In frequency-to-voltage mode the 9400 delivers an output proportional to the product of all four quantities — option 1:

\(V_{o}=V_{ref}\,C_{ref}\,R_{int}\,F_{in}\)

How the circuit produces that. Each input cycle causes a fixed packet of charge to be dumped into the integrator:

\(Q=C_{ref}V_{ref}\)

With \(F_{in}\) such packets per second, the average current delivered is

\(I_{avg}=Q\,F_{in}=C_{ref}V_{ref}F_{in}\)

That current flows in the integrating resistor \(R_{int}\), and Ohm's law gives the output directly:

\(V_{o}=I_{avg}R_{int}=V_{ref}C_{ref}R_{int}F_{in}\)

— the charge-balance principle, and the reason the relation is exactly linear in frequency.

A dimensional check confirms it. \(C\) in farads times \(R\) in ohms gives seconds, and seconds times \(F_{in}\) in s⁻¹ is dimensionless, so \(V_{ref}\) multiplied by a dimensionless number is a voltage. Option 2 divides a frequency by a voltage, option 3 leaves seconds in the denominator, and option 4 divides by \(V_{ref}\) — none of them yields volts.

The same chip runs the conversion backwards. In V/F mode the input voltage sets the current into the integrator and the output frequency adjusts until the charge balances, giving \(F_{out}=V_{in}/\left(V_{ref}C_{ref}R_{int}\right)\). Charge balance in both directions is why a single device serves as both a V/F and an F/V converter.

Where it is used : an F/V converter turns a tachometer's pulse train into a speed voltage, and the V/F direction gives a cheap, highly linear ADC when the output pulses are simply counted for a fixed gate time — with excellent noise rejection, since counting over an interval averages the input.

Hence, V_o = V_ref C_ref R_int F_in.