Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Help with sine wave inverter charging

Joeadeoye

Junior Member level 2
Junior Member level 2
Joined
Jun 16, 2016
Messages
24
Helped
1
Reputation
2
Reaction score
3
Trophy points
1,283
Activity points
1,522
I made a pure sine wave inverter using a low frequency transformer and full bridge mosfet configuration and it's working fine. Now I want to include a charging function in the same inverter.

The transformer used is 7v-220v and 4 IRF3205 mosfets were used to form the full bridge.

Talking about the charging, if 220v AC mains is applied to the transformer, the secondary side will give 7v AC which is too small to charge a 12v battery.

I have seen some commercial inverters do this, but I don't know how the voltage was boosted from 7v to 13.8v needed to charge the battery.

I have seen some threads discussing the same issue I am having but none of the answers in the thread worked for me.



The above links discuss the same issue I am trying to resolve. Please I need help
 
I used a mower which had electric start built-in (Lawn Boy brand). The battery was 6V and the charger read something like 5V with my old-fashioned VOM. The system worked for years.

For a long time I puzzled that the charger could make do at charging on that low voltage. Eventually I realized it had no smoothing capacitor. It produced peaks which were higher than the battery voltage therefore the charger's peaks were sufficient to do the job.

With 7VDC input you can do likewise by a voltage multiplier or boost converter. One is based on capacitors, the other is based on an inductor. You can build either type to double 7V and produce sufficient voltage to charge a 12V battery.
 
A voltage multiplier will not work for me because it can't produce the required current to charge battery. A boost converter on the other hand can do the job, the leakage inductance from the transformer can be used, but the question is how?
 
In sine wave inverter design, we use pwm signal s too switch the H-bridge alternately thats why we use 7v transformers because its SPWM signal and is not 50% duty cycle, during charging to High side mosfet will go off and the two low side mosfet oscillate which in turn makes transformer to work as boost converter and by doing so the voltage can go up to 14v, note :when the High side mosfet go off during charging it act as a Rectifier.. Hope this helps..
 
In sine wave inverter design, we use pwm signal s too switch the H-bridge alternately thats why we use 7v transformers because its SPWM signal and is not 50% duty cycle, during charging to High side mosfet will go off and the two low side mosfet oscillate which in turn makes transformer to work as boost converter and by doing so the voltage can go up to 14v, note :when the High side mosfet go off during charging it act as a Rectifier.. Hope this helps..
Thank you for the reply. I want to know,

1. The low side oscillate together or one after the other? I mean both will ON and OFF at same time, or one will ON while the other will OFF

2. What frequency will they oscillate?
 
The challenge in making your converter bi-directional is to design a step-up ratio that makes up for losses, then also making the identical transformer have a step-down ratio that makes up for losses in the other direction. Either direction causes losses. To find the correct winding ratio seems like the impossible dream.

To create the charging function, it might merely require winding a few more turns on the low voltage winding. Just sufficient turns to change the ratio and obtain a higher voltage produced by the battery side. The rest depends on how your transformer behaves with 50 Hz AC or PWM.
 
Hello
Why you not look online for a UPS service manual where is a schematic diagram of the whole circuit, I am into UPS field and try all the time get manuals from the internet
The small capacities UPS have data online
Look at this link, there are many types similar to design you work o I think

Look at the circuit and see how the designer made the charging circuit and use as a reference for your design
I talk about the transformer based UPS, small capacity, as there are many types and different capacity

I have a manual for some machine at work, I’ll try to photo and send you here
--- Updated ---

They often use relay to be triggered when ac plug in
 
The challenge in making your converter bi-directional is to design a step-up ratio that makes up for losses, then also making the identical transformer have a step-down ratio that makes up for losses in the other direction. Either direction causes losses. To find the correct winding ratio seems like the impossible dream.

To create the charging function, it might merely require winding a few more turns on the low voltage winding. Just sufficient turns to change the ratio and obtain a higher voltage produced by the battery side. The rest depends on how your transformer behaves with 50 Hz AC or PWM.
Thanks for the reply, the transformer I am using is salvaged from an old pure sine wave inverter which has charging function. The inverter PCB board was damaged by thunder lightening, so I decided to use the transformer to rebuild another inverter.

So I assume the transformer is already made for charging purpose, cos it charges well before.
 
It seems my earlier reply disappeared into the ether - if you have an H bridge of LV fets you can operate these as a booster by turning on the bottom pair and then releasing to go from the low o/p volts of the Tx to the higher volts on the battery - some knowledge of power electronics and control is needed to do this symmetrically to avoid load side pull of the BH loop in the Tx when powered from the mains side.
 
