Power Circuit Help

Other Hardware General Electronics
41

But once you press one of the buttons, wouldn't Vgs go to 0?
I'm looking at this for reference

42

For the Q9 cct.

  • The BTN_POW switch closes.
  • The Q9 gate goes to GND.
  • Q9 turns on.
  • The voltage on Q4 gate (normally sits at 9V) goes to 0V.
  • Q4 turns on.
43

If the drain of Q9 is on GND, and the gate goes to GND when the button is pressed, there is no way that the source stays at GND, because then all 3 pins are GND and there is no voltage to keep the gate open.

44

  • As long as you closed the BTN_POW switch, the Q9 gate is at 0V, agreed ?

  • As long as the Q9 gate is 0V, the Q9 transistor is ON, agreed ?

  • As long as Q9 is ON, the resistance source to drain is 0 ohms, agreed ?

  • As long as the Q9 resistance is 0 ohms, the gate on Q4 is 0V, agreed ?

Do you understand the gate of Q4 sits at 9V when Q9 and Q8 and Q5 are OFF ?

45

Agree with 1, but do not agree completely with 2 because it is not necessarily that the Q9 gate is 0v, it's that Qsource - Qgate has to be greater than the threshold voltage (Vgs). And since they are both 0v, it will be off or not function as intended.

46

image
The way it is wired is like this except that when wired like this, the gate voltage has to be negative voltage so Qsource - Qgate is a positive number greater than the threshold. A small amount of current has to travel from Qsource to Qgate to keep the MOSFET on.

47
48

Yea, when the Gate and Source were both +5v, no current flowed. Same if they are both 0V.
Note how the Source in that video stays at +5V. What you are suggesting is that goes to GND.

49

Do you have a solderless breadboard and a P channel MOSFET that can plug into it ?

50

No, I only have n channel MOSFETs. This is a nice debate though. Sadly, I have to go and finish my chemistry homework, so maybe see if you could put it on a breadboard to see what happens.

51

This was breadboarded as below.

2023-01-16_23-39-32
2023-01-16_23-39-321297×840 101 KB

  • When the the switch was opened, Q2 was OFF (actually Vsd = 9V which we can say is a HIGH). Q1 is OFF and the the motor was stopped.

  • When the the switch was closed, Q2 was ON (actually Vsd = 1V which we can say is a LOW). Q1 was ON therefore the motor was ON.

Summary

  • when Q2 gate is at 9V (switch open) Q2 Vsd is HIGH (at 9V).

  • when Q2 gate is at 0V (switch closed) Vds LOW ~1V. (I did say 0V, more correctly should have said LOW)

52

Ok, that makes sense. I'll probably just stick with what I have so the voltages go to either extreme. One extra MOSFET is only like one cent so it's alright. Thank you for your help btw, I learned a lot about MOSFETs today. Have a nice day

1 Like
53

:+1:

54

Here is a modified circuit that is often used in similar situations.

This version has your two switches that you wanted to turn the power on to a load.

This circuit is considerably less complicated too !

This has been breadboarded and works 100%.

EDIT

Updated schematic

2023-01-17_13-31-42
2023-01-17_13-31-421102×814 93.5 KB

55

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