Your 470 ohm pull-up resistor should be connected to the 'high' output voltage you want, in that case 15V. You don't need diode after the output.

It will result in output current 15V/470mA of almost 32 mA ( the limit is 60mA) and that would cause significant heating. I would put something like 4.7Kohm..10Kohm as pull up to restrict max output current to 3.2mA or less..

Dear Madam,

- I assume the diode is a light emitting diode.

- The way it is corrected, is not preferred. I mean, the way it is connected, if the diode is not

lit, the current is flowing through the 470 ohm resistor, and the output stage of the LM393 to ground.

- In order to avoid this, you could :

a. Reverse the polarity of the diode

b. Put it in series with the resistor,

c. Connect both in series from the Vdd to the output of the comparator.

d. Invert the inputs to the comparator to get the same logical operation.

- The voltage on the high-voltage side of the 470 resistor (or on the high-voltage side of the series

connection of diode and resistor) can be anything up to 36 Volt according to the LM393 datasheet.

It need not be the same voltage as the 5 Volt power supply.

- Typically, (to save power).

a. Calculate the current you need though the diode.

b. Look up the voltage on the output of the opamp at this current, and the voltage over the diode

at this current.

c. Select the power supply for the end stage and the resistor such that you more or less have

this current, and the current is stable. (Need a few volt eg 1-2, minimum over the resistor.)

Best Regards,

Henri.

You should understand "open collector outputs" - see picture.

"Open collector" makes the transition between analog levels.

After all, we are able always add "open collector output".

Silpa Baburajan

As mentioned by others, the LM393 has a so-called open-collector output.

You can think of this as a switch that makes contact to ground when the voltage on the negative input is higher than the voltage on the positive input.

This means that it cannot actively drive a positive voltage or current. The advantage of this is that you can tie multiple outputs together, and that the LM393 supply voltage can be different from any voltage on the output.

As Nikolay Pavlov already explained, you can get +15 volts on the output by taking a +15V supply voltage, and placing a resistor between this voltage and the output. The output will then show either 0V or +15V, depending on the input voltages. As the resistor 'pulls up' the voltage on the output, it is commonly called a pull-up-resistor. This is assuming that there is nothing else on the output, so no LED or other load.

The potential (geeky pun intended) problem here is that this output voltage may drop depending on the current you draw, i.e. the circuitry connected to the output. With the 470 ohms pull-up-resistor shown, the maximum current is 15 / 470=32 mA (as Nikolay already calculated). And with these values, the resistor can't be a standard 1/4 watt type, as it will almost draw 1/2 watt at this voltage and current. Smaller ones will become too hot and may eventually go up in smoke.

So you may want to use a higher value pull-up-resistor (e.g. 1K @ 1/4 watt), but that completely depends on what is connected to the output.

A possible solution is to use extra components to create a positive-going open-collector, as shown in the attached schematic. This circuit works exactly the other way round as the LM393 output: it can source a current/voltage (up to 15 volts here), but it can't actively pull the voltage to ground (called 'sinking'). Also note that this circuit inverts the polarity of the LM393 output, so you may need to swap the inputs.

So the best answer to your question is: it depends - on what you want to connect to the output of the LM393. If you can tell us more precisely what you want to build, we can provide a better answer.

Also note that there are comparators that can both sink and source current - search for so-called push/pull comparators. An example is the TLC3702. But if you want 15 volts on the output of this one, you also need to use +15V supply voltage for the comparator (and not more, as it is specified for 16V max.)

I hope that this helps.

Using a DC-DC/AC-AC booster, after achieving output voltage from the comparator, you can get any of the required voltage

Hello Silpa,

If your problem is to obtain 0-15V on the output of the LM393 comparator, the open collector structure allows you to do this with the pullup resistor tied to 15V instead of 5V. You have to recalculate it (470 Ohm is not correct) allowing a current of less than 10mA to flow in it. In this case the low level will be of a few 100mV and the high level, exactly 15V, if LED is not present. The LED limits high voltage to 2V approximatively, so you have to take it out.

Regards

FD