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"Addressing Voltage Drop Challenges in Inverter Circuits: Troubleshooting and Solutions

 "Troubleshooting and Solutions for Addressing Voltage Drop Challenges in Inverter Circuits

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.

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 utilising 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 centre 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.

Combining the two centre tap trafo halves, we get a result of 6V + 6V = 12V for the winding's two parts.

This 12V multiplied by the maximum current capability of 5 amps gives us 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 transfomer'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.

Analysing an Inverter Output Voltage Based on PWM

The power mosfets are already operating with a 50% duty cycle from the main oscillator, as was previously discussed. Now assume we apply a PWM chopping across the gates of the power mosfets, say at a rate of 50% duty cycle.

The average voltage across the mosfet gates is now reduced to: This again means that the previously computed 6V average is now additionally impacted by this PWM feed with 50% duty cycle.

Despite the peak being 12V, 6V multiplied by 50% equals 3V.

By combining the 3V averages for the two winding halves, we obtain

3 + 3 = 6V

This 6V multiplied by 5 amp gives us 30 watts.

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:
Using PWM chopping, we must ensure that the transformer winding matches the average voltage value given by the source,

 so that the output AC does not decrease when under load, and the current of the winding must be specified properly.

Consider the case in the example above where the addition of a 50% PWM reduced the input to the winding's rating to 3V. To reinforce and address this situation, we must guarantee that the winding of the trafo is rated at 3V as well. The transformer in this instance must therefore be rated at 3-0-3V.

Current Transformer 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 neutralise 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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