Follow the envelope, get the message back.
Ten minutes of reading. Move the sliders, then every waveform on the KL-93002 will make sense.
Demodulation: the opposite of modulation
An AM signal is a high frequency carrier whose amplitude follows the audio. Demodulation (detection) takes that audio back out.
An asynchronous detector needs no copy of the carrier. The diode (envelope) detector is the classic one: simple, cheap and in every AM radio.
A synchronous detector multiplies the signal with a copy of the carrier at the same frequency and phase. The product detector is one.
The envelope
The carrier is far faster than the audio (200 kHz against 3 kHz in Table 1, so about 67 carrier cycles fit in one audio cycle). On the scope the carrier looks like a filled band and only its outline, the envelope, is visible. When m stays below 1 the envelope is an exact, shifted copy of the audio. A detector that follows the positive peaks therefore gives the audio plus a DC level.
The diode detector
On the KL-93002 the AM signal first passes two inverting amplifiers (U1 and U2, R2/R1 = R4/R3 = 22 k/1 k). Two inversions give a non inverted signal with a gain of 22 × 22 = 484, enough to lift a few tens of millivolts to several volts so the diode switches cleanly.
D1 conducts on the positive peaks and charges C2. Between the peaks C2 discharges slowly into the load. R5 with C3 smooths the carrier ripple. The result at LPF O/P (TP7) is the positive envelope sitting on a DC level. C4 blocks that DC, so Detector O/P (TP8) carries only the audio.
Choosing the RC time constant
RC must be long compared with a carrier cycle, so C2 holds the peak between carrier peaks (small ripple), and short compared with an audio cycle, so the voltage can fall as fast as the envelope falls. For Table 1: 1/fc = 5 µs and 1/fm = 333 µs at 3 kHz. On the module C2 and C3 discharge mainly through R6, so RC ≈ 4.7 k × 0.02 µF ≈ 94 µs.
If RC is longer than this, the output cannot follow the falling envelope and cuts straight lines across it: diagonal clipping. The limit gets tighter for a deeper m and a higher fm.
| Case | Largest RC | Module RC |
|---|---|---|
| Table 1, m ≈ 0.5, 1 kHz | 276 µs | 94 µs: follows |
| Table 1, m ≈ 0.5, 2 kHz | 138 µs | 94 µs: follows |
| Table 1, m ≈ 0.5, 3 kHz | 92 µs | 94 µs: some diagonal clipping |
| Table 2, m ≈ 0.83, 3 kHz | 36 µs | 94 µs: some diagonal clipping |
Over-modulation and DSB-SC
VR1 of the AM modulator sets how much carrier is added to the audio. At one end the carrier is largest (m smallest, maximum AM amplitude: manual step 3). Towards the centre the carrier shrinks and m grows. Above m = 1 the envelope folds back at zero (over-modulation) and the detector output is no longer a clean sine.
At the centre the carrier is cancelled completely: DSB-SC. Its envelope is |cos 2π fmt|, so the diode detector gives mostly twice the audio frequency. This is why DSB-SC and SSB need a synchronous detector.
The product detector
The product detector multiplies the AM signal by a local copy of the carrier and keeps only the low frequencies:
It works for AM, DSB-SC and SSB, and needs no diode knee, so it also works for tiny signals. Its weak point is the carrier copy: the output falls as cos φ and vanishes at φ = 90°. On the KL-93002 you feed the modulator carrier into Carrier I/P of the product detector and turn Carrier Phase to see this.
The KL-93002 module: what connects to what
The KL-92001 powers the module (+12 V and −12 V for the op-amps; its rail lamps light with POWER). Each generator lead has a red (or purple) signal clip and a black ground clip: always put the black clip on a GND socket.
Using the instruments
Function generator
- Two channels. CH1 is the carrier, CH2 the audio. Type the values in the Generator tab or use CH1/CH2, FREQ and AMPL on the front.
- Amplitudes are peak to peak: 250 mV p-p is 125 mV peak.
- Nothing comes out until OUTPUT is ON for that channel.
Oscilloscope
- DC coupling for this lab (manual step 5). Press AC/DC per channel.
- Amplifier O/P is about 15 V p-p: use 2 V/div or 5 V/div. AM O/P is only tens of mV: 5 mV/div.
- Trigger on the detector output (CH2) to get a steady envelope picture at 100 µs/div.
Analyzer and meter
- Analyzer on AM O/P (200 kHz preset): the carrier with one sideband on each side, fm away.
- Meter on V DC at LPF O/P reads the DC level of the envelope; at Detector O/P it reads about zero.
What you should see
| Where | Table 1 (fc 200 kHz, 150 mV p-p audio) | Table 2 (fc 1 to 2 MHz, 250 mV p-p audio) |
|---|---|---|
| AM O/P (VR1 at an end) | AM, about 30 mV p-p, m ≈ 0.5 | AM, about 37 mV p-p, m ≈ 0.83 |
| Amplifier O/P | the same AM shape, about 14.5 V p-p | about 17.7 V p-p, the envelope dips close to zero |
| LPF O/P (DC coupling) | audio of about 4.8 V p-p on a DC level near 4.5 V | audio of about 6.9 V p-p on about 5 V DC, slightly distorted |
| Detector O/P (DC coupling) | clean audio, about 4.8 V p-p, centred on 0 V, at 3, 2 or 1 kHz | 3 kHz, about 6.9 V p-p; the falling edges are straighter (diagonal clipping) |
| Product detector O/P | clean audio, about 0.6 V p-p | clean audio, about 1 V p-p |
Levels are those of the simulator model. The manual gives no figures for this module, so treat them as the expected order of size. The model treats the op-amps as fast enough for 2 MHz; a real µA741 (1 MHz gain bandwidth) could not give a gain of 22 at these carrier frequencies, so on a real bench the Table 2 amplitudes can come out lower.
Check yourself
Q1Why is the scope set to DC coupling in this lab?
To see the DC level of the rectified envelope at LPF O/P and to see that C4 removes it at Detector O/P. AC coupling would hide the difference.
Q2fc = 200 kHz, fm = 3 kHz. Give the range for RC.
1/fc = 5 µs ≪ RC ≪ 1/fm = 333 µs. With m = 0.5 the diagonal clipping limit is about 92 µs.
Q3What happens at the detector output when RC is too large?
Diagonal clipping: the output cannot fall as fast as the envelope, so the troughs of the audio are cut by straight lines.
Q4VR1 is turned to the centre. What frequency leaves the diode detector for a 1 kHz tone, and why?
Mostly 2 kHz. With no carrier the envelope is |cos 2π fm t|, which repeats twice per audio cycle.
Q5The product detector output disappears when Carrier Phase is at 90°. Why?
The output is proportional to cos φ, and cos 90° = 0. The local carrier must be in phase with the received carrier.
Q6Why does the circuit amplify the AM signal before the diode?
A silicon diode needs a few tenths of a volt before it conducts. The AM O/P is only tens of millivolts, far below that knee.