Switching scheme in an inverter

[cnh1995]

I am studying single phase and 3-phase inverter circuits and I have a few questions.
In textbooks, while explaining how the inverter functions, they specifically mention the type of load on the inverter (R, RL, RC or RLC). The switching duty of the transistors changes according to the type of load and values of load components.
e.g. To get square wave ac output in a full bridge VSI feeding a purely inductive load, the transistors conduct for half the time period and freewheeling diodes (transistor body diodes) conduct for the other half.

If the load is RL type, diodes conduct for a time corresponding to the RL power factor angle and transistors conduct for the remaining time (out of 360 degrees). If the load is purely resistive, the diodes don't conduct at all. Basically, what I understood from their analysis is that the switching duty of the transistors depends directly on the load power factor.

I get the mathematical analysis and why the switching duty should change with the type of load.
But is that what happens in practice? Does a practical inverter (can consider simplest and cheapest home inverter) have any smart arrangement to "know" the power factor of load as soon as it gets connected?
If no, how is transistor-switching managed in the inverter when the load pf keeps changing?


Thanks in advance.

 

[scottdave]

It's been awhile since I've looked at one of these circuits, but from what I recall from power electronics, the transistor "knows" when to conduct because of the bias voltage on the base, determines the current flow.

 

[DaveE]

Your typical inverter isn't "smart" in that way. They are designed to operate over a range of load conditions. They will monitor the load (or switch) current either for protection (i.e. fault conditions), or often as part of the control law for the transistor drive. However, the controller won't know why the current is doing what it is doing, it will just tolerate it or protect it's self.

 

[DaveE]

Of course, the freewheeling diodes aren't controlled, they conduct, or not, solely based on the current direction imposed on them from the rest of the circuit (the load, mostly). For reactive loads you could have a MOSFET turned on at the gate, but the current may be flowing backwards through the body diode; this will happen automatically if that is the current polarity at the time.

 

[cnh1995]

this will happen automatically if that is the current polarity at the time.

Yes, I understand that the freewheeling would occur automatically based on reactive current direction at that instant. Please consider the following scenario:
When my home inverter got switched on due to mains power failure, I had only light bulbs turned on. So the inverter "adjusted" its MOSFET duty ratio accordingly and started feeding the light bulb-load.
After a few minutes, I turned on a fan.
Now the inverter has to detect this change and re-adjust ('reduce' in this case) its duty ratio to account for the reactive power (by allowing freewheeling action).
In both the cases, the frequency has to be maintained at 60Hz.
Does it have any special circuitry for this re-adjustment?

 

[Baluncore]

Does it have any special circuitry for this re-adjustment?

No.
The transistors are switched many times per half cycle of the output.
The output of an inverter has an LC low-pass filter for each phase.
The high frequency output switches are always driving the inductive LPF.

The transistors are “hard commutating” controlled switches.
The diodes are “soft commutating” automatic switches.
When a “hard” output switch is turned off, the inductive load current will continue to flow through one of the “soft” diode switches provided.

The controller does not need to know the power factor of the load.

When needed, the controller increases the output voltage of the LPF by turning on an available hard output switch, then “free-wheels” through the soft diodes once the output voltage is too high.

The switch duty cycle is therefor an indirect effect of the load PF.

 

[DaveE]

Does it have any special circuitry for this re-adjustment?

No, it just has to adjust (a little bit) to provide the desired output voltage waveform. The current will do what the load determines.

 

[cnh1995]

The transistors are switched many times per half cycle of the output.
The output of an inverter has an LC low-pass filter for each phase.
The high frequency output switches are always driving the inductive LPF.

I am not sure I understand this circuit. Do you have any circuit diagram for this? Or any link?
The one I am studying looks like this.

It is a basic single phase full bridge VSI. The control scheme is such that the output voltage should be a square wave of amplitude Vs and frequency 50Hz.
As per my textbook, if L=0 in the load (purely R load) switching is done such that T1 T2 conduct for 10ms, T3 T4 conduct for next 10ms.

