Inductive Power Transfer Onsite Lab Instruction

Introduction¶

In this onsite lab practicum you are going to implement, investigate and test a inductive power transfer system. The lab practicum is divided into several steps:

Investigate the possibility of transmitting power with the air core transformer.
Build a air core transformer and test the power transfer capability.
Learn how to compensate the air core transformer to enable it to transmit a considerable amount of power.
Test and evaluate the complete power transfer system with compensation and measure the efficiency.

Figure 1:Main IPT components available for the practicum.

The hardware components shown in Figure 1 are available on the lab table. In the following assignments you are supposed to build-up and test an IPT system step by step using the given hardware.

Assignment 1: measurements on an inverter feeding a passive load¶

Before carrying out this assignment please read Theoretical background: contactless power transfer section of this instruction related to gate drivers.

In this assignment you will get familar with a pre-implemented H-bridge inverter and measure its performance feeding a resistive load.

You are asked to measure and evaluate the following signals:

the PWM switching signals of the H-bridge inverter using the scope.
measure the output current and voltage of the inverter.

You may think about the following questions during the measurements:

Explain why a differential measurement (with two voltage probes with a common ground) should be used to measure that specific signal;
Observe the current going via the inverter and how it changes with changing frequency, and give the reasons why it changes in this way. You can obtain the current by measuring the voltage across the shunt resistor;
Explain why the switches on the same leg (Q1 and Q2, Q3 and Q4) cannot be turned on simultaneously;
Calculate the deadtime (the time interval between one switch turns off and the other switch on the same leg turns on) of the switches by measuring the waveforms.

Assignment 2: design and evaluating the air-core transformer¶

Please read Theoretical background: contactless power transfer section of this instruction for background knowledge to carry out this assignment.

In this assignment you are asked to:

Design and build an air-coupled transformer for the IPT system.
Determine the coupling coefficient and other circuit parameters of the air-coupled transformer with various separation distances.
Analyse the power transfer characteristics of the air-core transformer system. Evaluate the operation of the IPT system with the air-core transformer.

Each group will design and wind their own coils. First consider the design constraints:

The inside diameter of both coils should not be smaller than 5 cm. A small inside diameter not only limits the ability of the coil to capture flux but the inner winding also contributes little to the inductance.
The total inductance of the larger coil should be in the region of around 100 µH and 20 µH for the smaller.

Please use the given Litz wire bundles to implement the two coils. A air-core indcutance calculator can be found in the Winding of the Coils subsection of this instruction to help you determine the coil geometries. After winding the coils using the calculated geometries, you may follow the guidelines below to complete the rest of the assignment:

Measure the coil resistances, self-inductances of the air-core coils, and the mutual inductances between them for a separation distances of 2, 4 and 6 cm using the LCR meter. The method to measure the mutual inductance is given in the Measurement methods subsection of this instruction.
Analyse the power transfer of the circuit. Calculate the power supplied to a load of $10~\Omega$ , the efficiency and power factor of the circuit using the method given in the theoretical background. You may assume a 20 V, 100 kHz source.
Place the air-core transformer into the IPT circuit with 2 cm distance between the coils. Measure the operational waveforms of the converters: the primary current, secondary voltage, ouput current and observe how much power can be delivered to a $10~\Omega$ load.

Think about the following questions if you have successfully performed the experiment and studied the required theory:

Why is the open and short circuit test for a normal transformer not well suited for the air core transformer?
Discuss the method used to determine the mutual inductance using the series opposing and series aiding method.
How does the leakage and magnetising inductance change with an increase in separation distance? Explain.
The problems with delivering power with an uncompensated air core transformer.

Assignment 3: compensation of the air-core transformer¶

Please read Compensation subsection for background knowledge to carry out this assignment.

In this assignment a method of compensating the unwanted characteristics of the air-coupled transformer is to be analysed and implemented. When this compensated system is operating under certain conditions the system might fail due to overcurrents. This failure mode and the systems protection circuitry against this error mode is described and must be tested.

