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Ohms to trigger a switch

Alan P

Mar 12, 2015
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The sender has its own power supply as it is connected to a gauge. If I'm reading this right, it's going to put 12v into the sender which I don't need. I'm slightly confused by 135R. Is that a resister? Assuming that 12v represents +12v and the bottom right line end is ground, the sender wires connect between full and empty? If so, can I just cut the +12v at the sender since it has its own ps?
 

ADRT

Nov 25, 2014
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Ok, so the sender unit actually sources the output? This means that there is a voltage not a resistance reading on the output? If this sources and is not a sink output then that's a whole different ball of wax.
 
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davenn

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Sep 5, 2009
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Ok, so the sender unit actually sources the output? This means that there is a voltage not a resistance reading on the output? If this sources and is not a sink output then that's a whole different ball of wax.

indeed
 

ADRT

Nov 25, 2014
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Do you have the model number for the sending unit? Mybe a manual or schematic?
 

Alan P

Mar 12, 2015
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I probably should have been clearer. There is a voltage reading coming off the sending unit. At empty I read 5.0v dc. My trigger voltage is about .8v dc. .8v is almost full but still leaves me a safe buffer in case there is any delay. The gauge has a delay of about 20 seconds to read. The sender on its own is pure resistance, however There is a voltage which I did not initially expect powering the gauge.
 

ADRT

Nov 25, 2014
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Ok so can you separate the unit from the power supply? Use only the resistance part of the unit?
 

ADRT

Nov 25, 2014
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The other option would be to use the power supply from the gauge to also power the comparitor circuit.
 

Alan P

Mar 12, 2015
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The only option would be to use the existing voltage to power the circuit. I was trying to figure out how to modify the transistor circuit to use my existing power since I already have the parts to make it, but I've since ordered the comparator parts so I might have to go that direction unless I can figure out how to separate the 12v relay power from the sender power.
 

Alan P

Mar 12, 2015
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I'm working with the two schematics you guys posted here. The link you posted and the transistor circuit from Colin. I ordered a few different comparators,

IC COMP R-RINOUT QUAD 16-SOIC LT1721CS#PBF-ND
IC COMP PUSH-PULL 1.8V SOT23-5 MCP6561T-E/OTCT-ND
IC COMPARATOR R-R 1.1V SOT-23-5 497-14339-1-ND
IC COMPARATOR 2.4V REF SOT-23-6 MCP65R41T-2402E/CHYCT-ND
 

Alan P

Mar 12, 2015
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Hey Guys,

I've put together a test circuit, Unfortunately, it's not working. I have a constant +12v output, the relay never shuts off. The schematic is posted below. For simplification, the input trigger voltage is shown as simply .9v dc. When working properly, the goal is to be able to adjust the pot to the target voltage, once the input trigger matches that target voltage, the relay is triggered. I this case, .9v.

Circuit.jpg
 

duke37

Jan 9, 2011
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You should fit a pull-up resistor, try 4k7.

To find out where the +12V is coming from, disconnect R2 and R3.
Connect the input to 1.5V.
Does it switch when you twiddle?

The correct operation depends on the trigger input, a high impedance at the input will let R2 override the input.

Does the comparator use inputs down to 0V?
 

Alan P

Mar 12, 2015
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Also is this a single comparator IC? If it has multiple comparators you need to tie the unused + inputs to vcc and - inputs to ground.

This is a single comparator. The Comparitor has 5 pins. I've used 3 of them, skipping vcc+ and -. Not sure if that was correct or not. The base schematic that I used only showed those 3 connected. The comparator is linked here:
http://www.digikey.com/product-detail/en/TS881ILT/497-14339-1-ND/4576432

Does the comparator use inputs down to 0V?

No, it's .85 to 5.5v.
 

duke37

Jan 9, 2011
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If all else fails, read the instructions !

The comparator must have a power supply, not exceeding 5.5V not less than 0.85V.
The output is push/pull so does not need a pull up resistor.
The inputs can go between ground and the power supply.

If you have run the comparator without a power supply, then any input will be above the power supply and you may have popped it off.

Still do not know where the 12V was measured or where it came from.
 

Alan P

Mar 12, 2015
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Still do not know where the 12V was measured or where it came from.

The 12v is secondary voltage and is used only to power the relay which is switched at ground. If I'm misreading something, please let me know.

The inputs can go between ground and the power supply.

Can you elaborate on that, not sure what you mean. For the sake of this experiment, I have a 12v power supply which I've reduced to 5v to power the comparator and further reduced to .9v via a voltage divider to act as my trigger. I'm using a solid state relay. 12v to power the relay, 5v to power the comparator and .9v as my trigger. Is the only fault in my circuit the missing power supply to the comparator, or am I still incorrect somewhere?
 

duke37

Jan 9, 2011
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The comparator must have a power supply, 5V is good, positive to P5, ground to P2.
The inputs should not be outside the power supply range i.e. between 0V and 5V. If the inputs are outside this range, you may damage the device, depending how the input is designed. Sometimes exceeding the supply will feed current into the device which then may have its limit voltage of 5V exceeded.

Whether this is the only fault I know not.

Post details of the SS relay, It may be possible to drive it directly from the LT497.
 
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