It seems my earlier reply disappeared into the ether - if you have an H bridge of LV fets you can operate these as a booster by turning on the bottom pair and then releasing to go from the low o/p volts of the Tx to the higher volts on the battery - some knowledge of power electronics and control is needed to do this symmetrically to avoid load side pull of the BH loop in the Tx when powered from the mains side.
I don't really understand this, as some of your abbreviations got me confused. I understand the first part where you said the low side mosfets should be turned ON, but I don't understand the part that said "releasing to go from low output volt of transformer to high volt of battery"

I also don't understand the part you said "to avoid load side pull of thr BH loop in the transformer"

Can you please clarify? Thanks..
 
You have an H bridge of fets, if you turn on the bottom pair ( or the top ) you short out the Tx and current and energy builds up in the leakage inductance, lets say we turn on for 2uS say, when we release ( turn off ) the voltage will fly up to the level needed to keep the current going into the load - through one of the diodes in the top fets and one of the diodes in the bottom fets - in this case current flows to the LV bus with caps and the battery you are trying to charge. Note we could gate ON the diodes that are conducting backwards here if we are clever enough - however this is not a simple control matter - as - if you want to draw a nice sine wave off the mains you must modulate the ON & OFF times to do this over each half cycle and for current limiting and voltage limiting - and further you must ensure that the loading of the transformer on the LV side is the same for pos and neg half cycles - if not there will be flux walking in the transformer core BH loop which may lead to saturation and then the mains side fuse will blow - this is load side pull of the BH loop.
 
You have an H bridge of fets, if you turn on the bottom pair ( or the top ) you short out the Tx and current and energy builds up in the leakage inductance, lets say we turn on for 2uS say, when we release ( turn off ) the voltage will fly up to the level needed to keep the current going into the load - through one of the diodes in the top fets and one of the diodes in the bottom fets - in this case current flows to the LV bus with caps and the battery you are trying to charge. Note we could gate ON the diodes that are conducting backwards here if we are clever enough - however this is not a simple control matter - as - if you want to draw a nice sine wave off the mains you must modulate the ON & OFF times to do this over each half cycle and for current limiting and voltage limiting - and further you must ensure that the loading of the transformer on the LV side is the same for pos and neg half cycles - if not there will be flux walking in the transformer core BH loop which may lead to saturation and then the mains side fuse will blow - this is load side pull of the BH loop.
Thank you for the explanation. So, if I'm to toggle the low mosfets ON and OFF, it has to be over each half cycle of the mains AC wave form. Am I correct?

If this is the case, I might need to first detect the zero crossing of the mains AC before switching the mosfets so it would fall within each cycle. Am I correct?
 
you need to do more than just toggle the lower mosfets - and you need to modulate at a proper switching frequency, i.e. > 30kHz as well as the 100Hz part for the half cycles - really this is a job for an experienced power electronics engineer . . .
 
you need to do more than just toggle the lower mosfets - and you need to modulate at a proper switching frequency, i.e. > 30kHz as well as the 100Hz part for the half cycles - really this is a job for an experienced power electronics engineer . . .
You said I need to do more than just toggle the lower mosfets, then tell me what I need to do please.

Switching at 30Khz as well as 100Hz, I'm a little confused. How can I switch at 30Khz and still switch at 100Hz at the same time?
 
I'll contribute my explanation. I believe Easy's method is a clever equivalent to operating the transformer at a lower frequency. By letting through two positive waveforms (or two negative waveforms) It lets greater current (or flux field) build, thus rising to that much greater voltage at the secondary when the flux reverses. (To compare, if we take a power transformer which is designed to work on 60 Hz, and run it from 50 Hz mains, we create the risk it might overheat unknownst because it's operating beyond its specs. However we can safely do the reverse, to run a 50 Hz transformer on 60 Hz mains, and be within its operating specs.)

The 30 kHz is a carrier frequency, producing PWM. The pulses are to limit voltage and current waveforms entering the transformer. as necessary to charge the battery at proper current throughput.
 
Last edited:
@Joeadeoye you need to read the classic data sheet and app note for the UC3854 - this will teach you about modulation to draw a sine wave from the mains with a booster - yours is the H bridge version of the same thing run on the LV side of a transformer. This is a difficult project if your current knowledge of power electronics and control is low.
 

LaTeX Commands Quick-Menu:

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top