Please follow the guidelines below to complete the assignment:

Design the compensation circuit according to the theoretical background.
Draw the equivalent circuit, derive the transfer function and analyse it. Calculate the power supplied to a load of $10~\Omega$ , efficiency and power factor of the circuit.
Select capacitors from given components according to the calculated compensation, solder them to the board.
Test the compensated transformer in the inverter circuit. Adjust the frequency to the optimal value. Measure the power supplied to the load and compare it to the uncompensated system.

You may think about the following questions after successfully performed the experiment:

Why you can estimate how the air core transformer will react when connected to the inverter by investigating its operation with a 100 kHz sinusoidal source.
How to compensate an air core transformer and why this works.
The potential problem with this compensated system.
How the inverter circuit is protected against faults.

Assignment 4: efficiency testing¶

Once the system is completed and successfully tested, you may bring the setup to the lab instructor to test its DC to DC efficiency with the power analyser. The test can be done at different misalignment conditions.

Theoretical background: contactless power transfer¶

Power electronics¶

Figure 2:The DC-DC converter for contactless power transfer.

The power electronic subsystem of the contactless power transfer system as shown in Figure 2 consists of three main parts, a DC to AC converter (inverter), a high frequency air coupled transformer and an AC to DC converter (rectifier). Furthermore, there are additional sub-circuits for driving the switches, control and circuit protection (not shown in the figure). You must read the preparation documents and design all three components and its subcircuits.

Gate Driver (IRS2001PBF)¶

So far in our analysis of power electronic circuits it was assumed that all switches are ideal. This means that they receive some control signal that has no delays and requires no effort to generate. Furthermore ideal switches can switch instantaneously from the on-state to the off-state and vice versa.

Let’s first consider the control signal to the transistors. A MOSFET requires a voltage to be generated between the gate and the source terminals, when this voltage is above a certain level the transistor will switch on and below this threshold the transistor will be in the off state. Although it is true that the transistor can operate in the linear region between the fully on and fully off state (as used in linear audio amplifiers) in power electronic circuits transistors are operated in either the fully on or the fully off state to minimize losses. The MOSFET will be in the fully off state when the voltage between the gate and the source is 0 V and in the fully on state when (according to the datasheet) the voltage is higher than 10 V.

The first problem with the generation of this voltage is the fact that although it is relatively easy to generate the voltages for the two switches on the bottom (since their sources are connected to ground) it is much more difficult to do so for the top switches since their source terminals are not connected to ground and can therefore be at any voltage. A special circuit is needed that can deliver and keep a voltage of 15 V between the source and gate terminals of the topside switches under all operating conditions. This includes the worst case scenario where the source terminals are effectively connected to the positive side of the source (since the top switches are turned on).

Although it is possible to build this gate drive circuit using discrete transistors it is much more convenient to use a dedicated gate drive integrated circuit such as IRS2001PBF. The datasheet of this gate drive circuit is found on Brightspace. This gate drive circuit is able to deliver and keep the required voltage under all operating conditions. Furthermore, the gate drive circuit can deliver a current of 350 mA to the gate of the MOSFET transistor.

Switching Times¶

The second assumption concerning the switching elements was that they switch instantaneously from the on-state to the off-state. Ten years ago this assumption would have been pretty bad even at the low power levels that we are investigating. However, technology improved much during the last years. The transistors that can be used for your circuit can switch from the on-state to the off-state in less than 5 ns. That is pretty quick. Consider that light does not even travel 1 meter in 3 ns! This is however not true for all power electronic switching elements, some of the large switches such as the 6.5 kV and 500 A IGBT (a type of transistor) can have a switching time of between 4 and 7 $\mu \mathrm{s}$ . However, for your circuit it is safe to assume that the switching time is ideal.

You will have several options available for choosing the switches for the inverter circuit. They differ in switching times and losses when conducting. Think about which is the best choice to keep the losses as low as possible and thus efficiency of the whole circuit as high as possible.

