circuit design for breaker

[shreysuman]

Sir, i want to design a circuit that can break Power supply (Constant current source) when cell voltage drops below 450 mV

 

[AnalogKid]

What voltages are available to run the circuit?
What is the current range for the constant current source?
Can you post a schematic or sketch of the system you have so far?

ak

 

[alec_t]

What type of cell? 450mV seems a very low limit if cell damage is to be avoided.

 

[shreysuman]

What voltages are available to run the circuit?
What is the current range for the constant current source?
Can you post a schematic or sketch of the system you have so far?

ak

Sir I have a setup to test a small size fuel cell.In this hydrogen and oxygen are used as fuel.In the absence of fuel ,cell potential starts decreasing and if load (Current source) is connected to the cell then cell can damage.The OCV for this cell is 1 V (approx).But if by any means there is a starvation of fuel then cell voltage starts decreasing rapidly and if load (current sourse )is connected to the cell then cell can damage.It is seen that if cell voltage is above 450 mV then cell is safe but as soon as it goes below 450 mV then cell damage .To protect the cell in the absence of fuel this protection circuit is required by disconnecting the current source(Load) from cell.schematic is attached ..

 

[shreysuman]

What type of cell? 450mV seems a very low limit if cell damage is to be avoided.

Sir I have a setup to test a small size fuel cell.In this hydrogen and oxygen are used as fuel.In the absence of fuel ,cell potential starts decreasing and if load (Current source) is connected to the cell then cell can damage.The OCV for this cell is 1 V (approx).But if by any means there is a starvation of fuel then cell voltage starts decreasing rapidly and if load (current sourse )is connected to the cell then cell can damage.It is seen that if cell voltage is above 450 mV then cell is safe but as soon as it goes below 450 mV then cell damage .To protect the cell in the absence of fuel this protection circuit is required by disconnecting the current source(Load) from cell.schematic is attached ..

 

[AnalogKid]

Is the load a resistor as shown in your drawing, or is it a true constant current source (sink) circuit?

ak

 

[shreysuman]

Is the load a resistor as shown in your drawing, or is it a true constant current source (sink) circuit?

ak

sir it is constant current source ,we called constant current source as load,

 

[alec_t]

So, to help determine the type of cut-out needed, what is the value of this constant current? Microamps? Many Amps? ....?

 

[shreysuman]

So, to help determine the type of cut-out needed, what is the value of this constant current? Microamps? Many Amps? ....?

Sir it is amps range 0 to 20 amps.

 

[alec_t]

Given those voltage and current figures I think the most practical cut-out would be an automotive relay. That would need a 12V supply and a comparator circuit set to trigger at about 450mV, probably with a bit of hysteresis.

 

[crutschow]

A voltage reference, such as a TL431, a comparator such as an LM339/393, a transistor such as a 2N2222, a diode such as a 1N4148, and a relay, should do what you want.

 

[alec_t]

How dependent is the cell voltage on load current? Any comparator hysteresis involved would need to take this into account.
Once triggered, should the cut-out latch until manually reset? If not, under what conditions should reset occur?

 

[ronsimpson]

break Power supply (Constant current source) when cell voltage drops below 450 mV

OCV for this cell is 1 V

amps range 0 to 20 amps

It looks like the only source of power is the 1 to 0.45 volts from the cell. This makes most approaches not work. I think the voltage will collapse to zero for a short time. While the breaker is opening the voltage is between 0 and 0.45.

Please confirm that there is no other source of power.

 

[alec_t]

It looks like the only source of power is the 1 to 0.45 volts from the cell.

If that's the case then the only suggestion I have would be to modify a solenoid valve of the type used in flame-failure sensors. That type has a very low resistance coil powered by a thermocouple, so that even a fraction of a Volt can drive an Amp or so through the coil to hold the valve open.

 

[ronsimpson]

It looks like the only source of power is the 1 to 0.45 volts from the cell.

If this is true; I can make a boost power supply. Vin 0.5V to 1V, Vout 5V with a super cap to hold for a while. I think we need a latching relay.

 

[alec_t]

I see 1V>5V boost converters are available off-the-shelf. Might strike lucky and find they work at a tad under 1V.

 

[ronsimpson]

I see 1V>5V boost converters are available

I think it needs to work down to 0.45.

Here is a LED boost that can be modified for constant voltage not constant current. It works down to 0.50V.
I have used other parts that work lower but many are not in production now.
----edited----
The MCP1643 needs 0.65V to 0.8V to start. A unloaded cell will do that. Once started it will work down to 0.5. I think most worked to 0.45 if the load is light.
----edited----
Found boost PWM down to 0.1V but the price is 10X. "energy harvester"