Transformer feedback winding

[kathmandu]

Hello,

I'm building a sinewave inverter and I'm using a low frequency transformer at its output. As I need an isolated feedback signal to control the output voltage, I was wondering if I could make an extra winding (few turns) to get a proportional voltage.

The question is: will it be accurate enough? I mean, will it be proportional enough with the main secondary voltage (230V)? If the main load (output current) is changing (thus the main voltage will change too), will the feedback voltage closely follow the main voltage?

 

[treez]

You say you are putting transformer at inverters output..why?…..the power supply which feeds the DC to the input of your inverter will be isolated, so you don’t need a transformer at the output.

If you want to know the output voltage then I’d use a rectifier and divide down the output of that and interrogate its voltage…..use rc filter if you want to know average voltage aswell (of the rectified sne)

 

[chuckey]

Your feed back winding will reflect the magnetic flux. It will not show the I X R loss in the main output.
Frank

 

[c_mitra]

You can take the voltage from the most heavily loaded output and use a resistor divider network to select the feedback.

We hope that if one voltage is regulated the others will also fall in line.

You have not explained the reason why you want another winding for feedback.

 

[kathmandu]

You say you are putting transformer at inverters output..why?…..the power supply which feeds the DC to the input of your inverter will be isolated, so you don’t need a transformer at the output.

It's the classical sinewave inverter (a Mosfet full-bridge with a low frequency transformer as load). The input voltage comes from a lead acid battery bank (24V) so no isolation at all.

If you want to know the output voltage then I’d use a rectifier and divide down the output of that and interrogate its voltage…..use rc filter if you want to know average voltage aswell (of the rectified sne)

I need an isolated feedback signal to feed the uC ADC input. I might look for a linear output optocoupler but I was thinking of a cheaper/easier solution.

You can take the voltage from the most heavily loaded output and use a resistor divider network to select the feedback.

We hope that if one voltage is regulated the others will also fall in line.

I only have one output voltage (230V).

You have not explained the reason why you want another winding for feedback.

I needed an isolated feedback signal and I thought that winding few turns of magnetic wire could solve my problem.

Your feed back winding will reflect the magnetic flux. It will not show the I X R loss in the main output.
Frank

Thanks, Frank.. that's what I was afraid of. So I must read the output voltage, one way or another.

 

[mtwieg]

Correct, a sense winding won't take into account the voltage drop due to winding resistance. I'd recommend using a differential amplifier made from an op amp and precision resistors. Alternatively, using a separate, smaller low frequency transformer could be use to feed back the secondary voltage to the MCU.

 

[Warpspeed]

Its a pretty similar situation to any mains powered transformer with two secondaries.
If you only load one secondary, the voltage on the unloaded winding will drop, but not as much as the voltage on the loaded winding.

It could still work reasonably well, but it will not offer perfect output voltage regulation.

With an inverter, its usually the battery voltage that falls most, and provided the transformer efficiency is reasonably high, your idea is certainly workable, but the results will not be perfect voltage regulation.

A better solution might be to use a completely separate small transformer wired between the secondary of the main inverter transformer to provide the isolated feedback voltage. That will sense the real actual inverter output and should work much better.

 

[treez]

I need an isolated feedback signal to feed the uC ADC input. I might look for a linear output optocoupler but I was thinking of a cheaper/easier solution.

if you want simple isolated feedback of analog signal , use a voltage to frequency converter and pass the frequency through a digital isolator, and then put it back through a frequency to voltage converter to get the voltage back....linear.com do a chip that can converter v to f and reverse.

 

[c_mitra]

You can use both voltage and current transformers at the 230V output and get a fairly decent result. If you are using a microcontroller, you can also do a lot in software.

 

[kathmandu]

Okay, I'll go with a separate low-power transformer. Now the problem is how to get a proportional/accurate DC voltage out of its secondary AC voltage.

I know: bridge rectifier, smooth capacitor, voltage divider. But how to calculate them to get a ripple-free signal but still a fast changing one (to closely follow the 230V main voltage)?

 

[Warpspeed]

Several alternatives depending on your priorities.

One pretty good solution would be a true RMS chip.
https://www.analog.com/en/products/linear-products/rms-to-dc-converters.html

 

[chuckey]

One way could be to use an opto isolator driven via a resistor from the AC. you need to connect a diode in parallel with the LED to protect it from the reverse voltage. So if you use .1mA as your peak input current then you need a 230 X1.4/10^4 resistor ~3.3M . So you can now compare the output from the opto transistor with a reference.
I am curious about how you are going to reduce your output voltage without dissipating too much power.
Frank

 

[kathmandu]

Do I really need a RMS value? Being an almost perfect sine wave (50 Hz), wouldn't be enough to use a peak detector, for example?

