Op amp question

I'm not an experienced electronics person, and so have a "stupid question". I have a device that provides an "on" signal intermittently at 2.7 volts DC.

When the "on" of 2.7 volts is available, I need to have a "sub-circuit" receive an "on" of 5.0 volts. I have a source of 5.0 volts on the board. I've been told I need to use an op amp to do this, so I bought two NTE 928M op amps. I've tried to get this to perform as I need, but can't. HELP please!!! I put resistors in somewhere, but where and how big? tanksalot S.F.

Hi Stanley,

So what you may need is a simple TTL-Input. TTL level is defined as "HIGH" i every input voltage greater than 2.1 V. Take a little look on the datasheet of your input. Perhaps you didn't need anything.

Oh no, You need a comparator, not a opamp (if you need really a well defined swtchinglevel). Look about LM311 or similar. Define your reference voltage on inverting input and sense on noninverting input. The output of the comparator is then High or Low in digital means. Opamps are designed for Output voltages between the rails, not digital values.

Marte

He did...

--- Regards, Bob Monsen

The chief aim of all investigations of the external world should be to discover the rational order and harmony which has been imposed on it by God and which He revealed to us in the language of mathematics.

- Johannes Kepler

If current consumption is not a concern, and the "on" signal has decent drive strength, and you don't need ultra-fast switching, the easiest way to handle this is to use two transistor inverters like this:

5V 5V | | / \\ 10k \\ 10k / / \\ \\ / | | 100k /c +-------/\\/\\/\\---b-| npn 100k /c \\e signal----/\\/\\/\\----b-| npn | \\e GND | GND

The transistors can just be any old npn transistors. mmbt2222 or 2n3904 would both work fine, for example.

There are lots of other ways to level shift a digital signal, and some of them might be more suitable for your application, but you haven't provided many details.

--Mac

You might need a resistor b-e of the first transistor. I read the OP to imply presence/ABSENSE of the +2.7V, so leakage could be an issue. Also the 100K is superfluous, unless you're sweating power consumption... if so, just change the first 10K pull-up to 100K.

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
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Right! Here's another way for a higher current Vout and with diodes to prevent switching until input rises > ~2v

5V 5V Signal | | | \\ 1k | [D1] / | | \\ | [D2] / | | | 10k /e [D3] +-------/\\/\\/\\---b-| PNP 2N3906 | 10k /c \\c +--------/\\/\\/\\---b-| npn 2N3904 | \\e +---- Vout | \\ | / 150 | \\ | | Common ground ----------+-------------------+

But I'm wondering if he needs switching at precisely 2.7 v, as another poster suggested. And someone told him he needs an op amp - perhaps the reason is that his input signal varies only slightly.

For the OP: Here's a circuit you could use, based on the NTE928M:

+5------------+-------------+---------------------+ | | | | +------- | ------+ | | | | | | | | |\\ | | | | | | \\ | [100K] | | | | \\ | | | | | | \\|Pin8 | | \\ | | \\ | | / | | \\ | | 5K \\

Thanks for the critique, Jim. I guess I was just sort of mindlessly instantiating a stock transistor inverter for the second stage without thinking about it.

--Mac

Error in the drawing - change the 10K to the base of the

2N3906 to 1K. Ed