What resistors should be used for R1,R2,R3 from E12 series for this circuit?

Thread Starter

Lonasora

Joined May 21, 2019
2
Im new here! I want to know what resitors (R1,R2,R3) from E12 series i should use to be able to make Led light up when Vin is 4.4V>Vin > 5,7.


map.PNG Map1.PNG
 

ErnieM

Joined Apr 24, 2011
7,993
Here is one easy way to compute the resistors. Start by assuming R1 + R2 + R3 = 15K.

By inspection you get R3 = 4.4K, R2 = (5.7-4.4)K, and R3 = 15K - R2 - R3.

Next I open Excel and create an equation to compute the two voltages given the resistor values.

Finally I play with real world resistor values to see how close to the ideal I can get with real parts I can get.
 

WBahn

Joined Mar 31, 2012
24,692
Is this for some kind of school project?

I think the first thing you need to do is get a different circuit.

Consider that each of the comparators will be HI for all signals on one side of its threshold and LO for all signals on the other side. That means that if a given comparator is LO within the window, it will be low outside of the window on one side of it. Since it only takes one comparator to be LO to turn on the LED, this circuit will turn on the LED for signal levels outside of the window. The best you can hope for is to get the LED to be off within the window.
 

crutschow

Joined Mar 14, 2008
23,334
E12 resistors will give you low precision of the trip points.

What voltage accuracy do you want?
 

Thread Starter

Lonasora

Joined May 21, 2019
2
Here is one easy way to compute the resistors. Start by assuming R1 + R2 + R3 = 15K.

By inspection you get R3 = 4.4K, R2 = (5.7-4.4)K, and R3 = 15K - R2 - R3.

Next I open Excel and create an equation to compute the two voltages given the resistor values.

Finally I play with real world resistor values to see how close to the ideal I can get with real parts I can get.
Can you explain about the inspection? Thanks!!
 

WBahn

Joined Mar 31, 2012
24,692
Can you explain about the inspection? Thanks!!
He chose a sum of 15 kΩ so that the current through the string of resistors is 1 mA. That means that the voltage across each resistor will be 1 V per kΩ.

But the bigger problem is that the circuit will work the opposite of what you are wanting. The lower comparator will output -15 V (actually, something more like -13 V) for all input voltages below the lower threshold while the upper comparator will output -15 V for all voltages above the upper threshold. Between the thresholds both comparators will output +15 V.

Perhaps someone else can confirm that, since I have been in the hospital since Saturday and just came out of surgery a few hours ago.
 

MrChips

Joined Oct 2, 2009
19,273
Two things I would change.
1) You do not need -15V supply. I would use a single supply circuit.
2) You do not need 15V supply. I would choose a lower supply voltage such as 5V.
 

WBahn

Joined Mar 31, 2012
24,692
Two things I would change.
1) You do not need -15V supply. I would use a single supply circuit.
2) You do not need 15V supply. I would choose a lower supply voltage such as 5V.
If he runs it at 5 V he will need to attentuate all of his signals since his window is from 4.4 V to 5.7 V.

At 5 V the input common mode voltage of the 324 only goes up to 3 V or 3.5 V, depending on the flavor. The output voltage swing only goes to 3.5 V (though it does essentially go down to zero). That's getting near the point where even when the outputs are high they might turn the LED on since that's not a minimum rating, so it could be lower.

Also, the typical sinking current at 5V supply is only 8 mA. The setup the TS current has is aiming for an LED current of around 20 to 25 mA, depending on the LED Vf.

So unless he cuts his LED current in a third (probably doable), he needs to use a different opamp anyway.

Does no one else see the same issue with the basic circuit topology that I've pointed out twice now? Are the drugs I'm on really that good?

I would run the LED between the two comparator outputs and configure them so that they behave the same (either start low and transition high at the threshold or the reverse) so that within the window one is high and one is low. Then bridge the outputs with the LED, the current limiting resistor, and (if the differential supply voltage is more than the reverse breakdown voltage of the LED) a protection diode.
 
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