Design a circuit that would turn off output between two input voltage levels

Question

The circuits for the three stages of input voltage can be drawn easily, but I am confused about how to make the circuit "identify" the voltages.

For eg, How can I turn off output exactly between -3V and +5V. Please help me out!

Answer 1

Take a look at Zener diodes to provide the +5 and -3 V references. You can arrange them to subtract from the input voltage. For example, a 5 V zener can be used to create (Vin - 5).

After subtracting off the 5 V and 3 V offsets, you can use resistor dividers to get any slope < 1.

Added:

It's been long enough and the OP has apparently checked out, so here is the answer:

Here D3 and D4 are ideal diodes. If you actually want to build this, then use something like 1N4148 for D3 and D4, and make the zener voltages about 600 mV less to compensate for the diode drops.

Added in response to comments by EM Fields

The OP asked for switching at "exactly" -3 and +5 V. That's no spec at all. If he truly meant exactly, then no circuit suffices. Since everything by necessity will violate that spec, we have no way to know how far is OK. You don't know if a few 100 mV is good enough or not. Again, "exactly" is a meaningless spec, so can only be ignored.

To your second point, I think you are interpreting the graph incorrectly. The graph is quite clear in that Vout should be 0 for Vin from -3 to +5 volts. Specifically, here is the function described by the graph and that the circuit above realizes:

Vout = Vin + 3          (Vin < 3)
Vout = 0                    (-3 <= Vin <= 5)
Vout = (Vin - 5)/2    (Vin > 5)

You say the OP's drawing is wrong, but I don't see how you can know that since it's the only spec we've got, and it is quite clear by itself.

Answer 2

This kind of question normally comes up in a circuit theory class, where we can pretend that things like ideal op-amps and ideal diodes exist. Since you haven't said otherwise, I'm going to assume that's the context of your question.

While it won't hurt you to learn what a window comparator (mentioned by several people in the comments) is, that's likely not the solution your instructor is looking for.

Something more like this can be a building block in an answer that will likely satisfy your instructor:

^{simulate this circuit – Schematic created using CircuitLab}

D2 and D3 are zener diodes. Depending on where you are in your course, you might be expected to place ideal voltage sources in these locations instead of zener diodes.

D1 and D4 are ideal diodes (0 forward voltage). They're there to prevent the zeners from conducting from anode-to-cathode (or to prevent current the "wrong way" through if you've substituted voltage sources for the zeners).

OA1 is an ideal op-amp. In particular, this op-amp has no limits on input or output voltage due to its finite supply voltage.

You can see that in this solution, you don't do anything special to supress the output when the input voltage is near zero. You simply arrange that no output is produced until the input voltage exceeds some threshold in the negative or positive direction.

Answer 3

If your instructor expects you to use opamps, then use them as comparators.

Following is a circuit which uses comparators to do what what you want, but you could just as easily substitute opamps. Here's how it works:

V1 is the 10 volt supply and V2 is the input signal and, having no information to the contrary, I've assumed, for the sake of the argument, that V2 it can vary from - 10 volts to + 10 volts.

U1A and U1B are the two comparators comprising an LT1018, and their inputs aren't allowed to go more than 300 millivolts below ground, so if a 10 volt supply is used for V1, then with R4 and R5 both equal to 10k, when V2 is at -10 volts the voltage at the R4-R5 junction will be zero volts.

Then, since V2 can only go more positive than -10 volts, the problem of driving the comparators' inputs below ground when V2 is negative is averted.

Next, the switching points must be determined, and they'll be the voltages at the R4-R5 junction (V4) when V2 is equal to minus 3 volts and when it's equal to 5 volts.

In order to find them we can write:

$$ V4 = \frac{(V1-V2)\times R5}{R4+R5} +V2 $$

Then, With V2 equal to minus 3 volts we'll have:

$$ V4 = \frac{(V1-V2)\times R5}{R4+R5} +V2 = \frac{13V\times 10k\Omega}{20k \Omega} -3V = 3.5\text { volts} $$

Similarly, when V2 is equal to 5 volts, V4 will be equal to 7.5 volts

In order to generate these voltages we use the resistor string R1,R2,R3 as a voltage divider, arbitrarily force 1 milliampere through it, and select the resistors to yield the trip point voltages we need for U1A and U1B.

Since V1 is 10 volts and we need 7.5 volts for our high trip point, R1 needs to drop the difference, 2.5 volts, with 1 milliampere through R1. Using Ohm's law we can write:

$$R1 = \frac{V1-7.5V}{1mA} = 2500 \text{ ohms}$$

Selecting the closest E96 value, we'll choose 2490 ohms for R1

With the bottom of R1 at 7.5 volts and our next switching point at 3.5 volts, we'll need to drop 4 volts across R2, and with 1 milliampere through it, the E96 value closest to 4000 ohms will be 4020 ohms.

Finally, with the remaining 3.5 volts left to be dropped across R3, the E96 value closest to 3500 ohms will be 3480 ohms.

Now, as V2 slews from -10 to +10 volts, when it starts, U1B+ will be more positive than U1b-, forcing U1B's output high. That'll turn S2 ON, connecting V2 to R10. Also, since U1A+ will be less positive than U1A-, U1A's output will be forced low, turning S1 OFF.

This state of affairs will continue until V2 goes more positive than -3 volts, at which point U1B- will go more positive than U1B+ which will force U1B's output low and turn OFF S2. At this point S1 and S2 will both be turned OFF, and Vout will be pulled down to ground through R10.

As V2 continues to go more and more positive, eventually it'll get to 5 volts, at which time U1A+ will go more positive than U1A- and U1A's output will go high, which will turn S1 ON and connect the junction of R8 and R9 to U2+.

Since V2 and V3 vill now both be at 5 volts, U2+ will also be at +5 volts, and since U2 is a unity-gain voltage follower, its output will be at 5 volts, which will be connected to R10 through the now closed S2.

R8 and R9 comprise a 2:1 voltage divider because R8 and R9 have equal resistances, and with its tap connected to U2+, as V2 rises above 5 volts, the voltage on U2+ will rise at half that rate, satisfying the requirement for the slope in Vout above 5 volts.

So, the result of all this is that V2 will be connected to R8 from the time V2 is between -10 volts and -3 volts. After that, when V2 goes more positive than -3 volts, R8 will be disconnected from V2 and Vout will snap up to zero volts until V2 gets to 5 volts, when R8 will be connected To U2's output and will snap upt to +5 volts. Thereafter, V2 will stay connected to U2's output for as long as V2 stays more positive than 5 volts.

The schematic follows, and if you want to play around with the circuit, here's the LTspice .asc file so you can simulate it.