If R=0, L is finite: T1, T2 conduct for first 5 ms, then they turn off. Load voltage in this duration is +Vs. Freewheeling diodes D3, D4 conduct for the next 5ms, load voltage during this time is -Vs. In these 10ms, inductor has charged and discharged once, with a triangular current waveform. Same cycle repeats with T3 T4 and D1 D2 in next 10ms.

So this inverter is adjusting transistor conduction time as per the load.
Is this inverter in my diagram practically designable with a transistor control logic that changes the switching time dynamically as per the load, ensuring square wave output of magnitude Vs and frequency 50 Hz ?


Or is this just a simple introductory model designed to illustrate freewheeling and commutation concepts?

Sorry if this sounds repetitive. In my book, there is so much focus on type and nature of load while analysing these basic inverter configuations and switching schemes. So I couldn't help wondering about how it is taken care of in the real world, where you don't know the nature of the load.

 

[DaveE]

Or is this just a simple introductory model designed to illustrate freewheeling and commutation concepts?

Yes.

 

[Baluncore]

If R=0, L is finite: T1, T2 conduct for first 5 ms, then they turn off. Load voltage in this duration is +Vs. Freewheeling diodes D3, D4 conduct for the next 5ms, load voltage during this time is -Vs. In these 10ms, inductor has charged and discharged once, with a triangular current waveform. Same cycle repeats with T3 T4 and D1 D2 in next 10ms.

So this inverter is adjusting transistor conduction time as per the load.

Not if the output is a 50 or 60 Hz sine wave. The switching frequency will be over 1000 times higher than the sinewave generated. The switch conduction times will be more like 5 usec.

The output of an H-bridge will be a high frequency rectangular wave, but the low frequency voltage and load current after the LPF will be determined by the external load impedance.

Your simplified circuit ignores the output LPF and RF suppression.
Here is an H-bridge inverter with the LPF.
https://www.researchgate.net/figure/Full-H-bridge-and-driver-circuits_fig3_261459665

 

[DaveE]

Not if the output is a 50 or 60 Hz sine wave. The switching frequency will be over 1000 times higher than the sinewave generated.

Yes. This is what I think of as an inverter. At least a good inverter.

However, there are cheap inverters, and good AC motor drives, that output square waves, or crude approximations of sine waves that don't switch at high frequencies. Like the drawing below.

If, for example, you know that your load is a motor, which is inductive and doesn't respond quickly to voltage steps, then there isn't too much to gain by providing a sine wave (the motor is the LPF). This, of course, would be a bad approach for a capacitive load.

My point is that both methods exist, and you need to understand which sort of approach you are analyzing. For the most simplified schematics, it can be hard to tell the difference. They both have transistors switching the load on and off.

 

[cnh1995]

Thanks @DaveE and @Baluncore for the informative replies.

 

[Windadct]

The dynamics of a Square Wave (Direct) inverter and a PWM regulated Sinewave inverter are pretty different.

For a Square Wave inverter, the Output and Switching frequency are the same, Fsw=Fout. And for just the reasons you mention, estimating the losses and how they balance between the switching device and the free-wheeling diode is dependent on the current waveform.

So the focus you mention in your textbook makes sense, also note that this topology is used extensively in DC/DC converters where the load parameters are known to the designer since the high frequency transformer is part of the design.

As for line frequency inverters of this type ( ~50 or 60 hz Fout), yes some cheap inverters may use this, but in the motor drive market for a voltage source inverter they are pretty much all PWM-sine wave inverters. ( Current sourced motor drives are another animal entirely). The Square wave pulse to the motor is very hard on the motor and it's windings.

For PWM Sine Wave inverters, the low voltage (< 300 VDC or so) inverters using MOSFETs the Switching Frequency ( Fsw) can be quite high, as mentioned 1000x the output frequency ( Fout), but for higher voltage IGBT systems the ratio can be pretty low about 20x ( Fsw = Fout * 20) , to help manage the switching losses.

