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I´m running out of ideas here.

I have a device that should always output a 5v signal, with a maximum tolerance of 100 mV (4.95 to 5.05 V).

I´m building a "tester" device. My tester will compare the received 5v signal with a minimum and maximum reference voltage (Uref Min=4.95 V Uref Max=5.05 V )

If my device is faulty I want my tester to emit a signal, this signal will be connected to a binary counter that will be telling me if my device failed and how often it failed.

On the plan below, if I understood it right, the LM139 should be closed (0 V) when my device is =>4.95 V or =<5.05 and open (~6 V) if <4.94 V or >5.05 V. Given that I want a single signal from my LM139, I built an OR-Flip Flop with two transistors (2N2222). Joining the two outputs together was impossible to make it work.

The last picture are the connections I soldered on my LM139, I tried grounding the others inputs but it didn´t change anything, I tried many things but nothing worked I can´t get this guy to work properly...

Observations:

MinV: Below DeviceU (5 V) the output2 will be almost 0 V. From deviceU (4.96 V) the voltage will slowly rise till it reaches DeviceU (5 V) and the output2 will emit a solid 6 V.

MaxV: The voltage stays stable at the output1 (6 V) when I drop the deviceU bellow 4.95 V. Everything above, the input1 will throw almost random values...

Surveillance circuit

Real connections

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  • \$\begingroup\$ That is not a "flip flop" you've built with those two transistors. It is a funky kind of upside down "OR" gate. \$\endgroup\$ – JRE Dec 6 '18 at 11:44
  • \$\begingroup\$ Ups! thank you, do you still think it´s a reliable solution ? \$\endgroup\$ – Pepslight Dec 6 '18 at 14:29
  • \$\begingroup\$ All you need is a diode OR. \$\endgroup\$ – JRE Dec 6 '18 at 16:58
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You are trying to use the LM139 outside of its Common-mode input voltage range:

enter image description here

I deduce from your question that you are using a supply voltage of 6 V. That means that the Common-mode input voltage range is between Vcc - 1.5 V = 4.5 V and 0 V. Your input voltage are higher than that at around 5.0 V

So you will need to increase the supply voltage (I would choose at least 7 V).

Or you use a voltage divider to lower the 5 V input voltage to for example half of that: 2.5 V just to keep things simple. Then of course you will also need to halve your reference voltage.

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  • \$\begingroup\$ This is awesome, I thought about before, but I wasn´t sure where to look at. Thank you so mutch Bimpelrekkie, I will increase the input voltage and see what happens, Luckily I´m doing this on my workplace meaning I have almost an unlimited amount of components to burn ^^´ \$\endgroup\$ – Pepslight Dec 6 '18 at 10:47
  • \$\begingroup\$ It solved my problem!!! thank you so mutch, I increased my voltage and it started working without problems, I will now build my funky upside down "Or" gate and see what happens. \$\endgroup\$ – Pepslight Dec 6 '18 at 14:31

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