Cause a voltage drop of about 1V

It really depends on the accuracy needed, and how stable (temperature affects most electronic devices and can cause them to vary)

There are three ways to create a voltage reference:

• Resistor and diode (least stable and accurate, but least expensive)
• Voltage regulator (more expensive stable in the ~1mV range depending on the IC)
• Voltage reference (built to provide a voltage reference that is stable and accurate below 1mV and sometimes into the uV's)

If you need something that is more stable than 1mV, I'd use a voltage reference.

This article goes into more detail: https://www.nutsvolts.com/magazine/article/build_a_01_accurate_voltage_reference

You are in need of a voltage regulator.
A diode drop is not ideal (no pun intended). It's only 0.7V for a certain current for a certain diode make for a specific diode. i.e. Given several diodes, they will all measure a little different.

I would suggest adding another battery in series, then use a buck regulator to turn it into the low voltage (5V or 3.3V) that you need. Buck regulators are very efficient nowadays and can commonly be found in the upper 80%.

If you don't need 5V, then you can skip the extra battery.

A diode is the easiest place to start. A 1N4004 drops about 0.8 V, and a 1N4148 drops about 0.7 V (both dependent on the actual current through the diodes). Those calculate out to above 85% efficient.

The downside comes when the batteries are partially discharged. Now you don't need the voltage drop, and it is subtracting from the useful battery life. An LDO (low drop-out) regulator circuit can get you the regulation at higher voltages, but only about 0.1 V drop as the batteries discharge. However, it is a more complex circuit.