From BJT to MOSFET: design question

[earckens]

In the attached design I would like to replace Q3 and Q4 by MOSFETS. Q3 = P-channel, Q4 = N-channel. G, D and S are marked on the drawing in pencil.

Is the proposed setup correct?

I wonder if I should/could remove R6 and if not, if the values for R6 and R7 may remain the same?

Q1 and Q2 I would replace with 2N5551 (>100Vceo and 600mA Ic): good choice?

 

[dknguyen]

You can remove R7 or keep but it but greatly reduce by 100 times or more (to slow down turn-on time for noise and ringing issues).

Do not remove R6. It's a pull-up that works together with Q2 to switch Q3. Without it, Q3 is not guaranteed to turn off and stay off when Q2 is not conducting. R6 can remain approximately the same and would probably work just fine if it was anywhere between 1K to 10, depending on what V1 is due to power dissipation.

 

[earckens]

You can remove R7 or keep but it but greatly reduce by 100 times or more (to slow down turn-on time for noise and ringing issues). Do not remove R6. It's a pull-up to make sure Q3 turns does not conduct when Q2 does not conduct. R6 can remain approximately the same.

Thks. If R7 is removed how is the gate of Q3 driven?

Edit: sorry, you probably mean "replace by short"?

Edit2: and add cap in parallel over R7?

 

[dknguyen]

Yes, remove R7 is replace by short. I've refined my original post to try and make things a bit more clear.

(Removing a resistor = removing the resistance = 0hms. Sort of like how removing a voltage source is removing the voltage = 0V (short) while removing a current source is removing the current = 0A (open-circuit).

 

[dknguyen]

What I said for R7 applies to R4 as well if you replace Q2 with a FET.

 

[ronsimpson]

Sorry but viewing side ways is hurting my neck.

I keep the Gate resistors. R5, R7 but make then 5 to 10 ohms.
You need R6 to turn off Q3. 10K is fine.

 

[MikeMl]

Why does the valve need to be driven from both ends?

 

[audioguru]

The Mosfets you selected need 10V gate-source voltage to turn on fully. If the control signal level and the supply voltage are less than 10V then some of those Mosfets will not work. You might need "logic level" Mosfets.

 

[dr pepper]

If you really didnt want to go logic level fets, you could use a ferrite torroid and make a gate drive transformer from a 3e5 torroid, you could redesign the circuit and drive both gates with it, if you use a 7555 Q2 might be dispensable.

Would that be a moog hydraulic valve?

 

[ronsimpson]

We need to know V1 and V2 because it effects Q3 Gate.
If V1 is larger than 20V we need to rethink Q3 Gate circuit. If V1 is below 10V then Logic Level is probably better.
I would use LL-MOSFET for Q4 and not make a transformer.

Why does the valve need to be driven from both ends?

The idea is to have 3 different voltages on the valve. V1, V2, 0V.

 

[earckens]

Why does the valve need to be driven from both ends?

Sorry for the late reply, just now I saw these other messages.
The electrovalve gets V1 (the higher voltage) during less then a second; V2 is then applied: this is the minimum voltage needed to keep the valve activated.

 

[earckens]

The Mosfets you selected need 10V gate-source voltage to turn on fully. If the control signal level and the supply voltage are less than 10V then some of those Mosfets will not work. You might need "logic level" Mosfets.

Thanks, I wasn't aware they were not logic level MOSFETS. Will be adapted.

 

[earckens]

If you really didnt want to go logic level fets, you could use a ferrite torroid and make a gate drive transformer from a 3e5 torroid, you could redesign the circuit and drive both gates with it, if you use a 7555 Q2 might be dispensable.

Would that be a moog hydraulic valve?

No issue with logic level FET's. The circuit works fine, and I use it to drive irrigation waterfeed electrovalves.

 

[earckens]

Here is my actual schematic; for some reason I dropped Q3 as MOSFET and use a darlington; I think it was because I could not get it to switch properly at 12V, and I have plenty of these TIP125 lying around. And since it switches just for 20ms there is virtually no heat dissipated.