Double inverter

In electronics, a double inverter is an electronic circuit that can convert an electrical direct voltage into another electrical direct voltage. The amount of the output voltage can be both smaller and larger than the level of the original input voltage. As with the simple inverse converter , the output voltage has a negative sign compared to the input voltage, the output voltage is inverted. The double inverter is thus included in the group of inverting DC voltage converters .

construction

Block diagram of the double inverter with switch.

As the Cuk converter , the Zeta converter and the SEPIC converter is one of the double inverter with three active energy stores , two coils and a capacitor , in contrast to simple switching converters such as the boost converter , the buck converter and the buck-boost converter, a higher to a group of DC-DC converter order . As with any DC / DC converter, the converter’s energy stores are continuously charged with energy and discharged cyclically with the aid of semiconductor switches such as transistors and diodes .

In contrast to the functionally identical Ćuk converter, there are semiconductor switches on both the input side and the output side of the converter. The current flow from the input and output is thus constantly interrupted, so both the output current and the input current are discontinuous. Accordingly, large backup capacitors have to compensate for the associated voltage ripple . In practical implementations, the double inverter has only a subordinate role and can be replaced by the better converter topology of the Ćuk converter .

Mathematical description and function

Possible realization of the double inverter.

For the mathematical description of the double inverter it is assumed that all capacitors are very large and that the converter is in the steady state with a constant key ratio .

The current in the inductances fluctuates around a certain value Δi per switching period and is therefore zero on average.

The voltage across an inductance results from:

${\displaystyle u_{\mathrm {L} }\left(t\right)=-L\cdot {\frac {\mathrm {d} i_{L}\left(t\right)}{\mathrm {d} t}}}$

Accordingly, the mean values ​​of the voltages across the two inductances per switching period must also be zero.

The mesh equation over the inductance L1 for the view from the input circuit results in:

${\displaystyle -U_{\mathrm {in} }+U_{\mathrm {L1} }=0}$

${\displaystyle U_{\mathrm {L1} }=U_{\mathrm {in} }}$

The mesh equation over the inductance L1 for the view from the output circle results in:

${\displaystyle -U_{\mathrm {L1} }+U_{\mathrm {C} }+U_{\mathrm {out} }=0}$

${\displaystyle U_{\mathrm {L1} }=U_{\mathrm {c} }+U_{\mathrm {out} }}$

If the duty cycle d is now taken into account, the respective voltage is applied for duration d (input side, transistor conducts) or for duration 1-d (output side, diode conducts). Accordingly, adding both equations, taking into account the duty cycle, results in the mean voltage across the inductance, which must be zero.

${\displaystyle U_{\mathrm {L1} }=U_{\mathrm {in} }\cdot d+(1-d)\cdot (U_{\mathrm {C} }+U_{\mathrm {out} })=0}$

The mean voltage across the capacitor C is thus:

${\displaystyle U_{\mathrm {C} }=-U_{\mathrm {in} }\cdot {\dfrac {d}{1-d}}-U_{\mathrm {out} }}$

The voltage at the inductance L2 results analogously to:

${\displaystyle U_{\mathrm {L2} }=d\cdot (U_{\mathrm {in} }-U_{\mathrm {C} })+(1-d)\cdot U_{\mathrm {out} }=0}$

The voltage across the capacitor C can also be expressed from the second equation :

${\displaystyle U_{\mathrm {C} }=U_{\mathrm {out} }\cdot {\dfrac {1-d}{d}}+U_{\mathrm {in} }}$

If you now equate the two equations of the capacitor voltages, you get the output voltage of the double inverter depending on the input voltage and the pulse width ratio:

${\displaystyle U_{\mathrm {out} }=-U_{\mathrm {in} }\cdot {\dfrac {d}{1-d}}}$

conclusion

If you now insert the equation for the output voltage into the first equation for the capacitor voltage, you get:

${\displaystyle U_{\mathrm {C} }={\bigg (}-U_{\mathrm {in} }\cdot {\dfrac {d}{1-d}}{\bigg )}-{\bigg (}-U_{\mathrm {in} }\cdot {\dfrac {d}{1-d}}{\bigg )}}$

As can be seen, the average voltage per switching period across the capacitor is zero. Since the capacitor was assumed to be very large, it can also be assumed that the voltage will only change very slightly during a switching period. If this fact is simplified, the voltage across the capacitor is zero at all times, which means that the capacitor can be omitted.

It turns out that the two inductors are now connected practically in parallel and act as if it were a single inductor. Accordingly, the double inverter can be regarded as an ordinary inverse converter.

Since the double inverter thus has the same properties as the inverse converter and offers practically no advantages, this converter topology is of no importance in practice and therefore represents more of a theoretically possible converter topology.

See also

• Flyback converter
• Single ended flux converter
• Push-pull flux converter

literature

• Franz Zach: Power Electronics: A Manual. 2 volumes, 4th edition, Springer-Verlag, Vienna, 2010, ISBN 978-3-211-89213-8