Circuit Design About 0.75V voltage reference

MikeMl said:

When using a resistive divider to create a lower reference voltage from a higher one, say 0.75V from a 1.25V reference, you need to consider that the resistive divider makes a reference voltage source which is degraded because the resistors used in the divider effectively put a resistance in series with the lowered reference voltage.

Look at the sim attached. I purposely choose low value divider resistors that draw 10mA from the 1.25V input reference source. I set up a test where a 10K load is switched to the tap of the voltage divider at the 1s point in the simulation to see how the 10K load effects the voltage at the tap. Note that connecting the 10K to the tap causes a -2.2mV downward shift in the output voltage.

You could ask what the jump would be as a function of the divider resistance, the load resistance, and if the load resistance varies with time. If the load resistance is steady (time invariant) then you might be able to compensate for the effect of the load by just making R2 slightly bigger, such that when R2 is paralleled with 10K, the voltage at the tap is exactly 750mV.

If the load does vary, then to determine the variation at the tap voltage it is helpful to replace the circuit with its Thevinin equivalent, which calls for creating a voltage source with a voltage source in series with source resistance. Note the alternate circuit in the simulation. Note that the behavior is identical to that of the original circuit; open circuit voltage is 750mV, which drops to 747.75mV when the 10K load is connected.

To replace the original source/voltage divider with its Thevinin Equiv, reduce the source voltage from 1.25 to 0.75, and replace the divider with a resistor which is R1R2/(R1+R2).

Note that you could predict the drop by doing the following calculation: The 30Ω Thevinin source resistance and the 10K form a secondary voltage divider: The output voltage is = 0.75*10000/(30+10000) = 0.75*0.997 = 0.7477.

This shows the general approach how to calculate this for any similar requirement. Rules of thumb. Make R1,R2 as small as possible consistent with power requirements so that the Thevinin source resistance is low compared to the load resistance.

If you need better regulation than can be achieved with just resistors, there are ways to use an opamp to make a precision reference source derived from a higher one.

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eeguy said:

When you design a circuit or system with a specific IC or ICs, the first thing you need to do is look up the datasheet(s) for the components. In your case, the "555" datasheet will either have minimum and maximum frequency parameters or minimum and maximum periods given. I'm pretty sure it lists frequencies, but if not, the periods would consist of the "on" time and "off" time...so the frequency would be Fosc = 1/(Pon+Poff). If your 120 Khz square wave needs to be anywhere as accurate as your voltage reference, you may want to seek other options besides the 555...or, at the very least, use quality resistors and capacitors in the timing circuit.

It sounds like you're making a class D amplifier, DC power supply, or some sort of pulse-width modulation circuit. What are you working on, exactly, if you don't mind?

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