Hi,
I have a PSU that outputs 13.8VDC with a max load when all attached equipment is being used of 6.65A.

I have been reading a lot about over voltage circuit protection & this is something that I wish to add to my setup.

I've read about many different configurations & seen many, many different diagrams.

As a novice I really need some help here.

I am looking to protect my equipment by shutting off the supply if the voltage rises to 14VDC.

How do I calculate the values for the various components required in say, this setup: http://www.electronics-tutorials.ws/power/transient-suppression.html
under the section:
Zener Crowbar Clamping Circuit

for my given input voltage & load?

Any help would be appreciated.

Regards..,

 

Welcome to AAC!

Post a schematic for the power supply. Can it tolerate a short?

What is the nature of the load that it can't tolerate a variance of 2%?

 

OK, I have just stumbled across this: http://www.hwmakers.eu/lt-spice-iv/compact-overvoltage-undervoltage-reverse-protection-with-ltc4365/

It uses an IC LTC4365 that seems to do everything that I require.

Someone does take issue about 'transient voltage protection' on the input side: http://hackaday.com/2012/01/04/power-protection-circuit-tutorial/

but then I found the datasheet: http://cds.linear.com/docs/en/datasheet/4365fa.pdf

and that is covered in fig20.

So, please, what would be the components needed for my setup, how do I calculate the values for the resistors, diode etc in the fig20 on the PDF.

Regards..,

Kirk

 

Welcome to AAC!

Post a schematic for the power supply. Can it tolerate a short?

What is the nature of the load that it can't tolerate a variance of 2%?

Hi,
I cannot post a schematic of the PSU, it's an off the shelf purchase.

The connected equipment can use a range of voltage from 12VDC to 14.5VDC, I just picked a arbitrary number higher than my PSU's output but lower than the top end.

 

Still don't understand what problem you're trying to solve.

All power supplies have a tolerance. Single chip linear regulators often have a spec of +/- 4%.

If the power supply won't tolerate a short, you can't use a crowbar. If you don't know if it will tolerate a short, it's unwise to contemplate shorting the output.

 

Let's say for instance that the PSU does not have protection or perhaps this inbuilt protection fails & let's also say that the PSU will tolerate a short, that said & all disclaimers in place what would be the specs for the various components as required to build a protection circuit with the details I have given.

Regards..,

Kirk

 

Let's say for instance that the PSU does not have protection or perhaps this inbuilt protection fails & let's also say that the PSU will tolerate a short, that said & all disclaimers in place what would be the specs for the various components as required to build a protection circuit with the details I have given.

Google is your friend...

From http://www.radio-electronics.com/info/circuits/scr_overvoltage_crowbar/scr_over_voltage_crowbar.php

When the voltage gets high enough, the zener starts to conduct. When the voltage across the series resistor gets large enough to turn on the SCR, it will short the supply; resulting in the fuse blowing.

If you want more precise control, you can replace the zener+resistor with a comparator.

 

Google is your friend...

From http://www.radio-electronics.com/info/circuits/scr_overvoltage_crowbar/scr_over_voltage_crowbar.php
View attachment 128791
When the voltage gets high enough, the zener starts to conduct. When the voltage across the series resistor gets large enough to turn on the SCR, it will short the supply; resulting in the fuse blowing.

If you want more precise control, you can replace the zener+resistor with a comparator.

Thanks for replying, indeed Google can be your friend, that is how I found All About Circuits when I needed more help on this subject after searching and finding numerous sites similar to the one you have just listed.

I do not know the ratings of the components needed for my particular setup or in fact how to calculate them.

Of particular interest to me relates to my post #3 in this thread, regarding the LTC4365 setup.

If I have 13.8VDC in & I wish to protect the circuit from over voltage of 14VDC using the LTC4365, where or how do I calculate the values for the resistors & diode in the schematic on fig20 of the pdf?

I am willing and able to learn if pointed in the right direction.

Regards..,

 

If I have 13.8VDC in & I wish to protect the circuit from over voltage of 14VDC using the LTC4365, where or how do I calculate the values for the resistors & diode in the schematic on fig20 of the pdf?

Overkill for what you want. Use the circuit with the SCR, zener, and resistor. If that doesn't give you sufficient control, use the comparator.

 

Overkill for what you want. Use the circuit with the SCR, zener, and resistor. If that doesn't give you sufficient control, use the comparator.

OK, thanks for the tip, although I still need pointers on how to calculate the values for the components I would require.
Seems very difficult to get answers, I need info more in layman's terms please, thanks.