There is one other factor to consider and that is the fact that the control signal takes a certain time to travel from the control board to the MOSFET and again a certain time for the MOSFET to react on the signal. It is possible to calculate this time delay by adding the delay times of theMOSFET and the gate driver together. This data is available on the datasheets.

The PWM Generation and Control Circuit¶

Figure 3:Generating the PWM switching signals for the full bridge converter.

There are several methods to generate the PWM signals for a full bridge converter. The easiest method is shown in Figure 3. Again although it is possible to build a circuit to generate these waveforms using transistors and operational amplifiers it is easier to use a dedicated integrated circuit. We will use UC3525 manufactured by Texas Instruments, again the datasheet for this circuit is found on Brightspace.

Figure 4:A contactless power transfer coils.

Figure 5:The magnetic field distribution in the air core transformer.

Air-coupled transformer¶

For the contactless power transfer an air-core transformer will be used. As with any transformer the system consists of two coupled inductors. However, the main difference is the absence of a welldefined magnetic circuit. This lack of a low-reluctance path for the magnetic flux to couple the primary and secondary windings together results in a magnetically loosely coupled system.

Winding of the Coils¶

The basic shape of the winding that will be used is shown in Figure 4. Having this flat shape allows for a little more flexibility in the setup of the contactless power transfer. If the coils are made long and slim instead of flat and wide (as the one shown in Figure 4) then the alignment of the coils would be more difficult. To illustrate this point, consider Figure 5, here two coils of the shape of Figure 4 are placed above one another and the distribution of the magnetic flux is superimposed on the image. Although it can be seen that there is a fair amount of flux generated by the coil in the bottom that does not couple with the coil on top, it can also be seen that the relatively flat shape of the coil helps to capture as much flux as possible.

Each group will design and wind their own coils. First consider the design constraints. The inside diameter of both coils should not be smaller than 5cm. A small inside diameter not only limits the ability of the coil to capture flux but the inner winding also contributes little to the inductance. The total inductance of the larger coil should be in the region of around $\mathrm{100~\mu H}$ and $\mathrm{20~\mu H}$ for the smaller. A special type of high frequency wire, called Litz wire, that consists of many small strands bound together will be supplied for the coils. The total inductance and the length of wire needed can be estimated using the air-core inductance calculator located at http://www.pronine.ca/spiralcoil.htm using the parameters of the supplied Litz wire. You can now calculate the size of the inductor for different inner diameters and inductance values.

Measurement methods¶

Consider the coupled inductors shown in Figure 6. Let $L_1$ and $L_2$ be the self inductance of the two coils and $M$ be the mutual inductance of the two coils. You already know that we can model this circuit with

\begin{bmatrix} v_1\\ v_2 \end{bmatrix} = \begin{bmatrix} L_1 & -M\\ M & -L_2 \end{bmatrix} \cdot \frac{d}{dt} \begin{bmatrix} i_1\\ i_2 \end{bmatrix}

(1)

where

M = k \sqrt{L_1L_2}.

(2)

Figure 7:Series-aiding (a) and series-opposing (b) connection of coupled inductors.

Consider the circuits shown in Figure 7. The inductors are connected in series in both circuits, however, in the one instance the induced voltages are in the same direction while in the other they oppose one another. Measuring the inductance of these circuits as well as the self-inductances of $L_1$ and $L_2$ gives enough information to calculate the coupling coefficient.

Lets define $L_{s}$ to be the total inductance of the two coils connected in the series aiding configuration. Finally, let $L_o$ be the total inductance of the two coils connected in the series opposing configuration.

