so, i am back to just using a 7555 and driver fet's.
i think all this can be powered by 5v
the low power ST version of lm393
and the fet's are n & p in single package (FDC6333C) and they can reach full Ids at 5v Vgs. this unit is a tad hefty for my needs, but works. the fets have really fast on/off times too.


your thoughts

 

So you decided to leave the 555 driver off. OK, but you will need a better driver than the comparator.

MOSFETs require a hefty current surge when switching states. This is because the gate looks like a capacitor. Once charged, it draws practically no current at all, but you have to charge that capacitor as quickly as possible.

This is why my two drawing either used a transistor driver or the 555, and did not try to use the comparator directly. You can try it if you want, I prefer to be a bit conservative and not create potential problems up front.

The reason a resistor is used on the gate of a MOSFET is also the capacitance. Since the gate looks like a capacitor, and the wire going to the gate is an inductor, there will be ringing when it is switched quickly. This is turn creates an on / off cycle for the MOSFET, which can be unwanted at best and destructive at worst.

You need one resistor per MOSFET, as close to the gate as possible.

This is explained in Chapter 10, Transistor Drivers.

 

all valid points in post 42..

i can control the gate current down to within reasonable amount of mA to not exceed the 393. the gate charge is very small so the charge and discharge times will be low.

the fet package is gated 4.7nC(n),4.1nC(p)
5v across 500ohm = 10mA
1A = 1C/sec
10mA = 10^-3C/sec
1sec/10^-3C * 4.7^-9C = 470x10^-9sec = 470nsec = 0.47usec
*2 for two gates = 0.94usec
~1MHz is 0.94usec
200Hz is 5ms

anyone checking my math? it seems off in the timing calc, that time is how long it takes for 4.7^-9C to move at a rate of 10mA.

if we use the 5-period calculation for a capacitor:

using 1C=1Fx1v
the fets in question list gate charge as 4.7nC @ 10Vgs
to normalize divide by 10
@ 1v the capacitance is found to be 0.47^-9F
with this cap RC = 0.235^-6, then 5 periods later *5 = 1.175^-6sec

i am not a fet guru, thus i am not sure if the gate charge is linear to 1C=1Fx1v, nor do i know if a fet charge acts the same as a charge in a electrolytic cap, etc, i would need to go read up on that. if the fet gate acts like a std cap then as the charge bleeds off the voltage drops, thus the current will also drop off, thus the 10mA i initially stated is a transient #.

well geez, the spec sheet shows it rather darn linear, with an offset based on Vds. so, @5vgs the charge is half of 4.7^-9C the spec sheet noted @10vgs, thus the time to drain off the charge at 5vgs is ~0.5875^-6sec.

one resistor per gate? i did not read chap 10 (yet) but i have tied gates together in the past and handled ringing (if any) after scoping the gates (i provision some tracks in the pcb to accommodate a snubber if needed). i am in the 200Hz max area, so ringing is less of concern. since the fets are a single package, the track connecting the gates will be very very short, probably a direct line between the gates, and the resistor will be right on a gate pin.

but ok, it's on the fence of needing a bjt between 393 and fets. i can bread it w/ and w/o the bjt to see how it performs.

 

Try it and see. I've smoked my share of parts learning, and electronics does have a way to teaching if you make a mistake.

 

hey Bill, does this seem better

R12 10k
R13 2k
R14 10
Q1 2N2222

 

You need to discharge and charge it quickly, which is why we had a active pair feeding it. It is internal to a 555 too.

Again, try it and see. Finding out what you can get by with is half the fun for me.

 

You need to discharge and charge it quickly, which is why we had a active pair feeding it. It is internal to a 555 too.

Again, try it and see. Finding out what you can get by with is half the fun for me.

i would change it back to post #17 using a FMB2227A or similar, keeping my pair of fets.

this will allow me to remove one resistor from post #45.

 

I would strongly consider using a 7555 myself.

 

I would strongly consider using a 7555 myself.

i'm with you.... the ALD7556 looks attractive for my needs, really simplifies things. two discrete IC's vs four to do the same job.

 

If you are ever interested I found (and lost) an old drawing showing how to do something very similar with a conventional 555 using 5V. For this application the 7556 is a much better choice though.

I draw a lot.

 

here's the change.

question, whats the diff between A and B (per the way R6 is used in the circuit)?

i will move R17 to the anode side of that opto, makes it easy to scope using two channels, etc.

 

Pots are wired as in example A to protect circuits if the wiper becomes inoperative. When wired like "A" instead of an open, you get full resistance of the pot.

 

question, whats the diff between A and B (per the way R6 is used in the circuit)?

The difference is what happens when the wiper intermittently loses contact with the substrate.

 

in pot A the current has two paths to take, so does end-to-end of that pot look like two resistors in parallel ?? kinda like a shunt circuit? as you can probably tell, me just being lazy, i have pots sitting in front of me, just too lazy at this sec to bend the pins over and hook my fluke to it. i shall test this, but need some morning joe 1st.

 

There are two paths, but one has "infinitely" more resistance than the other, so by far the majority of the current will pass through the wiper directly. While the wiper is touching the substrate, A and B are essentially identical. It's only when the wiper loses contact that the circuits are different.

 

i mapped a 26x360° 1M linear pot. A & B were identical when i graphed the data.

 

As predicted. Now add a goober onto the substrate, to force the wiper off as it passes over the goober. Or find a dirty old pot and try it again. A better test than mapping would be to "listen" as you turn the knob while the pot is in a volume or tone control circuit. Old pots get scratchy.

 

Think of it as a failsafe.

 

so in post #51, R9 pot, what happens to comp output if the wiper lifts off?

 

Unknown, it floats. If you add a 10Mohm to ground it will either go 100% or 0% if the wiper fails.