see the full sequence.

Where the equations come from

Sw1 closed. V_s is across P1; the same volts per turn appear on every winding. The secondary half S1 delivers V_sN_s/N_p through D1, and the open switch Sw2 blocks 2V_s. Sw2 closed is the mirror image through D2.

Both open. The secondary voltage is zero and L_x sees −V_o:

(Δi_Lx)_closed = {{[V_s(N_s/N_p) − V_o]DT|L_x}} (Δi_Lx)_open = −{{V_o|L_x}}({{1|2}} − D)T

The inductor sees the pulse twice per period, so volt-second balance gives

V_o = 2D V_s {{N_s|N_p}} (D < 0.5, the duty ratio of each switch)

The factor of 2 appears because the filter behaves like a buck converter running at twice the switching frequency with duty ratio 2D. The same reasoning gives the other results:

L_x,min = {{(1 − 2D)R|4f}} {{ΔV_o|V_o}} = {{1 − 2D|32L_xCf²}} v_sw = 2V_s

The output can approach V_sN_s/N_p as D → 0.5 but never exceed it. The ripple is at twice the switching frequency, which is why a modest capacitor does a good job. The ESR contribution is Δi_Lx·r_C.

Core balance

With ideal, equal switches, the magnetizing current rises while Sw1 conducts, rests while both are open, falls while Sw2 conducts, and returns to its start. If one switch conducts longer than the other, the volt-seconds do not cancel and the flux walks away. In practice controllers use current-mode control, or a series capacitor, or other methods to keep the two halves balanced.

Bridge relatives. The half-bridge and full-bridge converters work like the push-pull, with the same V_o = 2D V_sN_s/N_p in the full-bridge (and half of that in the half-bridge, because each capacitor holds V_s/2). Their advantage is that each open switch blocks only V_s, which is why they dominate high-voltage and high-power designs.

Push-pull lab

The starting point is Example 7-6: 30 V in, N_p/N_s = 2, D = 0.3. Try D near 0.5, then use the on-time mismatch slider and watch the last graph.

Activities

What happens when…

Engineers keep an intuition for how a converter reacts when something changes: a lighter load, a different turns ratio, a mismatched switch. Each card below changes one thing. Make a prediction first, then see what the model does, with before-and-after numbers and graphs. You can also open any scenario in the lab and explore around it.

Flyback scenarios

Forward scenarios

Push-pull scenarios

Choosing a converter

In theory any converter can be designed for any application. In practice some suit particular jobs much better. The usual deciding factors are the power level, the voltage stress on the switches, the part count, and the size of the transformer.

Power and stress

Drag the power slider to see which converters reach it, using the typical ranges given by Hart. Then drag the source voltage to see the stress on an open switch for each. Reduced stress matters a great deal when the input voltage is high.

Which converter reaches this power? How much voltage does the switch block?

Typical use of isolated converters
Converter	Typical power	Voltage across an open switch	Strengths	Weaknesses
Flyback	to about 150 W, most often 10 W or less	V_s/(1 − D), 2V_s at D = 0.5	Low parts count, simple, no output inductor	Transformer core grows quickly with power; high switch stress
Forward	to about 500 W	V_s(1 + N₁/N₃)	One switch; smaller core than a flyback	High switch stress; extra output inductor; D limited by reset
Double-ended forward	to about 500 W	V_s	Lower switch stress, good for high voltage	Two switches, one with a floating drive
Push-pull	to about 1000 W	2V_s	Drives share a common point; core driven both ways, so it is smaller	High switch stress; core saturation if the halves are unbalanced
Half-bridge	to about 500 W	V_s	Similar to push-pull, with low switch stress	Needs two capacitors; floating drive
Full-bridge	to about 2000 W	V_s	Often the choice for high power; low stress	Four transistors and floating drives

Beyond this module

Multiple outputs. Extra secondary windings give extra outputs (for example 12 V, 5 V, and −5 V from a common ground). The duty ratio and each turns ratio set each output, but only one output can be regulated by the feedback loop; the others follow according to the duty ratio and their own loads.

Resonant converters switch at voltage or current zeros, reducing switching loss, so they can run at higher frequencies with smaller components. Power factor correction uses a boost-type stage after the rectifier to make the ac line current nearly sinusoidal; its output (around 400 V) then feeds an isolated converter such as a forward converter. A complete off-line supply is an ac line filter, a power-factor-correction stage, and an isolated dc-dc converter with feedback. These topics, and compensation of the feedback loop, are in the remaining sections of Hart’s chapter.

Worked examples

Each example gives a design or analysis problem. Set the answer sliders, watch the checks and graphs respond, and use Change the givens to ask “what if?”. Show the worked solution when you are ready, and use the question panel to ask why.

Design challenges

Now put it together. Each challenge gives a specification. Choose the turns ratios, the switching frequency, the inductance and a capacitor. Test my design checks every requirement across the operating range, including the hard cases (light load, high line, low line) where a design is most likely to fail.

Check your understanding

A short quiz, three reflection prompts, and a place to collect your responses and any questions you want to bring to class.

Quiz

Reflection

Questions for your instructor

If the question panels in the worked examples could not answer a question, you can save it here.

Download your responses

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Formula reference

Ideal components and steady state. D is the duty ratio of the switch (of each switch for the push-pull converter), T = 1/f.

Equations for the three converters
Quantity	Flyback (N₁/N₂)	Forward	Push-pull
CCM output	V_s{{D\|1 − D}}{{N₂\|N₁}}	DV_s{{N₂\|N₁}}	2DV_s{{N_s\|N_p}}
Key inductor	L_m (magnetizing)	L_x (filter)	L_x (filter)
Current swing	{{V_sD\|L_mf}}	{{V_o(1 − D)\|L_xf}}	{{V_o(1/2 − D)\|L_xf}}
Average current	I_Lm = {{V_o²\|V_sDR}}	I_Lx = {{V_o\|R}}	I_Lx = {{V_o\|R}}
Minimum inductance for CCM	{{(1 − D)²R\|2f}}(N₁/N₂)²	{{(1 − D)R\|2f}}	{{(1 − 2D)R\|4f}}
DCM output	V_sD[[{{R\|2L_mf}}]]	buck DCM with V_sN₂/N₁	buck DCM at 2f, duty 2D
Open-switch voltage	V_s + V_o{{N₁\|N₂}}	V_s(1 + {{N₁\|N₃}})	2V_s
Core	diode resets it	D ≤ {{1\|1 + N₃/N₁}}	balanced halves
Ripple (ideal C)	{{D\|RCf}}	{{1 − D\|8L_xCf²}}	{{1 − 2D\|32L_xCf²}}
ESR ripple	I_Lm,max{{N₁\|N₂}}r_C	Δi_Lxr_C	Δi_Lxr_C

Three rules behind every equation

Volt-second balance on every inductor, including the magnetizing inductance: the average voltage is zero. Capacitor charge balance: the average capacitor current is zero, so the average diode or inductor current feeds the load. Power balance: with ideal parts, the source power equals the load power.

For the source of these results see Hart, Power Electronics, Chapter 7, Sections 7.2–7.8 and 7.10.