inverting comp behaving strange

I am using an LT1716 comparator in the inverting mode. So, I am simulating it in LTSpice. I understand that the Vout is Vcc when Vin,at the inv terminal is less than Vref at the non inv terminal and vice versa. But in my simulation its not so. I am attaching the screenshots with this post. For eg / the Vref is 4V, Vin is 1V. The Vcc to the comp is 7V and the Vss is ground.

But in this case, ideally I should be getting Vout of about 900mV. Or do I need the whole hysterisis compensation resistors to make it work _

3 Answers

Datasheet says

Output: The output stage of the LT1716 can drive loads connected to a supply more positive than the device, the same as comparators with open collector output stages.

And indeed LT1716 has very feeble current source capability, but decent current sink:

You are using it [wrong basically] as a current source.

It's not very clear what you asked in that comment, but for an open-collector comparator (like LM339), the load goes between its output and the positive supply rail. The LT1716 can be used even with a (lower-resistance) 1K load (at 7V), but the 1K load has to go between its output and the positive rail.

The LT1716 actually has a push-pull stage but the (high-side) PNP transistor is weak (perhaps because it is a lateral one). You can use the LT1716 with a low-side load, but it has to be of fairly high resistance (greater than 100K or so for 7V supply).

While a LM339 (which has a true open-collector output) is usable in the same manner with a high-side load:

It is not usable at all with a low-side load:

EM Fields raises a fair point that the currents/voltages on the high vs. low side load resistors are in anti-phase. This is shown below (at the top of the graph) with a (5V) CMOS [push-pull] comparator that can sink and source current almost equally well. Also shown on the below it is the same circuit but with the comparator replaced with an open-collector one (LM339). The currents are also in antiphase, but that comparator has no ability to influence the output when it's (sole) output transistor is "off", so the external circuitry dictates the off point, which in this case is at midrail because it's an equal voltage divider. The LT1716 is almost like the open-collector one because of the large difference (2-3 orders of magnitude) between the currents it can source vs. sink. Hope this helps.

• thank u very much. I guess the same would hold true for the LM139&239 as it can also sink a max of 16mAmps, right. Hence/forth,I will need a pull up on the output pin. Are my assumptions right. – Board-Man Nov 18 '15 at 21:23
• @VUK: Your assumption is incorrect if you want Vcc across your load, since the pullup will be in series with your load when the comparator is OFF. I've posted a solution for you as an answer. – EM Fields Nov 18 '15 at 21:40
• Your high side load logic isn't quite right, since the OP stated that for a low side load Vref was 4 volts, and was connected to the + input of the comparator. To migrate from that to a high side load, the connections to the comparators' inputs have to be reversed, and what was formerly the one volt signal now becomes the signal which has to swing above and below 4 volts in order to switch the comparator's output ON and OFF. – EM Fields Nov 18 '15 at 22:43
• @EM Fields: I was just illustrating the difference between connecting the load on the high vs load side of the comparators output. (Unlike you) I'm not trying to fix his application because I don't know what that actually is... with all the talk of pull-up and so forth. – Fizz Nov 18 '15 at 22:46

The LT1716 has very low output power capability, increase R1 to 100k

As already noted, the LT1716 has a very weak internal pullup to Vcc, (to save from having to install one externally) and its output stage is only designed to sink current. In order to get around that, if your load can be tied to the high side rail you can do this:

or this:

depending on whether Vout is supposed to go low when Vin is higher than Vref or when Vin is lower than Vref.

Note also that even disregarding the comparator's output stage's voltage drop to GND, with Vcc equal to 7 volts and the load equal to 10k, the highest possible current into the load will be $I{_L} = \frac{Vcc}{R{_L}} =\frac {7V}{10k\Omega} = 700\mu\text{A}$ , not the $900 \mu\text{A}$ you originally gave.