DC/DC Converter Help

[alec_t]

how can I add this?

Connect a reverse-biased diode, rated for at least 3A, across the motor (which has considerable inductance, so is likely generating voltage spikes and causing the drop-out effect you are experiencing).

 

[LukeKnepp]

across the motor

Can I just put this diode across the entire load?

 

[LukeKnepp]

Never mind, I believe I figured it out!!!!!!!!!!!!!

 

[alec_t]

Can I just put this diode across the entire load?

The sewing machine motor (possibly with snubber components) is the intended load, as I understand it. Rload in my circuit is just for simulation purposes.

 

[LukeKnepp]

That low voltage shutoff circuit is not working...

It just turns partially off as the voltage decreases, and doesn't kick off...

When I asked for a reset-able cutoff, what I meant was a way to electronically reset the circuit below, instead of manually shorting the source and drain of the MOSFET...

Thanks!!!!

 

[alec_t]

That low voltage shutoff circuit is not working...

Which one (I've posted three, but it looks like you are using the first one)?
Have you tried the post #92 circuit with the modifications I suggested in the edits?
Have you got the diode across the motor?
Does the motor need to reverse in use?
Can you post a schematic of the complete circuit as you currently have it assembled?

 

[LukeKnepp]

Whoops, I blew up my sewing machine!!!!!!!!!!!

Complete schematic of the sewing machine board-

Schematic of low voltage cutoff circuit #1 -

Works EXACTLY how I want it, except I need a way to reset it after it shuts off, instead of manually shorting the source and drain...

Schematic of Low Voltage Cutoff #2 -

This one only slows down the load, starting at about 16.6V, and doesn't shut off until like 15.5V...
It also turns back on when I put the voltage back up, which would not work with a large load...

 

[alec_t]

You haven't answered the questions in post #106. If the motor gets reversed then the diode should NOT be connected across it.
With cutoff ~2 shown above, connection to downstream circuits should be from the two terminals of R7 (the simulated load); NOT from between 'out' and the ground symbol. Depending how you've connected that circuit to the rest of your system might account for its non-latching behaviour. The ground symbol shown is for simulation purposes and perhaps would be better placed below R7. Apologies for any confusion. The added capacitor (post #98) might also be responsible for the non-latching behaviour.
Re cutoff #1, connecting a 1meg resistor across C2 should allow the circuit to reset quickly once power source V1 has been disconnected.

 

[LukeKnepp]

You haven't answered the questions in post #106. If the motor gets reversed then the diode should NOT be connected across it.

I did not connect the diode...

With cutoff ~2 shown above, connection to downstream circuits should be from the two terminals of R7 (the simulated load); NOT from between 'out' and the ground symbol.

Correct, That is how I did it...

The added capacitor (post #98) might also be responsible for the non-latching behaviour.

Which capacitor???

 

[alec_t]

Which capacitor?

I don't know; but in post #98 you said you had added one!
BTW, glad you got rid of those invisible blue components .

 

[LukeKnepp]

in post #98 you said you had added one!

I had taken it back off when it changed nothing...

BTW, glad you got rid of those invisible blue components .

I had no idea how hard I was straining to see until I changed it!!!

 

[LukeKnepp]

Also, every time I hook up a larger load, the board shuts off, so I have to have the load connected and then short the source and drain, and then it works great?????
Do I just have to build in a manual reset switch to short the source & drain?

 

[alec_t]

Also, every time I hook up a larger load, the board shuts off

Not surprising. Large loads often pull the supply voltage down.

Do I just have to build in a manual reset switch to short the source & drain?

That's an option. A momentary switch (button) would do it. But doesn't the 1meg resistor drain the cap (C2, cutoff #1) once the supply V1 has been removed from the circuit? C2 can't discharge if V1 is left connected.

 

[LukeKnepp]

But doesn't the 1meg resistor drain the cap (C2, cutoff #1) once the supply V1 has been removed from the circuit? C2 can't discharge if V1 is left connected.

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!YYYYYYYYYYYYYEEEEEEEEEEEEEEEEEEESSSSSSSSSSSSSSSSSSSSSS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

I did not get what you were saying!!!!!!!!
Thank you so much!!!

Two last questions(Maybe)

Large loads often pull the supply voltage down.

Is there a place to put a capacitor to delay the shutoff, so that when the voltage drops when I hook up a load, it would not shut off???

Also, how do I, or can I make so that it switches off even on a small load, instead of slowly dropping to zero??

THANKS AGAIN!!!!!!!!!!!!!!!!!!!!

 

[alec_t]

Is there not a place to put a capacitor to delay the shutoff

Increasing the value of C2 (cutoff #1) or C1 (cut-off #2) should help.
A properly chosen, well designed power supply should already include enough capacitance, or a regulator, so that its voltage doesn't droop below the cut-off trip point when you connect the intended load. Any connecting wire between the supply and the cut-off needs to be fat enough that its resistance doesn't cause a significant voltage drop.

 

[LukeKnepp]

MANY THANKS!!!!!!

How do I, or can I make so that it switches off even on a small load, instead of slowly dropping to zero??

 

[alec_t]

Define 'small' and 'slowly'.
Simulation shows that the cut-off should take only a fraction of 1mS, regardless of the load size, unless the load is highly capacitive. I'm not seeing the output 'slowly dropping to zero' for mainly resistive loads.
Since you have LTspice, can you post your asc file where 'small' and 'slow' demonstrate the problem?

 

[LukeKnepp]

Since you have LTspice, can you post your asc file where 'small' and 'slow' demonstrate the problem?

Not in LTSpice, when I actually built the circuit...

 

[alec_t]

So, what is a small load, and how slowly does the output drop in the real world? I'm struggling to see which circuit parameters may account for the difference between simulation and reality here.

 

[LukeKnepp]

The small load is about 250 mA, and the output drops from 16V to 0V in 15-20 minutes.(Very estimated guess!!)

I tested that the day before yesterday, I am testing it again now, with the added components, but I have to wait for the battery to drain that far first...

I will let you know late tonight or early Monday morning if I can, since I have an outing planned this weekend starting at 2:30 tomorrow morning...

Thank you so much for your help!