I need some information regarding the Power MOSFET switching.

1.what is the difference between high side & Low side switching
2.when we have to use High side switching
3.when we have to use Low side switching

Can anybody elaborate me with schematic and calculations !!!

Thanks in Advance !!!

 

First, you need to define what type of Mosfet you are using. Is it depletion mode or enhanced mode? Is the mosfet channel P-type or N-type? I will assume you are asking about N-type, enhanced mode mosfets; although, in your previous question about battery charging you used an P-type, enhanced mode device.

"Low-side" means the current travels from the load or device through the mosfet to ground (common). "High-side" means the current travels from the supply through the mosfet to the load and then to ground.

Another way to put that is:
Low-side = mosfet source to ground, drain to load, load to supply.
High-side = mosfet drain to supply, source to load, load to ground.

Low-side switching is most common for N-channel devices because the gate drives are simpler. High-side is used when you need to supply power to a device that is grounded.

As for specifics, there are hundreds of articles on gate drives. IR, Vishay, Fairchild, and TI have plenty of advice.

As for the gate drive of the two configurations. Remember that to turn on a N-channel device, the gate must be at a potential above that of the source. Often 10V ,e.g., Vgs =10V. That is easy to do with a low-side configuration, since the source is referenced to ground.

Now, what happens with a high-side configuration. The voltage drop across the mosfet is very small, so the source is effectively at the supply potential (i.e., the supply is effectively across the load). Now to get a Vgs =10 V, the gate must be at a potential higher than the supply. Some sort of circuit to provide a boost is needed, such as a charge pump.

Regards, John

 

But for high side switching you can use a P-channel enhancement mode MOSFET where the off gate voltage is equal to the load supply voltage and the on gate voltage is ground/0V. The only problem then is if the gate drive circuit works on a lower voltage than your load e.g. a 3.3V microcontroller driving a 12V load.

 

But for high side switching you can use a P-channel enhancement mode MOSFET where the off gate voltage is equal to the load supply voltage and the on gate voltage is ground/0V. The only problem then is if the gate drive circuit works on a lower voltage than your load e.g. a 3.3V microcontroller driving a 12V load.

I wanted to keep the discussion simple. So now that you have brought it up, with a P-channel mosfet, the Source (not drain) is connected to the supply side:



An easy way to keep that straight (a crutch, if you will) is to look at the body diode orientation of a P-channel device:



You do not want to forward bias that diode.

John

 

First, you need to define what type of Mosfet you are using. Is it depletion mode or enhanced mode? Is the mosfet channel P-type or N-type? I will assume you are asking about N-type, enhanced mode mosfets; although, in your previous question about battery charging you used an P-type, enhanced mode device.

"Low-side" means the current travels from the load or device through the mosfet to ground (common). "High-side" means the current travels from the supply through the mosfet to the load and then to ground.

Another way to put that is:
Low-side = mosfet source to ground, drain to load, load to supply.
High-side = mosfet drain to supply, source to load, load to ground.

Low-side switching is most common for N-channel devices because the gate drives are simpler. High-side is used when you need to supply power to a device that is grounded.

As for specifics, there are hundreds of articles on gate drives. IR, Vishay, Fairchild, and TI have plenty of advice.

As for the gate drive of the two configurations. Remember that to turn on a N-channel device, the gate must be at a potential above that of the source. Often 10V ,e.g., Vgs =10V. That is easy to do with a low-side configuration, since the source is referenced to ground.

Now, what happens with a high-side configuration. The voltage drop across the mosfet is very small, so the source is effectively at the supply potential (i.e., the supply is effectively across the load). Now to get a Vgs =10 V, the gate must be at a potential higher than the supply. Some sort of circuit to provide a boost is needed, such as a charge pump.

Regards, John

Thank you so much for your reply !!!
I tried simulating a simple circuit using N channel Mosfet-
A- Low side switching
B- High side switching

I have few doubts-
1. In the attached circuit load is running for the clock signal is high means there is no operation difference. So its difficult to decide which side of switching need to select the particular operation. How to distinguish if we have n channel EMOSFET. In my point of view we can go ahead either of them because circuit performs same .

