This is a charging-on indicator, charging current is set to 25mA but what I measured is around 23mA and I want to verify via calculation that the charging indicator draws the ~2mA offset.If what you intended was to simply switch the led on/off the led needs to be on the drain (resistor) side of the MOSFET.
The MOSFET here is turned on permanently by the 1Mohm resistor taking the gate -ve wrt to the source but...
The MOSFET turns on when Vgs > Vth, but as soon as current flows in the LED the voltage at the source falls from 3.8v due to the Vf of the LED reducing the MOSFETs Vgs. The device ends up in equilibrium where the MOSFET is turned on just enough to maintain a current in the LED where Vgs - Vf ~= Vth
Of course if Vf is close to the supply rail of 3.8v there may not be enough voltage for even that equality to hold. For a red LED @1.5v say that's probably OK.
Calculating that is tricky, the simulation uses some basic physics for MOSFET Id v Vgs and LED Vf v If. But how close to reality that is will only be known when you have the specific components on a breadboard.
If you have the curves for the specific LED Vf v If and the MOSFET's Id v Vgs you could get an indication.
What did you intend with this circuit.
I'm not so sure myself also why he did that. Anyway I'm just looking out where the current bleed occurs and it's in those led indicator.I will attempt to make this statement politely.
How does the designer expect the full 25 mA charging current to reach the battery, if there is a previous current bleed of 2 mA?
Also the note: "Change R19 to 1M for lower current" is completely false, as that resistor, connected as is, has no impact on either the LED or the battery's current.
Thanks for this Irving now I get it. My load is not inductive at all. Just for interface board. Then it's no issue to remove both.Read the data sheet. D4 and D5 are protection against back-emf on an inductive load. In normal use they are reverse biassed and have no impact. If your output is ~1v then I suspect D4 and D5 are in backwards...
If your load is not inductive you don't need them.
The reason your input gets to 4v before the output voltage appears is that the charge pump to turn the control FET on doesn't start up til 4v
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