This video provides a tutorial on how to implement a P-channel MOSFET current limiting circuit. More information is available at: www.apsolutions...
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@ducciom.gasparri97272 жыл бұрын
Wow, one of the best explanation I found on the internet, I love the way you go over the circuit, very clear and to the point
@R.Daneel Жыл бұрын
Really good teaching style. I'm a beginner, and not familiar with much of this - yet didn't feel lost at any point. Great job.
@BusyElectrons2 жыл бұрын
Both the video and the write-up on your web site are excellent, going into enough detail to fully understand the circuit. Thank you for taking the time to share this.
@louco211 ай бұрын
Thank you so much for taking the time to explain how the circuit works intuitively and also with detailed math justification! subscribed for more of these!
@stark_energy4 ай бұрын
Note this design is very inefficient when the load current (before limited) is way higher than the maximum. For example case above, if the limit is 1.4A, and without the limiter the load will draw 100A (near zero ohm resistance), then the Mosfet will need to handle almost 40V drop IN ITS LINEAR REGION. The heat dissipation formula P = I^2 x RdsON is not applicable when Mosfet is in linear region. What he said at the end of video that the Mosfet will dissipate at worst case 40V x 1.4A = 56W is correct. This Mosfet will dissipate that huge amount of heat if you put load that demands much higher than 1.4A. This design is not for continuous current limiter because the rest of the current are rather "consumed" by the Mosfet as heat instead of being limited. If you want true current limiter where your device will often pass higher than the current limit, then use BJT (Epitaxial Planar Power Transistor) instead. This Mosfet current limiter design is ideal for transient spike current or short circuit protection. It is also ideal in a case when the device MUST operate under certain current limit (where overcurrent is a failure). The advantage is that using Mosfet under Ohm region (using its RdsON) is more efficient than using BJT especially in higher current (>2A). And so the LED indicator in the schematic (and also beep/buzzer) is very important in order to notify people around it that there is overcurrent going on and this overcurrent issue must be fixed immediately.
@oynamalan5 ай бұрын
You're a great teacher. Thank you.
@ZeusandHades2 жыл бұрын
Nice video. Its easily understandable, and I could follow it well. I wish I could come up with such ideas on my own.
@markkennard8612 ай бұрын
Nice ! Im going to upscale the fet and try to set I limit to around 10A.. wish me luck. :)
@RexxSchneider2 жыл бұрын
Depending on your application, a latch-type solution can be an alternative. If you add another npn transistor cross-coupled with the 2N4923 (each collector drives the other's base via a high value resistor), you can then use the new transistor to sink the current in your 15K/33K bias chain instead of using the upper 2N3906 to drag up its mid-point (that transistor is then redundant). As soon as the lower 2N3906 turns on, the two npn transistors will flip state and turn off the MOSFET and the output supply entirely. You'd need a momentary push button between the 2N4923 base and ground to reset the circuit when the overcurrent condition has been cleared.
@davidtremmel895 Жыл бұрын
If you could find a way to post a schematic of that, I would appreciate it
@markdlehane2 жыл бұрын
Great design and brilliant explanation. Thanks for going through in detail, it's extremely helpful indeed.
@fedimakni120014 күн бұрын
thanks for the explanation. i have 2 questions: - does it work well with inductive load as well as resistive load? - what kind of issue we can face if the device always operate in the current limiting mode? thank you again.
@bayareapianist7 ай бұрын
Is there a way to make sure the MOSFET is either on or off is the power doesn't get dumped on it? The worse case was calculated to be 56W. But could be (1.4A)^2x(0.06 ohm)=0.1176W?
@psychedelicfungi9 ай бұрын
Good explanation 👍
@mike_oe3 ай бұрын
"Heatsunk".... That was my takeaway 🤣
@tc82574 жыл бұрын
Cool circuit, great explination.
@przoch58353 жыл бұрын
Great tutorial, thank you
@Guishan_Lingyou5 жыл бұрын
This is a very helpful video! Thank you for tanking the time to make it. There is one point I am confused on. When you calculate the power across the P-channel FET at 7:45, isn't it the power dissipated across the component that would be relevant to heat dissipation? And wouldn't that be (i^2)r, which would be .1176 Watts just before the current limiting kicked in? I get that the resistance goes up in the current limiting condition, but there is never going to be a 40 Volt drop across the FET, is there?
@kdkoadd5 жыл бұрын
Thanks Paul. Think about what must happen to the voltage drop across our load resistor as its resistance is decreased (assuming we keep out current at 1.4A)
@Guishan_Lingyou5 жыл бұрын
At the risk of looking stupid after having already displayed my ignorance...As the load resistance goes down in the current limiting condition, the voltage drop across the FET goes up corresponding to how far it goes down across the load?
@kdkoadd5 жыл бұрын
@@Guishan_Lingyou Correct. I have described the circuit behavior in detail on my website: www.apsolutions.com/mosfetcurrentlimit.html If you go to that page and scroll down to the graph on the bottom of the page, you can see a graph of a parameter-sweep simulation that illustrates what is happening in the circuit. You might also pull up the IRF5210 data sheet (I linked to it from my webpage) and take a look at the Id versus -Vds curves. That might help clarify also.
@leslieleslie3362 жыл бұрын
great video!
@akifkacar93 Жыл бұрын
Thank you for this video but is it possible to do same circuit with n-type mosfet? Why we use p-type?
@josephke20 Жыл бұрын
im not adequate in the understanding of math department but i would like to use this for a 12v supply, how do i calculate R1 and R2.... thanks
@rawandjaza59295 жыл бұрын
That is good.., i need design for adjustable current limiting 1-20A so can you made it or circuit diagram for it
@RobertoSuarez2444 Жыл бұрын
How could I make this work if I want to use only 5V and limit the current to 2.4 Amps?
@zirikh86084 жыл бұрын
How efficient is this circuit ?
@thghho4 жыл бұрын
Hello, Can you guide me for limiting the output current to 350mA of a power supply (with output 250vdc, max output 15Amp). Should I simply add 0.22 Ohm 10W (9 Resisters) in Parallel before load?
@tafadzwamapuranga4295 жыл бұрын
hey there thank you for the video it was very helpful ,i am trying to build the same circuit but to limit the current from 50 amps to 300 amps ,what modifications can you make to this circuit that can make something like that possible
@anthonycalia13172 жыл бұрын
Q2 is not dissipating 56W that is the load power dissipation. The MOSFET is dissipating its Drain Current squared times Rds ON which is far less...
@RexxSchneider2 жыл бұрын
Unfortunately, if the output is short-circuited, then Vgs reduces and Rds increases. Because the limiting load current of 1.4A will still be flowing (otherwise the 2N3906 would not be turned on), that means the 40V supply has to appear across the MOSFET, and it dissipates 56W. If you like, you can consider the 2N3906/IRF5210 to be an amplifier that will try to adjust the voltage at the MOSFET gate so that the voltage across the sense resistor is 0.7V. While the load is greater than 40V/1.4A = 28R, the MOSFET is in saturation ("ohmic region") and Vgs is driven as low as it can be. Once the load resistance drops below 28R, the circuit will raise the gate voltage until the MOSFET starts to move out of its ohmic region and it will hold the output current at 1.4A as the load resistance gets smaller, thus decreasing the voltage across the load and increasing the voltage across the MOSFET. When the load resistance is 0R, the current is still 1.4A, and the MOSFET has the full 40V.