mosfet connections in place of diode

Thread Starter

joewales44

Joined Oct 8, 2017
150
since i didn't get a suggestion, i made an attempt.
here is my thinking...please tell me where i'm wrong.
for Nfets Q4 and Q5, when switch is up, source is high and gate is low so it conducts.
when switch is down, source is low and gate is low, so it turns off.
can i use Nfets for this situation where source is high?
from previous directions for Pfets at LED1 and LED4, someone said i must use Pfets in this situation.
i tried to use Pfets for Q4 and Q5 but never could figure out how it would work.
do i need R6 and R7 or could i connect straight to ground?
is my logic correct or am i in the weeds here?
thanks
upload_2018-5-7_16-35-26.png
 

Thread Starter

joewales44

Joined Oct 8, 2017
150
those are way over my head. i'm mechanical, not electrical.
i don't have a clue on how to hook them up in my circuit.
cost is too high for my application and mouser or digikey don't have any.
are you saying Q4 and Q5 in my schematic won't work?
thanks
 

Sensacell

Joined Jun 19, 2012
2,718
It's not a stupid question, but your post is not terribly inviting.

It's no fun to decode what is going on from your messy schematic; it's a confusing layout with little context.
My eyes glaze over, and I move on...
 

Picbuster

Joined Dec 2, 2013
1,026
those are way over my head. i'm mechanical, not electrical.
i don't have a clue on how to hook them up in my circuit.
cost is too high for my application and mouser or digikey don't have any.
are you saying Q4 and Q5 in my schematic won't work?
thanks
Please write down what your circuit should do.
Each circuit has input(s) and output(s). ( output or input can be heat, light, power, voltage, current, pulse or frequency or a combination)
Describe them allowing us to understand what you want to create.

Picbuster
 

WBahn

Joined Mar 31, 2012
26,398
When asking for free help from strangers all over the world, the onus is on you to make it easy for them to help you.

You title indicates you are trying to replace a diode in that circuit with a MOSFET. Okay, which one? Why? What is it you are trying to accomplish?

Then you talk about a switch being up. But you have two switches. Which one are you talking about?

By this point, a lot of members will simply move on. You've made it too difficult to even attempt to offer any assistance.

And then when someone does offer a suggestion but doesn't rush back and answer your response in less than two hours (when it took YOU three to respond to their suggestion), you make a snarky comment. That does not encourage people to engage with you -- quite the opposite.
 

WBahn

Joined Mar 31, 2012
26,398
those are way over my head. i'm mechanical, not electrical.
i don't have a clue on how to hook them up in my circuit.
cost is too high for my application and mouser or digikey don't have any.
are you saying Q4 and Q5 in my schematic won't work?
thanks
Since cost is apparently a factor, might it not be a good idea to give a hint on what the limit is on the cost of a solution?

If you are limited to devices available from Digi-Key or Mouser, might it not be a good idea to indicate that?

Otherwise people are playing a guessing game and they will very quickly get tired of doing so and move on.
 

OBW0549

Joined Mar 2, 2015
3,566
is my logic correct or am i in the weeds here?
You're in the weeds. DEEP in the weeds.

This is just plain wrong:
for Nfets Q4 and Q5, when switch is up, source is high and gate is low so it conducts.
Stop what you're doing and learn the basics of how MOSFETs work:
  • For an N-channel MOSFET, the gate must be positive with respect to the source to turn it on.
  • For a P-channel MOSFET, the gate must be negative with respect to the source to turn it on.
As you have it now, your circuit cannot possibly function-- no matter what it was intended to do.

Speaking of "function", what the heck is this circuit supposed to do, anyway? Are we supposed to read your mind or something? In your first sentence you said "since i didn't get a suggestion, i made an attempt." That indicates to me that your post is somehow related to another thread, somewhere, but you don't say where. Instead of starting a new thread here, you should have just posted as a continuation of the previous thread so your questions would have some context.
 

Thread Starter

joewales44

Joined Oct 8, 2017
150
i was asking the difference how N and P mosfets work.
since then i learned what you said about Nfets needing higher gate than source.
the circuit works as it should, just needed to eliminate voltage drop across diodes.
i have very little electronics experience.
sorry i bothered everyone.
thanks any way.
 

ebeowulf17

Joined Aug 12, 2014
3,275
i was asking the difference how N and P mosfets work.
since then i learned what you said about Nfets needing higher gate than source.
the circuit works as it should, just needed to eliminate voltage drop across diodes.
i have very little electronics experience.
sorry i bothered everyone.
thanks any way.
I think the main issue here is context. I remember seeing this circuit in your earlier threads, but obviously others here didn't see it. If you could provide a link to the other thread, that would give people a way to see what you've done so far and how/why you've gotten to this point.

Also, the earlier posts didn't explain why you were switching from diodes to MOSFETs. Without knowing why you want to change, it's hard to comment on how effective the change will be. This post makes it clear that you're trying to reduce/eliminate the voltage drop, so we're one step closer now.
 

ebeowulf17

Joined Aug 12, 2014
3,275
i was asking the difference how N and P mosfets work.
since then i learned what you said about Nfets needing higher gate than source.
the circuit works as it should, just needed to eliminate voltage drop across diodes.
i have very little electronics experience.
sorry i bothered everyone.
thanks any way.
As for the specifics of this situation, it's early and I'm not thinking clearly, but I'm not sure you can use a MOSFET the way you want to here. If I'm not mistaken, you're trying to mimic the concept used at Q3 for reverse polarity protection.

