You need pull-up resistors on the outputs (or whatever load you want to drive).
Inside the ULN2003 there are darlington transistors with an open collector, so without some external circuit the output will not be pulled up to some higher voltage.
The COM pin is connected to a freewheeling diode for each transistor, so you can connect it to the positive rail to protect against overvoltage, for example for inductive loads.
This answer is general to processors and peripherals, and has an SRAM specific comment at the end, which is probably pertinent to your specific RAM and CPU.
Output pins can be driven in three different modes:
- open drain - a transistor connects to low and nothing else
- open drain, with pull-up - a transistor connects to low, and a resistor connects to high
- push-pull - a transistor connects to high, and a transistor connects to low (only one is operated at a time)
Input pins can be a gate input with a:
- pull-up - a resistor connected to high
- pull-down - a resistor connected to low
- pull-up and pull-down - both a resistor connected to high and a resistor connected to low (only useful in rare cases).
There is also a Schmitt triggered input mode where the input pin is pulled with a weak pull-up to an initial state. When left alone it persists in its state, but may be pulled to a new state with minimal effort.
Open drain is useful when multiple gates or pins are connected together with an (external or internal) pull-up. If all the pin are high, they are all open circuits and the pull-up drives the pins high. If any pin is low they all go low as they tied together. This configuration effectively forms an AND
gate.
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Note added November 2019 - 7+ years on: The configuration of combining multiple open collector/drain outputs has traditionally been referred to as a "Wired OR" configuration. CALLING it an OR (even traditionally) does not make it one. If you use negative logic (which traditionally may have been the case) things will be different, but in the following I'll stick to positive logic convention which is what is used as of right unless specifically stated.
The above comment about forming an 'AND' gate has been queried a number of times over the years - and it has been suggested that the result is 'really' an 'OR' gate. It's complex.
The simple picture' is that if several open collector outputs are connected together then if any one of the open collector transistors is turned on then the common output will be low. For the common output to be high all outputs must be off.
If you consider combining 3 outputs - for the result to be high all 3 would need to have been high individually. 111 -> 1. That's an 'AND'.
If you consider each of the output stages as an inverter then for each one to have a high output it's input must be low. So to get a combined high output you need three 000 -> 1 . That's a 'NOR'.
Some have suggested that this is an OR - Any of XYZ with at least 1 of these is a 1 -> 1.
I can't really "force" that idea onto the situation.
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When driving an SRAM you probably want to drive either the data lines or the address lines high or low as solidly and rapidly as possible so that active up and down drive is needed, so push-pull is indicated. In some cases with multiple RAMs you may want to do something clever and combine lines, where another mode may be more suitable.
With SRAM with data inputs from the SRAM if the RAM IC is always asserting data then a pin with no pull-up is probably OK as the RAM always sets the level and this minimises load. If the RAM data lines are sometimes open circuit or tristate you will need the input pins to be able to set their own valid state. In very high speed communications you may want to use a pull-up and a pull-down so the parallel effective resistance is the terminating resistance, and the bus idle voltage is set by the two resistors, but this is somewhat specialist.
Best Answer
You could use a zener diode and resistor.
simulate this circuit – Schematic created using CircuitLab
Pick a Zener voltage between 1.8V and 2.75V. Pay attention to the tolerances to make sure it won't sink current simply from the 1.8V output of the gauge. Also make sure it has a low enough Zener voltage to keep the gauge's pin at a safe level.
For example, pick a diode with a 2.2V Zener voltage. If you want to keep current less than 10mA, you'd need 110 ohm resistor. This assumes the MCU high output is 3.3V, and it can safely drive 10mA of course.
\$\ {3.3V-2.2V\over 10mA}= 110\Omega \$
EDIT: This assumes the gauge has some reasonable source impedance. You may want to add another resistor of the same value between the diode and the gauge A schematic/datasheet would be helpful. However, at that point, you're at the same component count as a simple, MOSFET level shifter.