Active Clamping Forward Converter: Full Guide

Active clamping is a method to protect power electronics switches against induced back-EMF energy when the switch going to be turned off in inductive loads.

The problem is that the switch is unable to be turned off immediately because the inductive current cannot be zero sharply. On the other hand, the switch should tolerate a considerable overvoltage due to sharp current disruption. Various techniques have been proposed to solve this problem, which have their pros and cons.

In this article, active clamping forward converter is discussed, and operation modes and waveforms are presented. The following sections explain its application and design briefly, and finally, a basic example is simulated as the sum up.

Forward Converter

Basically, when a transformer is inserted at the middle of a buck converter, a forward converter is produced. The transformer isolates the source side from the load side and mostly has a unity turns ratio. As can be seen in the next figure, there is :

• a MOSFET,
• a transformer,
• two diodes,
• an LC filter together with the load.

This circuit has two modes based on MOSFET (Q₁) conditions.

Mode 01 (Q1 is turned ON)

The current flows through the transformer primary, and the induced voltage is produced on the secondary side. In this situation, the diode D1 will be in forwarding bias, and the energy is stored in the inductor (inductor charging), and the required current for load is supplied, as shown in the next figure. The diode D₂ is turned off at this stage, and there is no current in the parallel diode.

Mode 02 (Q1 is turned OFF)

In this step, the switch Q1 and diode D1 are off, and the energy stored in the inductor L1 is discharged to the circuit, as can be seen in this figure:

All steps are clear, but this circuit has a significant issue owing to inductive current disruption. When the switch is going to be turned off, the inductive current in the transformer primary would be cut out, which imposes stress on the switch because the inductive current cannot jump to zero. As a matter of fact, the current has no path to go in the transformer primary. Many techniques have been presented to solve this problem, which will be discussed in the following section.

How to prevent sharp inductive current in forward converter?

Solution 01

One of the solutions is adding a tertiary winding to the transformer series with an anti-parallel diode. In this way, the stored energy in the transformer primary is fed back to the source by tertiary winding, and there will be no sudden change in primary current when the switch wants to be turned off.

However, adding a winding to the transformer increases the transformer volume, and more space is needed. Moreover, the switch voltage in this condition will go beyond the twofold input voltage, which is dangerous and may damage the switch significantly.

Solution 02:

Another method to reset the transformer is an active clamp forward converter. This technique needs a MOSFET and a capacitor and can be added to the circuit in two ways, including high-side active clamping and low-side one. Both configurations are illustrated in this article, but the low-side active clamping converter is discussed in detail.

The previous figure illustrates a high-side active clamp forward converter in which the capacitor and switch are in series, and they are connected to the primary winding in parallel. An N-channel MOSFET must be used in this configuration, which requires a floating gate drive signal. This situation will increase the circuit costs and complexity. Thus, we can move to the low-side active clamp forward converter.

Low side Active Clamp Forward Converter

A P-channel MOSFET is used in series with a capacitor in a low-side active clamp forward converter. The package of switch and capacitor is connected to the main switch (Q₁) in parallel, as shown below.

In P-channel MOSFET, when the gate is grounded, the switch is on, and when a voltage pulse is injected between gate and source terminals, the switch will be turned off. The conditions are opposite for an N-channel MOSFET.

Waveforms of Active Clamp Forward Converter

For better understating, waveforms are presented below:

• In the first step (from 0 to T₁), switch Q₁ is turned on, and its applied voltage is V_Gate(Q1), and the current flows in the transformer secondary as discussed before. Moreover, the voltage across the switch (V_DS(Q1)) is zero. The other MOSFET is turned off because the P-channel MOSFET voltage is V_Gate(Q2). In this condition, all currents go up linearly, as shown in the figure, because of the output inductor. Moreover, the output voltage is a constant DC voltage.
• In the next step, Q₁ is turned off, and the inductor discharges its energy to the circuit (inductor current decreases, but the output voltage is constant). In this situation, the transformer primary current flows through the clamp circuit and charges the capacitor. It means that from T₁ to T₂, the switch voltage will increase. Switch current also decreases to zero in this time slot. However, the primary current should charge the capacitor at first and then reach zero at T₃.
• After charging the capacitor, the polarity across the primary winding is reversed, and the capacitor starts to be discharged. Therefore, the primary current becomes negative, and the switch voltage reduces slightly until T₄. Then, switch Q₂ is turned off, and the waveforms continue as explained.