That MOSFET is configured such that as long as power is wired correctly, it always conducts, but if power is wired backwards, it interrupts the circuit ground.

What you want in the new location is a diode function that allows voltage to pass through to both rails when K1 C3 is powered, but doesn't allow voltage to pass through it when switch one is low, powering K1 C2. Since, in this new situation, the ground doesn't change polarity, you can't use the same circuit concept here.

There should be a straightforward way to use a MOSFET to do what you want, but I'm to sleepy to do it right now, and I'm not at a real computer where I can easily sketch schematics. If you don't get this figured today, I'll try to jump back in later and see if I can help.
 

Thread Starter

joewales44

Joined Oct 8, 2017
150
circuit works as it should thanks to CRUTSCHOW.
K1 is main power that turns on string 1 either with or without LED1 being turned on.
K1 also supplies power to K2.
K2 turns on string 2 with or without LED4 being turned on.
i just don't have enough overhead with 12 volts.
i got my head bit off because i didn't understand the difference in Ns and Ps.
i have zero electronics training.
i INCORRECTLY thought there only had to be a 10 volt difference in either direction which i will never forget.
i hope i'm not as grouchy when someone asks me about something in my field of expertise...lol.

i updated drawing to original parts and eliminated TVS diodes which seemed to create confusion.
how can i replace D3 and D4 with something that has very little voltage drop?
thanks
upload_2018-5-9_9-25-57.png
 
Last edited:

ebeowulf17

Joined Aug 12, 2014
3,275
circuit works as it should thanks to CRUTSCHOW.
K1 is main power that turns on string 1 either with or without LED1 being turned on.
K1 also supplies power to K2.
K2 turns on string 2 with or without LED4 being turned on.
i just don't have enough overhead with 12 volts.
i got my head bit off because i didn't understand the difference in Ns and Ps.
i have zero electronics training.
i INCORRECTLY thought there only had to be a 10 volt difference in either direction which i will never forget.
i hope i'm not as grouchy when someone asks me about something in my field of expertise...lol.

i updated drawing to original parts and eliminated TVS diodes which seemed to create confusion.
how can i replace D3 and D4 with something that has very little voltage drop?
thanks
View attachment 152165
One very simple way to reduce the voltage drop would be to choose a different diode with lower Vf. Many Schottky diodes have Vf around 0.2 instead of the 0.6-0.7 that's normal for typical diodes. I vaguely remember BAT42, BAT43, etc. being useful for me this way. I don't remember how much current they could handle, and the Vf increases with greater current flow, so you'd have to check the datasheets to see if they would get your voltage drop low enough.

Meanwhile, I'm brainstorming the MOSFET solution but don't have anything solid yet.
 

OBW0549

Joined Mar 2, 2015
3,566
how can i replace D3 and D4 with something that has very little voltage drop?
You could use a high-current Schottky rectifier diode for D3 and D4, such as an STPS10L25D (rated at 25V Vr(max) and 10A If(max), which has a forward voltage drop of around 450 mV at If = 10 A. The typical forward voltage curves give the drop at other currents:

Untitled.png

That's about as good as you're going to get, I'm afraid.
 

Thread Starter

joewales44

Joined Oct 8, 2017
150
i've been using ss14 schottkys.
i'm pulling 800mA through D3 when K1 is up and 400mA through D4.
when both switches are up is the problem.
i'll check both recommendations.
thanks
 
Last edited:

ebeowulf17

Joined Aug 12, 2014
3,275
I think this will work. Rail1 represents the high side of your top switch, and Rail2 represents the low side. I've had to simulate the switching differently, because I don't have models of SPDT switches, but the resulting behavior is the same.

When switch is set to power Rail1, Q1 is activated, pulling MOSFET gate low, activating MOSFET and powering Rail2. Voltage drop across MOSFET is a few mA, instead of hundreds with a diode.

When switch is set to power Rail2, R3 guarantees that MOSFET gate and source voltages are the same (or very nearly the same) so that MOSFET is inactive, leaving Rail2 powered, and Rail1 un-powered.

Hopefully some more experienced users will double check me here and make sure there aren't pitfalls that I'm overlooking, but I think it makes sense, and it plays nice in simulation!

MOSFET-as-Diode_01.png
 

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ebeowulf17

Joined Aug 12, 2014
3,275
thats over my head.
guess i'll give up.
thanks everyone.
It's only two more components than your MOSFET solution.

  • First you had a diode.
  • Then you proposed a MOSFET with one extra resistor.
  • Now I'm proposing 1NPN, 1P-MOSFET, and two resistors.
This isn't very complicated compared the size and complexity already in your project. If you need the voltage, why not implement a fix?

Incidentally, why does it matter if you lose half a volt through the diode? How much voltage drop can you tolerate? If it doesn't actually matter, I can understand not fixing it, but if your circuit is misbehaving with the voltage drop, I think we have a solution for you.
 
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