2. Now same operation can be performed by using P channel EMOSFET for low and high side.

3. I think if we have 4 available option so which one to select for particular applications basis.

4. I need one more thing why do we need to put low resistance in series with gate , lets say 10E.

I need detailed clarifications on that !!!

 

Hi mishra,

Your .doc attachment is blank when I open with Word 2007. What is its original format? Also tried .png -- which didn't work either.

John

 

Hi mishra,

Your .doc attachment is blank when I open with Word 2007. What is its original format? Also tried .png -- which didn't work either.

John

That doc is in "Microsoft Office Word 97 - 2003 Document (.doc)" format.
I have attached some other format as well "Microsoft Word Document (.doc)".

Note - with reference to above description circuit b not working !!!

 

Adobe Acrobat attempts to open the attached "Word" file when I try to view it. Very strange, so I won't try to open your attachments again. Perhaps you could take (a) screen shot(s) and embed them in your posts.

 

Still can't see any circuits. Can you try zipping it or do a cut (snip with Win 7) and paste.

John

 

Hi mishra,

Your .doc attachment is blank when I open with Word 2007. What is its original format? Also tried .png -- which didn't work either.

John

Hope this will be accessible !!!
Please follow post 5 to answer .

 

Your circuit for the low side switch will be fine as long as the light draws less than 1A (about the maximum current that MOSFET will give at 5V gate to source. Your high side switch won't work as you might expect. The gate-source threshold voltage is between 2V and 4V - that is the voltage needed to get the MOSFET to conduct at all. So the source will be between (5V - 2V) = 3V and (5V - 4V) = 1V and that is all the voltage your light will see.

To run a load at a higher voltage than the gate drive with a simple circuit you need to use the low side driver circuit.

 

Your high-side switch needs at least 15V, preferably 20V, at the gate relative to ground (using your schematic as reference).

Please re-read post #2. With N-channel mosfets, the gate typically needs to be higher than the supply referenced to ground.

John

 

Your high-side switch needs at least 15V, preferably 20V, at the gate relative to ground (using your schematic as reference).

Please re-read post #2. With N-channel mosfets, the gate typically needs to be higher than the supply referenced to ground.

John

Thanks for your reply .

Yes , i have given 20V - Vgs and light start illuminating !!!

I have gone through your below statement given in post #2 but still unable to clear with fundamentals .

"As for the gate drive of the two configurations. Remember that to turn on a N-channel device, the gate must be at a potential above that of the source. Often 10V ,e.g., Vgs =10V. That is easy to do with a low-side configuration, since the source is referenced to ground.

Now, what happens with a high-side configuration. The voltage drop across the mosfet is very small, so the source is effectively at the supply potential (i.e., the supply is effectively across the load). Now to get a Vgs =10 V, the gate must be at a potential higher than the supply. Some sort of circuit to provide a boost is needed, such as a charge pump. "

Could you explain me with some sort of calculation so that its easy to understand me for either cases !!!

Why do it need low voltage Vgs in low side and high voltage Vgs in high side .
Now still questions arises which one chose which applications if want to operate it by N channel MOSFET.

Lets try to understand in depth .

 

Hi,

Hope this helps:


John


Edit: Just noticed the at I used alt+175 instead of alt+247 (≈). Change all of the » to ≈
Need better glasses in the morning.

 

Hi,

Hope this helps:
View attachment 107168

John

Hello John.

Thanks so much for support .

I need to study more to visualize the clear pictures.
Why is this calculation valid for only high side switch and why isn't for low side switch.
As we have seen .
I need to understand with calculations.

 

The calculation is valid for both high and low configurations. In the low configuration, the source (S) is grounded. So, 10V + 0V = 10V. BTW, notice my edit. In the dim light of morning, I misread my cheat sheet for symbols and used » instead of the intended ≈ (approximately equal to) in the explanations attached to the image.

John