I would really appreciate if someone could give me advice on how to build circuit for this case. On my circuit input I have slowly arising voltage from 0V to 10V, and on my circuit I have 2 outputs. When input has 0V-5V, first output has to be active, but when voltage rises from 5V-10V, first output has to deactivate and second output become active. And vice versa, when voltage decreasing to less than 5V, second output has to deactivate and first become active. 5V is the point where output changes, depending on voltage. I was thinking to use 5V or 6V relay which gets active when voltage gets close to 5V. But relay coil will always use energy, which is not good in my case, as I need as much as possible voltage and current on outputs. Can you help to project switching between two outputs using transistors or comparators circuit? Thanks in advance.

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    \$\begingroup\$ Will the variable voltage also be used for the logic supply and how much current will your loads need and/or how much series resistance through the switch can you stand? \$\endgroup\$ – EM Fields Mar 22 '15 at 20:51
  • \$\begingroup\$ What is "output has to be active?". Is it a signal that goes high or low? What voltage? Power that turns something on? What voltage and current? What power supply is available to run the circuitry that activates and deactivates these outputs? You should be able to see for yourself that much context you are assuming hasn't been stated. \$\endgroup\$ – Olin Lathrop Mar 25 '15 at 22:21

Here's a solution using a LM397 comparator and a quad 4066 analog switch:

enter image description here

Because the input is slowly rising, I have added just a little hysteresis to the comparator input. The comparator trips on the input reaching 4.99v, comparator input, but doesn't turn off until the input falls below 4.89v. The trip points are controlled by the three resistors R1, R2, and R3. Here is the formulae used:

$$Vh =\frac{Vs*R2}{\frac{R1*R3}{R1+R3}+R2}$$


$$Vl = Vs * \frac{\frac{R2*R3}{R2+R3}}{R1+\frac{R2*R3}{R2+R3}}$$

Vs is the 10v supply Vh is the upper threshold, and Vl is the lower threshold.

Here are the same formulas suitable for input to a spreadsheet if you want to experiment with the values:

= ((A10*A2) / (((A1*A3) / (A1+A3)) + A2))

= A10 * ((A2 * A3) / (A2+A3)) / (A1 + ((A2*A3) / (A2+A3)))

where A1..A3 are the resistor values R1..R3, and A10 is the 10v supply. The final values used were adjusted to reflect actual 1% resistor values available.

I show the 10v for the reference (top of R1) tied into the main supply; you may want a separate more accurate reference. Also, I have not included decoupling caps for the two ICs; they should be added. Less important than digital ICs doing fast switching, but always a good idea.

Note I have used all four analog switches. When the analog switch for the selected voltage is off, then I grounded the line rather than leave it floating.

  • \$\begingroup\$ Hi,thank a lot for all information you sent. To make more clear I have drawn a plan of my circuit and hosted this picture here: postimg.org/image/hr0xq8jth \$\endgroup\$ – Mingas Mar 25 '15 at 20:47
  • \$\begingroup\$ I tried to find TL397 comparator, but I could not find it, I only found TL393. When I search for this model number on google it does not come up. Do you know where can I find it? Also, can you please explain how to include decoupling caps for two ICs? I appreciate your help, thanks \$\endgroup\$ – Mingas Mar 25 '15 at 20:58
  • \$\begingroup\$ @Mingas Should have been LM397, not TL397. My bad. Adding decoupling caps just means to add a capacitor -- typically 0.1 µF (or 100 nF, same thing) between the power supply pin and ground as close to the part as possible. \$\endgroup\$ – tcrosley Mar 25 '15 at 21:54

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