When the two coils are connected in series (irrespective of whether they are connected in the aiding or opposing configuration) they share the same current, $i$ . If $v_1$ is the voltage across $L_1$ and $v_2$ is the voltage across $L_2$ the following can be written

\mathbf{V_1} = j\omega L_1\mathbf{I}+ j\omega M\mathbf{I}

(3)

\mathbf{V_2} = j\omega L_2\mathbf{I} + j\omega M\mathbf{I}

(4)

when the coils are connected in the series aiding configuration. Using these equations one can calculate the voltage across the two coils when connected in the series aiding configuration, $v_a$ , as

\begin{align} \mathbf{V_a} &= \mathbf{V_1}+ \mathbf{V_2} \nonumber \\ &= (j\omega L_1 \mathbf{I} + j\omega M\mathbf{I} ) + (j\omega L_2 \mathbf{I} + j\omega M \mathbf{I} ) \nonumber \\ &= j\omega (L_1 + L_2 + 2M)\mathbf{I} . \end{align}

(5)

Since the voltage across the the two coils connected in the series aiding configuration is related to the total inductance of the two coils as

\mathbf{V_a} = j\omega L_s\mathbf{I}

(6)

you now know that

L_s = L_1 + L_2 + 2M.

(7)

You can prove that for the series opposing connection of the two coils the total inductance of the two coils are

L_o = L_1+L_2-2M.

(8)

Using these equations you can now calculate the mutual inductance as

M = \frac{L_s-L_o}{4}

(9)

and since

k = \frac{M}{\sqrt{L_1L_2}}

(10)

you can now calculate the coupling coefficient.

In this discussion you have made a number of simplifying assumptions, most notably you have ignored the winding resistances. Although this simplified model gives you enough information to understand the operation of the system it can not describe all the behaviour of the system, such as the efficiency for example. In the master’s course methods of analysing the resistance and describing it as a function of frequency as well as methods to generate detailed models of the magnetic characteristics will be addressed.

Power transfer characteristics¶

Due to the loose coupling of the inductors when the air core is used the operation of the DC-DC converter will be quite different. In a full bridge DC-DC converter with a ferrite transformer the coupling between the two coils is very high and one could ignore the transformer power transfer characteristics. Although it is very difficult to analyse the circuit operation in the time domain using the techniques you have used thus far when analysing power electronic converters you can use a sleight of hand to get you out of trouble. Remember that since the switching waveform of the inverter is periodic you can rewrite the time-domain waveform in the frequency domain using the Fourier series. You can now analyse the operation of the transformer circuit and assume that the source is sinusoidal. Although you theoretically have to calculate the currents and voltages for each and every component of the Fourier series to get the full picture of what will happen when you connect the air core transformer to the inverter you can get a very good idea of how well the transformer is working by investigating only the fundamental frequency component. If the transformer is able to deliver a sensible amount of power at the fundamental component chances are good that it will be able to do so with the full waveform.

Figure 8:Basic coupled inductor power transfer circuit.

Consider the coupled inductor circuit shown in Figure 8. Since you are interested in the fundamental component you can assume that the frequency of the source is 100 kHz. Recall that for the coupled inductors you can write the following equation, in the frequency domain,

\begin{bmatrix} \mathbf{V_1}\\ \mathbf{V_2} \end{bmatrix} = j\omega \begin{bmatrix} L_1 & -M\\ M &- L_2 \end{bmatrix} \begin{bmatrix} \mathbf{I_1}\\ \mathbf{I_2} \end{bmatrix} .

(11)

Using this expression you can write the Kirchhoff voltage loop law equations for the system in Figure 8 as

\mathbf{V_{in}} = j\omega L_1 \mathbf{I_1} - j\omega M \mathbf{I_2}

(12)

\mathbf{I_2}R = -j\omega L_2 \mathbf{I_2} + j\omega M \mathbf{I_1}

(13)

By solving these equations we is able to obtain the active power transfer to the secondary side. below 0.4.

P_2=\frac{V_{in}^2RL_2/L_1}{R^2/k^2+\omega^2(k-1/k)^2L_2^2}

(14)

We can see very little power is transmitted to the load resistance if the coupling coefficient is low, say below 0.4.