How to design a forward converter?

As mentioned before, the forward converter is a buck converter with a transformer. Therefore, the converter design is similar to the buck converter, and only a transformer must be designed additionally.

This section briefly explains a forward converter with a tertiary winding design. As can be seen in below, there is an additional winding in the transformer to create a path for inductive current in transformer primary and reset the transformer.

Tertiary turns number

Considering tertiary winding with N_t turns, the maximum duty cycle would be:

\frac{D_{max}}{1-D_{max}}=\frac{N_p}{N_t}

In this case, a trade-off must be done in the selection of N_t because small value will lead to higher D_max and smaller output inductance. Moreover, higher N_t will result in lower voltage stress across the switch as shown in the following equation.

V_{sw,max}=V_{in}\left( 1+\frac{N_p}{N_t}\right)

Hence, this value should be chosen carefully. A popular selection is N_t=N_p , and it proves that the switch voltage stress would be at least twice the input voltage as mentioned in previous sections.

Transformer turns number

Apart from the tertiary turns number, the transformer primary to secondary ratio (n)  is also important. The output voltage is a function of duty cycle and transformer turns ratio, and it can be written as:

V_o=nDV_{in}

By maximizing the duty cycle, the inductance value and switch current will be minimized, and lower losses will be generated. Hence, the transformer turns ratio can be calculated by:

n=\frac{V_o}{D_{max}V_{in,min}}

The abovementioned equation is for an ideal transformer. However, leakage inductance and winding resistances are available. For this reason, a margin factor is considered namely k_m and the equation is modified to:

n=\frac{V_o}{k_mD_{max}V_{in,min}}

where k_m is in a range of 0.9 to 0.95. Considering N_t=N_p, the maximum duty cycle would be 0.5.

Output Filter design

The filter is a combination of a capacitor and an inductor that determine the output voltage ripple. The inductor also limits the inductor current ripple, and its value will be determined by:

L=\frac{V_o(1-D_{min})T_s}{\Delta I_L}

where

I_{L,peak}=I_{o,max}+\frac{\Delta I_L}{2}

The capacitor value is selected based on the output voltage ripple, inductor current ripple and switching frequency:

C=\frac{\Delta I_L}{\Delta V_o}\frac{1}{8f_s}

Diodes Choice

Now, diodes must be chosen based on peak voltage, average and peak currents. For D₁, the parameters are:

V_{D_1, peak}=\frac{N_s}{N_t}V_{in,max}

and

I_{D_1, average}=D_{max}I_{o,max}

I_{D_1, peak}=I_{L,peak}

For the second diode (D₂), we have:

V_{D_2, peak}=nV_{in,max}

and

I_{D_2, average}=(1-D_{min})I_{o,max}

I_{D_2, peak}=I_{L,peak}

In this way, all needed parameters can be obtained at the design stage. It should be noted that each converter has its specifications, and the presented formulas must be modified based on the converter specifications. In the next section, an example of a forward converter is presented.

‌Basic simulation of a forward converter

This section presents a simple forward converter in Simulink/MATLAB. In this example, the input voltage is 170 V, and the switching frequency is 300 kHz. The secondary voltage and inductor current are 5 V and 5 A, respectively. Assuming the turns ratio equal to 12, the duty cycle would be 0.353 based on the equation below:

\frac{N_1}{N_2}=\frac{V_s}{V_o}D

Considering 40 percent inductor current variation, the required inductance would be equal to 5.39µH. The calculation can be done as below:

\Delta I_L=0.4*5=2A

and

L=\frac{V_o(1-D)T_s}{\Delta I_L}=5.93 \mu H

Similarly, assuming 1% output voltage ripple, a 400 µF capacitor can be selected. The simulated circuit is presented below:

The output voltage would be 4.17 V which can be seen below:

All in all, the forward converter is a buck converter with an inserted transformer in between. To prevent sudden change current when switch is in turned-off condition, active clamp method and tertiary winding can be implemented. The design of the forward converter is similar to the buck converter and has been explained in this article.