MOSFET not performing to datasheet spec (Vgs)

[ACharnley]

Thanks Ron, that's exactly what I came up with last night - though the 1K resistor seems extreme, I was thinking 100k.

Edit: you chose 1k as most the voltage will try to go via the capacitors supposing them are empty? If it were 100k the P-channel could remain on at this initial charge state damaging it?

 

[ACharnley]

Next mosfet challenge

For the bypass circuit shown, the outcome will only be the ~2.4v I had previously. Am I correct in saying I should use two back-to-back P-channel's and drive them low with an N-channel?

Old:

New:

 

[ronsimpson]

you chose 1k as

100k is fine.
I chose 1k because I turn MOSFETs on/off many times a second. I think you will turn it on only once a day. lol
There is a very small amount of capacitor inside the MOSFET. The 100k and capacitor make a RC time constant that sets how fast the transistor turns off. In your case if it takes 0.0001 seconds who cares.

 

[ACharnley]

Thanks, I had to check. Is my bypass circuit update ok?

 

[ACharnley]

Final one.

My P-channels arrive tomorrow so I can have a play.

 

[crutschow]

Thanks, I had to check. Is my bypass circuit update ok?

It should work.
But you could connect the MOSFETs source-to source instead of drain-to-drain and just use one resistor from the two sources to the gates.

 

[ci139]

there are two things concerning the original question

1. the supply capability - as enable enough power through the drain circuit for the drain load (if present) to form a major voltage drop
2. drain voltage versus drain current , somewhat covered https://electronics.stackexchange.com/questions/152345/what-precisely-is-r-dson

 

[ACharnley]

The buck doesn't work, the P-channel doesn't saturate fast enough.

I'm driving it with a 100KHz PWM. If I change the resistor to 470ohm it begins to behave (range still 4v+) but at 1k it's 4.7v+. I can't go much lower without eating into efficiency, is there another way?

 

[ci139]

? **broken link removed** - the efficiency is at greatest when you close your switch just at coil saturation and define/set(dynamically) your cycle duration (charge/+discharge inductor) at "a bit" greater time step than it takes

 

[kubeek]

You can use higher resistor and add two buffer transistors like this https://i.stack.imgur.com/himCV.png

 

[ronsimpson]

100KHz Ok you need speed. Here the turn off resistor is amplified by Q1.
Why 100khz? With the 22mH coil it takes 1mS to respond.

Another thing. When I draw; the highest voltage is on top/ground at bottom/ if negative voltage it will be at the far bottom. Input on left and output at right. Understanding a drawing is about flow.

 

[ACharnley]

Ok, now I know why MOSFET drivers exist. I've attempted to low-side the buck instead. It'll mean the voltage readouts are at Vcc when "off" but that's easy to deal with.

I'm unsure if the USB voltage readout would be accurate, it would be USB+NFET voltage but I'm driving the FET at a constant Vgs so I should be able to work it out?

 

[audioguru]

If the Vgs is constant then the current will still be changed if the Mosfet is replaced (each one is different even if they have the same part number) and if the temperature changes. You need plenty of negative feedback and maybe an error amplifier for the current to remain constant.

 

[ronsimpson]

I think I am going crazy.
I think there are two circuits. Maybe more.
1) Has a power resistor and charges a super capacitor.
2) Has a 22mH coil and does what?

One of the problems is that (if there is two circuits) some of us are working on (1) and some on (2) and some on (?).

Post #32 looks like another circuit.
I have lost track of that function(s) you want. Can not see the big picture.

 

[ronsimpson]

The buck doesn't work, the P-channel doesn't saturate fast enough.

For education:
Post #25 did not have Q1 etc. It only has a 47k resistor to turn off M1.
M1 has a gate turn off voltage of 1V. A gate turn off resistor of 47k with 1 volt across it results in 20uA of drive.
Adding a transistor to amplify this current helps.
This is how many ICs do it.
D3+C2 makes a 9V supply.
In this circuit I am getting 30mA of gate current at turn off. (1000x)
With out D4, Q1 turns on too hard and gets sticky/slow. I normally use a Schottky diode.
If the gate is off for a long time the 9V supply will run down. I added R2 (100k to 1meg) to keep the gate off over long times.