 

For the LTC4365, if you don't want under-voltage or reverse-connection protection then you can use the circuit in Fig. 13 which only requires one MOSFET.
The over-voltage point is where the voltage at the OV (OverVoltage) pin exceeds 500mV nominal so you pick R1 and R2 for a resistive divider that generates 500mV at the OV point you want.
For 14V you could use 10kΩ (arbitrary value) for R1, giving a value of 270kΩ for R2.

Pick an N-MOSFET with an ON resistance such that it dissipates (I²R) no more than about 1W at the maximum current output you will have.
Thus 6.65A maximum would mean an ON resistance of ≤22mΩ.
If you go much higher than that you would have to put the MOSFET on a heatsink.

Here's an LTspice simulation of the circuit:
Note that it cuts off the output at 14V.

 

For the LTC4365, if you don't want under-voltage or reverse-connection protection then you can use the circuit in Fig. 13 which only requires one MOSFET.
The over-voltage point is where the voltage at the OV (OverVoltage) pin exceeds 500mV nominal so you pick R1 and R2 for a resistive divider that generates 500mV at the OV point you want.
For 14V you could use 10kΩ (arbitrary value) for R1, giving a value of 270kΩ for R2.

Pick an N-MOSFET with an ON resistance such that it dissipates (I²R) no more than about 1W at the maximum current output you will have.
Thus 6.65A maximum would mean an ON resistance of ≤22mΩ.
If you go much higher than that you would have to put the MOSFET on a heatsink.

Here's an LTspice simulation of the circuit:
Note that it cuts off the output at 14V.

View attachment 128800

OK , brilliant that gives me some numbers to crunch. Thanks

 

Seems very difficult to get answers, I need info more in layman's terms please, thanks.

You'll get more out of this if you take the time to understand how the circuit works so you can calculate the values yourself.

For the zener/SCR crowbar, you need an SCR that has a hold off voltage (the voltage where it starts to conduct without being triggered) sufficiently above the voltages you want to allow. The zener needs to be lower than the crowbar voltage, but not so low that it will dissipate excessive power. The resistor value depends on Ohm's Law.

If you used a 1N5243 13V zener with a 2N5060 SCR with a trigger voltage of about 0.8V, ideally, the scr would trigger at about 13.8V. Unfortunately the devils is in the details. A 13V zener will conduct at a voltage of 12.35-13.65V and the trigger of the SCR is a maximum of 0.8V at room temperature; increasing with lower temperatures.

If you require a more precise crowbar, you could use a comparator.

If you want to learn how to design circuits, you need to find circuits you can study and understand. Most members on this site aren't willing to put more effort into your design than you do.

 

Hi,
I have a PSU that outputs 13.8VDC with a max load when all attached equipment is being used of 6.65A.

I have been reading a lot about over voltage circuit protection & this is something that I wish to add to my setup.

I've read about many different configurations & seen many, many different diagrams.

As a novice I really need some help here.

I am looking to protect my equipment by shutting off the supply if the voltage rises to 14VDC.

How do I calculate the values for the various components required in say, this setup: http://www.electronics-tutorials.ws/power/transient-suppression.html
under the section:
Zener Crowbar Clamping Circuit

for my given input voltage & load?

Any help would be appreciated.

Regards..,

The crowbar shorts the PSU output at overvoltage and you have to cut power to reset it - that doesn't sound like what you need.

Old British motorcycles used a dirty great 100W 15V zener to regulate the charging system, but that's more voltage than you asked for.

A DIY shunt regulator is probably the way so you can make it adjustable.

The published application notes describe how to add power transistors to a TL431, it gives a sharper knee than basing it on a Zener diode.

 

The TL431 is a brilliant simplification for many op-amp and comparator circuits, but the LT4365 is even better for this application.

 

....
The over-voltage point is where the voltage at the OV (OverVoltage) pin exceeds 500mV nominal so you pick R1 and R2 for a resistive divider that generates 500mV at the OV point you want.
For 14V you could use 10kΩ (arbitrary value) for R1, giving a value of 270kΩ for R2....

OK, is it possible to show the formula used to acheive this result? Thanks

 

Vmax x R4/(R3+R4) = 0.500V

 

Vmax x R4/(R3+R4) = 0.500V

Thank you

 

OK, if I've got understand this correctly and my maths stands up I've come up with this.

For a 13.8v supply to cut off 14.3v (Vin + 500mV) I could use 9.53Kohm + 267Kohm resistors as the voltage divider.

Is this correct?

Now I need to figure out how to calculate the R6 value & the C-out value in fig13 of the LTC4365 datasheet for my particular setup.

Thank in advance

 

OK, if I've got understand this correctly and my maths stands up I've come up with this.

For a 13.8v supply to cut off 14.3v (Vin + 500mV) I could use 9.53Kohm + 267Kohm resistors as the voltage divider.

Is this correct?

No quite.
My calculations give a cut off of 14.5V.