DC/DC Converter Help

[alec_t]

.model BZX79C5V1 D(Is=2n Rs=.5 Cjo=200p nbv=3 bv=5.1 Ibv=1m Vpk=5.1 mfg=Me type=Zener)

 

[ronsimpson]

mfg=Me

point of view

mfg=you

mfg=alec_t

 

[LukeKnepp]

What is L2 for????? why inductor from ground to Anode of diode??

 

[ronsimpson]

What is L2 for????? why inductor from ground to Anode of diode??

The "boost" supply that has only one coil; you have a 20V supply and on top of that you make a 13 supply to get 33V. The PWM is only a 13V x 1A = 13 watt supply. The output can not be less than VIN.

The supply with two coils; (transformer) starts out at ground. The output supply is 33V x 1A = 33 watts. The output can be 0 volts.

 

[LukeKnepp]

The PWM is only a 13V x 1A = 13 watt supply.

But I needed about a 3.5A???

 

[ronsimpson]

But I needed about a 3.5A???

Sorry. I have many thing to do.

 

[LukeKnepp]

alec_t,

Is there any way to set a voltage for this thing to kick back on?

Every time the battery dies, I have to short the two sides of the MOSFET before it will restart!!!

Thank you so much for your input, it is working great!!!!!!!!!!!!!!!!!!!!!!!!!!

 

[alec_t]

Every time the battery dies, I have to short the two sides of the MOSFET before it will restart!!!

Disconnecting and re-connecting the battery will also reset the circuit.
I thought a cut-off-and-stay-off arrangement is what you wanted, hence the latch formed by Q2/Q3. Without a latch a cut-off circuit will hunt and repeatedly turn off/on/off/on......, because battery voltage generally bounces back up once the load is disconnected.
If your aim is to prevent damage to the battery from over-discharge, why do you want it to switch back on? It's doable, but I don't see why you'd want that.

 

[LukeKnepp]

Disconnecting and re-connecting the battery will also reset the circuit.

I tried that, and it doesn't work... It does not reset unless I short the MOSFET's source and drain...

 

[LukeKnepp]

Do I have to somehow drain the capacitors' charge after shutoff before it will reset?

Notice the component changes I made, I am actually using a 1N5234B Zener diode, and the FDS4685 MOSFET...

 

[eTech]

Hello,

uh....why don't you select a switching regulator chip that has UVLO and SHDN already built in?

eT

 

[alec_t]

Do I have to somehow drain the capacitors' charge after shutoff before it will reset?

C2 needs to discharge. I made no provision for that as I assumed leakage current somewhere would do the discharge.
I've simplified the circuit. Try this:

Edit: You can connect a ~1meg resistor across C1 to ensure its discharge.
A series resistor Rx (say 4k7) between R1 and C1 would be advisable so that R1 and Rx form a potential divider to ensure Vgs of the FET is well below its maximum rated value.
Note that during the FET's transition from fully on to fully off it will dissipate several Watts of power (depending on the load), but as the transition time is short a heatsink is unnecessary.
In the real world you might find you need a capacitor (~1uF or more?) across R3 or the zener diode to reduce transients caused e.g. by the load.
R4 is probably redundant, but may help in stabilising the zener voltage.

 

[LukeKnepp]

Will build this soon, thank you!!!!

Works on simulator!!!!!!

Must leave, but thank you very much!!!

 

[alec_t]

Works on simulator!

I find your simulator colour scheme hard on the eyes, but I expect the printer ink/toner manufacturers would approve . Blue components on a black background are almost invisible!

 

[alec_t]

Here's an alternative version if you want to use a N-MOSFET instead of a P-MOSFET for the cut-off:

 

[LukeKnepp]

Which one of these would have less current drain after cutoff?

Also, what voltage will this turn back on?
Or does it only reset when the power is removed??? That is how the simulator shows it, and I am hoping that is how it works!!!!

Thank you so much for your help!!!

 

[alec_t]

Which one of these would have less current drain after cutoff?

If you connect things as suggested in the post #92 edits then both circuits would have the same current drain after cut-off, i.e about 20uA via the 1meg resistor, plus any leakage current through the MOSFET (hopefully trivial).
Both circuits are designed to need disconnection of the supply (or shorting the MOSFET drain and source) in order to reset.
The post #95 circuit is just an 'upside-down' (opposite polarity) equivalent of the post #92 circuit.

 

[LukeKnepp]

Back to this circuit, I thought I was ready for my production run when I hooked it up to my machine to test once more, and now every so often the light blinks and it quits running for about 1/2 second and then resumes. It does this about once every 10-20 seconds....

I added another output capacitor... Without results...

The only thing I changed was the inductor from this one to this one...

Edit - I also tried the other inductor and it made no difference...

 

[alec_t]

I hooked it up to my machine

Which machine?

the light blinks

Which light?

I added another output capacitor

To which output?

The only thing I changed was the inductor

Inductors generate voltage spikes (which can interfere with circuit operation) when the current through them is interrupted. What precautions have you taken to suppress spikes?

 

[LukeKnepp]

Which machine?

The sewing machine I am trying to run....

Which light?

The light on the sewing machine that is on when the machine has power...

To which output?

The 33V output...

What precautions have you taken to suppress spikes?

None that I know of, how can I add this?