Tricky Circuit

A four-diode full-wave bridge

Turn both halves of a wave upright

AC means Alternating Current: its direction alternates. DC means Direct Current: it keeps one direction, even when its size changes. A full-wave bridge uses four diodes, with two conducting in each half-cycle, to send both halves through the load in the same direction. Peak means the largest instantaneous value.

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EXPERIMENT BENCH

Change the circuit settings, predict the readings, then check a challenge. These are learning models, with the assumptions explained in each lesson.

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How it works

Only an isolated low-voltage source; never mains. Each conducting diode drops 0.7 V in this model. No smoothing capacitor: the output is pulsating DC. This is not a tested hardware design.

Output peak = max(input peak − 1.4 V, 0). Two conducting diodes account for 1.4 V. The average current is the waveform average divided by load resistance. The output pulse repetition frequency is twice the input frequency. The doubled repetition applies only while output pulses exist. At input peak ≤ 1.4 V the bridge never conducts, and the repetition meter reads 0 Hz.

Difficulty
  1. Build · Observe

    Power on at input peak 5 V, input frequency 50 Hz and load 1 kΩ. Read output peak 3.6 V and repetition 100 Hz.

    Need a hint?

    5 V peak; 50 Hz; 1 kΩ → 3.6 V peak; 100 Hz

  2. Explore · Adjust

    At input peak 5 V and input frequency 60 Hz, keep average load current 1.8–2.0 mA.

    Need a hint?

    5 V peak; 60 Hz; 1 kΩ → about 1.909 mA

  3. Solve it · Discover the limit

    At input peak 9 V and input frequency 50 Hz, obtain average load current 8–9 mA.

    Need a hint?

    9 V peak; 50 Hz; 500 Ω → about 8.798 mA

Extra reading and circuit recipes

Turn both halves of a wave upright

A useful connection

A 50 Hz input produces 100 output humps per second because both half-cycles contribute.

A surprising detail

The output still reaches zero around every input zero crossing. A bridge alone does not make a flat supply.

A useful trap to avoid

3.6 V is the peak here, not the average. Do not calculate average current by dividing the peak by resistance.

Input frequency doubles, but peak and load stay fixed. Average current…

  1. doubles
  2. halves
  3. stays unchanged
Read more

The ideal waveform repeats faster but keeps the same shape and amplitude, so its time average stays unchanged.

hyperphysics.phy-astr.gsu.edu