I'll try to avoid further chips (RMS to DC) as their output signal needs further conditioning (the ADC requires a 3.3V DC input). Beside, I need a fail-safe circuit (as the inverter output voltage relies entirely on it).

I was thinking of using discrete components (I guess I've seen such a circuit in a comercial UPS) but it requires some fine tuning to get a fast response yet a low ripple DC voltage.

- - - Updated - - -

I am curious about how you are going to reduce your output voltage without dissipating too much power.
Frank

Are you talking about the inverter output voltage? It is controlled by the MCU, by changing the PWM duty cycle according to an internal look-up table.

 

[KlausST]

Hi,

I don't think a true RMS converter is needed.
***
On the other hand you expect an almost perfect sine wave.
This may be true for no load or linear load conditions.
But in real word you should expect non sinuidal current and thus distorted sine. How much I don't know.

For your feedback almost any rectify/average circuig will do. No need for perfect linearity or precision.

Klaus

 

[kathmandu]

Thanks, Klaus. By averaging you mean the classical bridge rectifier/capacitor?

 

[BradtheRad]

Okay, I'll go with a separate low-power transformer. Now the problem is how to get a proportional/accurate DC voltage out of its secondary AC voltage.

I know: bridge rectifier, smooth capacitor, voltage divider. But how to calculate them to get a ripple-free signal but still a fast changing one (to closely follow the 230V main voltage)?

I built my own house voltage monitor, based on a 10V transformer. Rectification is one diode. I added a smoothing capacitor. A 3914 IC turns on one of 10 led's to show voltage.

I added a separate circuit which beeps a few times when a blackout hits. It even beeps in response to a few cycles of brownout. Two seconds worth of power is stored in a 2200uF electrolytic. It warns me if the mains is getting glitchy, which could cause my computer to shut off without warning.

The smoothing capacitor has to be the right value, so it maintains a reasonably smooth DC supply, and yet discharges a noticeable amount when there's a low cycle or two. This required a bit of experimentation, to find a value that was satisfactory.

 

[kathmandu]

I found this circuit in an Application Note (Texas Instruments):



This is a quote from that app note ("800VA Pure Sine Wave Inverter’s Reference Design"):

The Output Voltage in Inverter mode is sensed through the Auxiliary winding, which is filtered and rectified and given to the ADC of the microcontroller.

When fall in the output voltage is sensed with the increase in the output Load, the duty cycle of the H Bridge drive (from microcontroller) is multiplied by a constant greater than 1 so that the final Inverter’s output voltage is closer to the No Load Output Voltage (120V/220 VAC) and vice versa on moving from higher load to the lower load.

For example in this reference design, at no load condition, the duty cycle of the PWM drives given to the H Bridge are varied from 10 percent to 88 percent and when the load is constantly increased at the Inverter’s output, the duty cycle of the PWM is multiplied by a factor greater than 1 so that we can regulate the output voltage within allowable range. While decreasing the load, vice versa followed.

If the duty cycle is increased beyond a point, the output voltage will start clipping and hence results in higher distortion. Hence care should be taken while regulating the output voltage through a feedback.

As you can see, they're actually using an auxiliary winding, too.

Anyway, I'll use a low-power transformer for a better output voltage follow-up but I guess the rest of the circuit (from the image above) it's still applicable.

 

[chuckey]

I like that constant multiplier to regulate the PWM. As you generate the 50 HZ, why not sample my half sinewave pulses at their peak value. I have a problem with mains transformers and more with capacitor in the sampling circuits. When you switch a load onto the main winding, the reservoir capacitor takes time to discharge during which time the output voltage sags. Why not sample the rectified DC at the peak time.
Frank

 

[kathmandu]

Seems like they have chosen a pretty small reservoir capacitor (10uF) thus I expect small charging/discharging times.

The problem with peek detection is the false triggering due to output voltage spikes (e.g. when a big load is (dis)connected).

That's why I still prefer some kind of averaging (though I could manage it in software).

 

[c_mitra]

The problem with peek detection is the false triggering due to output voltage spikes (e.g. when a big load is (dis)connected).

A RC circuit acts like an integrator with a time constant that depends (obviously) on the RC values.

If you select a time constant equal to 1-5 cycles, then voltage spikes will effectively get averaged out. There is no great advantage in increasing the time constant to larger values and much is lost if you decrease the time constant to less than a cycle.