Voltage dividers

Sometimes you need 24 V at the panel but only 5 V at a sensor input. A voltage divider uses two or more resistors in series to produce a lower voltage at a tap point.

It is still a series circuit: one current, drops add to \(E_T\). The difference is intent—you care about the voltage at the junction, not only total current.

\(R_1\) on top, \(R_2\) on bottom, \(E_T\) across the pair. Current: \[ I = rac{E_T}{R_1 + R_2} \] Voltage at the tap (across \(R_2\)): \[ E_{ ext{out}} = I imes R_2 = E_T imes rac{R_2}{R_1 + R_2} \] The tap voltage is always less than \(E_T\) unless \(R_2\) is essentially the whole string.

For any resistor \(R_x\) in a series string from \(E_T\), the drop across \(R_x\) is: \[ E_x = E_T imes rac{R_x}{R_T} \] where \(R_T\) is the total series resistance including \(R_x\).

Because current is the same everywhere, each drop is proportional to its resistance share of the total. Rearranging this relationship gives the general voltage divider formula used throughout the book.

\(E_T = 120\,\mathrm{V}\), \(R_1 = 100\,\Omega\), \(R_2 = 50\,\Omega\), \(R_3 = 150\,\Omega\). \[ R_T = 300\,\Omega,\quad I = 120/300 = 0.4\,\mathrm{A} \] \[ E_1 = 40\,\mathrm{V},\quad E_2 = 20\,\mathrm{V},\quad E_3 = 60\,\mathrm{V} \] Check with formula on \(R_3\): \(E_3 = 120 × 150/300 = 60\,\mathrm{V}\). Match.

Tap between \(R_1\) and \(R_2\) is at 80 V above the bottom reference (60 + 20) or 40 V below the top—depending on reference.

In divider circuits, ground often defines the 0 V reference. All tap voltages are measured relative to that point.

Two technicians can disagree on a number if one references the bottom of \(R_3\) and the other the bottom of the string. Agree on reference before troubleshooting.

The formulas above assume no load on the tap (or a very high-impedance load like a meter input).

Connect a low-resistance load from tap to ground: it is in parallel with the bottom resistor (or part of the string). The divider ratio shifts. Always ask: what is hanging on the tap?

Control circuits with wet contacts or long cables may need stiffening (lower divider R, buffer amplifier, regulated supply).

1. Divider current \(I = E_T / R_T\) should be large enough that load current does not dominate—but not so large you waste power and heat resistors.
2. For a high-impedance input, \(R_T\) in the tens of kΩ may work; for a relay coil on the tap, the divider model fails—the coil is not a high-Z load.
3. Power check: \(P_n = I^2 R_n\) on each resistor; size wattage accordingly.

Look for two or more resistors (or a potentiometer) in series creating a reference or signal level:

• Speed reference pots.
• Thermistor/resistor networks in analog inputs.
• Bleeder strings on capacitors (with safety rules).

If you see a tap and series R, think divider first—then check for load on the tap.

Situation. A 48 V PLC input should see 24 V when a selector is in RUN. The print shows two fixed resistors in series across 48 V with the tap to the input. In RUN, the PLC sees only 18 V and faults.

How to think with this lesson.

• Unloaded calculation may predict 24 V at the tap.
• The PLC input has finite impedance and leakage; it loads the bottom leg.
• Measure tap voltage with the PLC connected.
• Fix: adjust ratio, lower divider resistance, or use an proper 24 V supply instead of a soft divider.

Conclusion: dividers are series math plus load awareness.