 

[cnh1995]

The dynamics of a Square Wave (Direct) inverter and a PWM regulated Sinewave inverter are pretty different.

For a Square Wave inverter, the Output and Switching frequency are the same, Fsw=Fout. And for just the reasons you mention, estimating the losses and how they balance between the switching device and the free-wheeling diode is dependent on the current waveform.

So the focus you mention in your textbook makes sense, also note that this topology is used extensively in DC/DC converters where the load parameters are known to the designer since the high frequency transformer is part of the design.

As for line frequency inverters of this type ( ~50 or 60 hz Fout), yes some cheap inverters may use this, but in the motor drive market for a voltage source inverter they are pretty much all PWM-sine wave inverters. ( Current sourced motor drives are another animal entirely). The Square wave pulse to the motor is very hard on the motor and it's windings.

For PWM Sine Wave inverters, the low voltage (< 300 VDC or so) inverters using MOSFETs the Switching Frequency ( Fsw) can be quite high, as mentioned 1000x the output frequency ( Fout), but for higher voltage IGBT systems the ratio can be pretty low about 20x ( Fsw = Fout * 20) , to help manage the switching losses.

Very helpful! Thank you.

 

[Windadct]

OH - I did forget a bit about BLDC motors - they are three terminal, you could say "phase" motors, that DO use the inverter in Fout=Fsw setup, these are typically servo type, high speed applications, and use a trapezoidal supply voltage (and resulting current). This may look the same as the IPM motors that are designed to use a sine input.

 

[cnh1995]

But is that what happens in practice? Does a practical inverter (can consider simplest and cheapest home inverter) have any smart arrangement to "know" the power factor of load as soon

May be I was overthinking this. Say you want a 50Hz square wave at the output.
The transistors can be gated at 10ms intervals. Whether they will conduct or not will depend on whether there is any current through the freewheeling diode or not.
So in lagging loads, even if the transistors are gated for 10ms, they will conduct for less than 10ms (depending on load power factor). The freewheeling diodes will conduct for the remaining time. A steady state is reached where fundamental componens of voltage and current have the desired phase difference and we get a 50Hz square wave without knowing the load power factor.

There will be unnecessary gate losses in the transistors when they don't conduct during freewheeling. This can be avoided by gating the transistors using zero-crossing detector for the load current.

 

[Windadct]

It is really about the current, and the load.

When T1/T2 turn off the current commutates to D3/D4, for this period energy is being delivered from the load back to the DC link, as this current reaches zero and then starts via T3/T4 now the load current is reversed.

Personally I think that the understanding that energy is flowing INTO the inverter helps with the understanding here.

So when the Transistors are turned on, the current remains flowing in the opposite direct through the FWD(bodydiode, APD, etc...), from the previous half-cycle.

As for Gate losses - I am only dealing with insulated gate devices(MOSFET and IGBT), so the "gate losses" are minimal(looks like a nice capacitor), and the switching losses are different.

 

[shjacks45]

Of course, the freewheeling diodes aren't controlled, they conduct, or not, solely based on the current direction imposed on them from the rest of the circuit (the load, mostly). For reactive loads you could have a MOSFET turned on at the gate, but the current may be flowing backwards through the body diode; this will happen automatically if that is the current polarity at the time.

When drain is inverted but gate is turned on (normal operation of a switching regulator) the current flows through the drain-source channel because the body diode has a higher forward voltage and higher forward resistance.

 

[Baluncore]

When drain is inverted but gate is turned on (normal operation of a switching regulator) the current flows through the drain-source channel because the body diode has a higher forward voltage and higher forward resistance.

To put it another way.
If gate voltage can be arranged to turn the MOSFET channel on, while the parallel diode is conducting, the channel effectively short-circuits or shunts the parallel diode and so can reduce voltage and power dissipation in the package by a factor of about ten. That is called synchronous rectification and can significantly increase the efficiency of high-current and low-voltage inverters.