When you include the winding resistances, as in the circuit of Figure 9 the efficiency of the system is very low. You can actually expect that since you can see that the power factor of both systems in Figure 8 and Figure 9 is very low.

Compensation¶

In the previous section, while analysing the air-coupled transformer, you have seen that the power transfer characteristics of such a transformer are poor, because the power factor of both sides in Figure 9 is very low. In this assignment, you need to look for a solution to this problem.

Figure 10:Basic compensated coupled inductor power transfer circuit.

A common approach to improve the power factor of the inductive circuit is adding compensation capacitors. As shown in Figure 10, capacitors C1 and C2 are added in series to the primary and secondary circuit respectively. Similarly to (11)-(13), by applying the Kirchhoff’s law, you can write

\begin{align} \mathbf{{{V}}_{in}}=j\omega {{L}_{1}}{{\mathbf{I}}_{1}}-j\omega M{{\mathbf{I}}_{2}}+\frac{{{\mathbf{I}}_{1}}}{j\omega {{C}_{1}}} \\ \mathbf{{{I}}_{2}}R=-j\omega {{L}_{2}}{{\mathbf{I}}_{2}}+j\omega M{{\mathbf{I}}_{1}}-\frac{{{\mathbf{I}}_{2}}}{j\omega {{C}_{2}}} \end{align}

(15)

The Thévenin’s equivalent impedance of the primary side is calculated from (15) as

\mathbf{{\mathbf{Z}}_{eq}}=\frac{{{\mathbf{V}}_{in}}}{{{\mathbf{I}}_{1}}}=j\left( \omega {{L}_{1}}-\frac{1}{\omega {{C}_{1}}} \right)+\frac{{{\omega }^{2}}{{M}^{2}}}{R+j\left( \omega {{L}_{2}}-\frac{1}{\omega {{C}_{2}}} \right)}.

(16)

To transfer maximum power with certain voltage and current, the Thévenin’s equivalent impedance should only have real component. Therefore, the two $j$ components in (16) both should be 0, which gives

\omega =2\pi f=\frac{1}{\sqrt{{{L}_{1}}{{C}_{1}}}}=\frac{1}{\sqrt{{{L}_{2}}{{C}_{2}}}}.

(17)

As you can see, $L_1$ and $L_2$ now resonant with $C_1$ and $C_2$ respectively at frequency $f$ . Then the primary side Th’evenin’s equivalent impedance becomes

\mathbf {{{Z}}_{eq}}=\frac{{{\mathbf{V}}_{in}}}{{{\mathbf{I}}_{1}}}=\frac{{{\omega }^{2}}{{M}^{2}}}{R}

(18)

Now let’s calculate the power transfer efficiency of the above fully compensated case, the primary current can be written as

{{I}_{1}}=\left| \frac{{{\mathbf{V}}_{in}}}{{{\mathbf{Z}}_{eq}}} \right|=\frac{{{V}_{in}}R}{{{\omega }^{2}}{{M}^{2}}}.

(19)

Substitute (19) and (17) into (15) you can get the secondary current is

{{I}_{2}}=\frac{{{V}_{in}}}{\omega M}.

(20)

The output power is calculated as

P_2 = I_2^2R=\frac{V_{in}^2R}{\omega^2M^2}

(21)

As we can see from (18) to (21), when the load and the input voltage are fixed, the current in the coils and the transferred power is determined by the switching frequency and the mutual inductance, hence the power transfer efficiency. In practice the switching frequency is often tuned according to the load to realize a high efficiency power transfer by the so-called ”impedance matching”.

The other problem for the fully compensated case is the misalignment. Consider what would happen if the secondary is removed from the primary or the two coils are heavily misaligned. The mutual inductance $M$ becomes close to 0. According to (18) and (19), the primary side current will be extremely high since the equivalent impedance is too low. Large current may damage the inverter circuit. Therefore a current sensor is added to the main current path. It measures the current and protects the inverter in case of overcurrent.