DC/DC Converter Help

[kubeek]

I think that looks correct. Under some voltage on Vin it stops switching, the output capacity slowly discharges. When you go above a treshold again it starts working. You can make the ramp slower to see it more clearly or wath che output pin of the controller.

 

[LukeKnepp]

Play with the resistor to get the switching point where you want.

I tried...

I cannot get it to shut off anywhere except 10V!!!!

 

[ronsimpson]

The LT1242 needs 16 volts to start and will work down to 10 volts.
Look at the "OUTPUT" pin. If it stops at 10v on the down ward side and restarts at 16 volts that is a function of the IC. With out the TL431.

If you want to study the LT431, just look at Vin and COMP pin. It should make a choice (turn on/off) at exactly 2.5V on pin-1 of the TL431.

Here the TL431 changes at 15.8V

 

[LukeKnepp]

With out the TL431.

So how do I get the TL431 to communicate and tell it to shut off??

The simulation above is the output pin, so essentially, the circuit for low voltage cutoff is doing nothing!!!!

What am I doing wrong???

 

[ronsimpson]

Trouble shoot. Brake it down into small pieces.
1) does the TL431 work? Look at VIN and TL431-2. Does the pin change level at about 15.8 volts? low=2.5V high = 8V
2) Does COMP get pulled to ground when VIN is below 15.8 volts?
----edited----
The first several TL431 models I tried did not work. Some had a reference of 1.25 volts not 2.5 volts. So I used a LT1431 as shown below.
The LT1431 comes with Lt Spice.

 

[LukeKnepp]

Pin 2 just hovers at 700mV, do I have an incorrect model for TL431?

 

[ronsimpson]

Pin 2 just hovers at 700mV, do I have an incorrect model for TL431?

look at your voltage divider. 25k & 2.5k ??? There will be 2.5V on the 2.5k and 25V on the 25k so 27.5V is the point. Please do some math and pick a different top resistor to find a voltage of 15.8 volts or what ever you want.

 

[LukeKnepp]

33V into a 30.5k and 2.5k resistor divider puts out 2.5V.

 

[alec_t]

I need a universal solution for low voltage protection that would shut off between 16.1V and 15.5V.

Although the arrangements above can shut off the IC when the supply voltage drops below the threshold, because of the boost regulator configuration a high load current will continue to be drawn. What is needed is another FET in series with the supply to cut off both the IC and the load. Some hysteresis around the control of that FET would be advisable to prevent repeated on/off switching when battery voltage recovers after the load is disconnected.

 

[LukeKnepp]

I am looking more for something to put between this board and the power source...

Would something like this work?

 

[LukeKnepp]

Although the arrangements above can shut off the IC when the supply voltage drops below the threshold, because of the boost regulator configuration a high load current will continue to be drawn. What is needed is another FET in series with the supply to cut off both the IC and the load. Some hysteresis around the control of that FET would be advisable to prevent repeated on/off switching when battery voltage recovers after the load is disconnected.

Exactly what I am trying to say!!!!!!!

 

[alec_t]

Would something like this work?

No. The series N-FET needs a gate voltage >> supply voltage.

 

[LukeKnepp]

There is no way to modify this circuit to use?

I also thought about trying this one...

 

[ronsimpson]

I am looking more for something to put between this board and the power source...

Let me guess. You are building a LED power supply. It would have been good to know that 73 post back.
Yes or No; LEDs. What LEDs and how many. (voltage and current)

 

[LukeKnepp]

No, it is actually a sewing machine with a dc motor, and I am wanting to power it off of a DeWalt Cordless tool battery...

Like these...

However, I will need this protection board for many products, such as fans and lights...

 

[alec_t]

Here's a suggested cut-off. The threshold voltage is set by R3/R4/D1. Current drain after cut-off is ~100uA.

Edit: To ensure the FET's Vgs is well below 20V (the usual absolute maximum), connect a resistor Rx (say 4k7) between R1 and the collector of Q1, thus forming with R1 a potential divider for the FET gate.
The FET dissipates several Watts of power (depending on the load) when transitioning from fully on to fully off, but the transition time is short so a heatsink should be unnecessary.
Further down this thread I've posted a simpler circuit.

 

[ronsimpson]

Attached is my spice file. Needs some work
33V @ 1A and current limit at about 2A.
If the PWM is shut down the output voltage will be zero. (current = 0)
L1, L2 is a dual coil from coilcraft.com or a transformer with 1:1 windings.

 

[alec_t]

Nice. What is the purpose of C7 and R11? Why is Q2 gate driven by U2 output?

 

[LukeKnepp]

If the PWM is shut down the output voltage will be zero. (current = 0)

What do you mean? When will the output shut down?

What is L2 for?

Thanks for all your help!

 

[ronsimpson]

What is the purpose of C7 and R11?

L1 rings (leakage inductance) I am trying to pass that energy out to the load. snubber

Why is Q2 gate driven by U2 output?

This moves the "PWM" to a "constant off time" controller.
All the MOSFETs are off for a time set by the RC on the RC pin.
The MOSFET on time is set by the input voltage and the inductance L1. Time= what ever time it takes to input the right amount of power.
At low voltage the frequency drips. At high voltage the frequency get higher.

could you share your spice model for BZX79C5V1 diode?

Edit the diode library. There is a line for BZX....6V2 diode.
Copy that line with cut and paste and make two line the same.
Edit one of the lines. Change the name to …..5V1, and the bv=5.1 (not perfect but close) If you want closer we can work on that.