PWM switch over with pic 18F4520

[srjth.m1]

Anybody know how this circuit works with pwm in pic 18F4520

 

[goldsmith]

Hi srijth
When PWM is incoming from the micro to the base of BJT it will be turned on and then gate of P- mosfet is grounded and then it will be turned on too . but for my opinion 6.8kohms is too large as RG decrease it as low as 10 or 6.8 ohms .
Best Wishes
Goldsmith

 

[srjth.m1]

So when i am disabling pwm signal Transistor goes off state and disable the battery supply?

 

[goldsmith]

So when i am disabling pwm signal Transistor goes off state and disable the battery supply?

Of course ! when transistor be off then mosfet will be off and then the load can't be supplied !
Good Luck
Goldsmith

 

[srjth.m1]

ok!!Thank you,but i have one more doubt that why pwm is using for this we can also able to do with other pins?

 

[goldsmith]

ok!!Thank you,but i have one more doubt that why pwm is using for this we can also able to do with other pins?

PWM ? because thus you can control the average voltage with controlling the duty cycle .
Best Luck
Goldsmith

 

[srjth.m1]

Ok understood Thank's again Goldsmith..!

 

[Tahmid]

In this case, by adjusting PWM duty cycle, you can adjust average charging current. Bear in mind that this is average charging current. Since there is no LC filter at the output, you are not really affecting instantaneous current. A P-channel MOSFET is used as it is easier to drive in high-side configuration compared to N-channel MOSFET.

Hope this helps.
Tahmid.

 

[srjth.m1]

Can you explain how the charging of battery is working here?In the circuit where i have to connect the battery,a.c and from where th output will take?

 

[Tahmid]

The mains AC is stepped down, rectified and filtered to provide the 16V input shown in the diagram. There's a point in your diagram labelled "BATTERY +". Going by that, that's where the battery + should be connected. The battery - should be connected to ground.

 

[srjth.m1]

So current will come in Reverse direction through MBRS320? or else will you mark it in diagram exactly?

 

[Tahmid]

There won't be any current through a diode if it's reverse biased.

Direction of current is from +16V to the battery to the place labelled "16V".

 

[srjth.m1]

Ok,so what is +16V given just below 16V and from where i have to connect to a output circuit?

 

[Tahmid]

Probably the same as the input +16V.

The overall thing looks like a switchover circuit.

When mains is there, +16V is there and output goes from there while battery is charged.
When mains is absent, output is run from battery.

 

[srjth.m1]

yes the same..as you mentioned itz a switch over circuit.

- - - Updated - - -

This is my doubt,pls check it

 

[Tahmid]

Yes. I've explained the working. Where exactly is your doubt?

 

[srjth.m1]

the douts are given as question mark why using two +16 volts and from where output is taking?

 

[Tahmid]

The two +16V are the same - they come from the same input source - probably after the mains is stepped down with a transformer, rectified and filtered.

The point labelled "16V" should be the output to the load.

The wire labelled "BATTERY +" is where battery + terminal should be connected.

Hope this helps.
Tahmid.

 

[srjth.m1]

can you mention in figure where i have to connect the battery am some what confusing now that's why

 

[Tahmid]

The anode of one diode is connected to the source of the MOSFET. Let's call this diode D1. The cathode of D1 is connected to the anode of another diode, which I'll call D2. Connect the battery at the cathode of D1, which is also connected to the anode of D2.