"Troubleshooting and Solutions for Inverter Voltage Drop Issues"
Inverter voltage drop becomes a significant problem whenever PWM is used in an inverter to enable a sine wave output, especially if the parameters are not calculated properly.
You may have seen a lot of concepts for sine wave and pure sine wave inverters employing PWM feeds or SPWM integrations on this website. The idea is excellent and gives the user the necessary sine wave equivalent outputs, but they appear to have problems with output voltage loss under load.
This post will teach us how to rectify this using straightforward reasoning and arithmetic.
First, we must understand that an inverter's output power is just a function of the input voltage and current being given to the transformer.
Therefore, in this case, it is crucial to confirm that the transformer is appropriately rated to handle the input supply in order to create the needed output and maintain the load without dropping.
In the discussion that follows, we'll attempt to calculate the best way to solve this problem by setting the parameters correctly.
Mitigating Inverter Voltage Drop: Causes, Implications, and Solutions
Square Wave Inverter Output Voltage Analysis
The waveform as illustrated below is generally seen across the power devices in a square wave inverter circuit, which delivers the current and voltage to the appropriate transformer winding in accordance with the mosfet conduction rate utilizing this square wave:
Here, we can observe that the waveform's equal ON/OFF times correspond to a peak voltage of 12V and a duty cycle of 50%.
To continue the analysis The average voltage induced across the appropriate transformer winding must first be determined.
If we use a center tap 12-0-12V /5 amp transformer and apply 12V @ 50% duty cycle to one of the 12V windings, we can determine the average voltage induced within that winding and MOSFET drain as follows:
12 x 50% = 6V
Since the oscillator applies a 50% duty cycle to the MOSFET gates, this also becomes the average voltage across the gates of the power devices.
We receive 6V + 6V = 12V for the two parts of the trafo winding (combining both half of the centre tap trafo).
When we multiply this 12V by the entire current capability of 5 amp, we get 60 watts.
Now that the transformer's actual wattage is 12 x 5 = 60 watts, it follows that the power induced at the transformer's primary is full and that the output will also be full. This will enable the output to operate without experiencing any voltage drops when under load.
This 60 watt is equivalent to the transfamer's real wattage rating, which is 12V x 5 amp = 60 watts. As a result, even when a maximum load of 60 watts is attached, the output from the trafo operates with maximum force and maintains the output voltage.
Analyzing an Inverter Output Voltage Based on PWM
Now imagine that we apply a PWM chopping across the power mosfets' gates, say at a rate of 50% duty cycle (which are already operating at a 50% duty cycle from the main oscillator, as mentioned before).
Once more, this suggests that the previously estimated 6V average is now also impacted by this PWM feed with 50% duty cycle, lowering the average voltage value across the the transistor gates to:
(The peak is still 12V, but 6V times 50% equals 3V)
This 3V average for all winding halves combined gives us
3 + 3 = 6V
We get 30 watts when we multiply this 6V by 5 amp.
This is, in fact, 50% less than the transformer's rated capacity.
Due to the 12V peaks, the output may appear to be 310V when measured at the output, but under load, this voltage may quickly decrease to 150V because the average supply at the primary is 50% lower than the stated amount.
We must simultaneously address two parameters in order to resolve this problem:
(1) Using PWM chopping, we must ensure that the transformer winding matches the average voltage value given by the source,
(2) and the winding's current must be specified appropriately so that the output AC does not decrease when under load.
In the case of the example we used above, where the addition of a 50% PWM led the input to the winding to be lowered to 3V, we must make sure that the winding of the trafo is rated at the same 3V in order to strengthen and address this problem. As a result, the transformer in this scenario needs to be rated at 3-0-3V.
Transformer's current specifications
We may require the primary of the trafo to be rated at 60 / 3 = 20 amps, yes, that's right, 20 amps, which the trafo will need to be to ensure that the 220V is sustained when a full load of 60 watts is attached to the output. This is because the output from the trafo is intended to work with a 60 watt load and a sustained 220V.
Always keep in mind that in these circumstances, if the output voltage is measured without a load, one may observe an unnatural increase in the output voltage value that may appear to be greater than 600V. The fact that the peak is always 12V even if the average value produced across the mosfets is 3V may explain why this occurs.
However, if you chance to notice this high voltage without a load attached, there is no need to be concerned because as soon as a load is connected, the voltage will instantly drop to 220V.
Having said that, if consumers find it unsettling to observe such elevated voltage levels without a load, this can be fixed by additionally implementing an output voltage regulator circuit, which I have already covered in one of my earlier postings and which you can equally successfully utilise with this idea.
As an alternative, you can neutralize the rising voltage display by connecting a 0.45uF/600V capacitor across the output or any capacitor with a rating similar to that. This will also help to filter out the PWMs and provide a sine wave with a gradually increasing frequency.
The Important Issue
In the previously mentioned example, we observed that using a 50% PWM chopping required us to utilise a 3-0-3V trafo for a 12V supply. This forced the user to buy a 20 amp transformer in order to acquire 60 watts, which seems illogical.
If 3V requires 20 amps to produce 60 watts, it follows that 6V would need 10 amps to do the same, and this amount appears to be pretty manageable. Alternatively, a 9V would enable you to operate with a 6.66 amp trafo, which appears even more feasible.
Since the average voltage is dependent on the PWM ON time, the aforementioned statement simply implies that in order to achieve higher average voltages on the trafo primary, you simply have to increase the PWM ON time. This is another alternative and efficient way to correctly reinforce the output voltage drop issue in PWM based inverters.
Use the comment section below to express your thoughts and ask any specific questions you may have about the subject.
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