Clipping Detector for an amplifier - Page 1
Forum section: electronics / projects

author: Shay
author_rank: Regular Contributor
timestamp: August 09, 2024, 05:21:37 pm
content: Hello,I am looking to ideas on how to design a clipping detector for a 15Mhz Amplifier.The goal of the design is to be able to light an LED on the front panel which indicate that the output waveform is clipped.Example:5Vp-p 15MHz sine wave signal is fed into the Amplifier that has a gain of 10. Output now should be 50Vp-p, however, the output voltage is limited to 40Vp-p, so the signal is clipped.What would be a simple circuit that can light an LED when the output is clipped like this? I was thinking of a fast comparator and some kind of RS latch but the issue is that the signal is very short in duration (~5ns) and it is not fast enough to trigger a basic RS latch like the CD4043.

author: MarkT
author_rank: Frequent Contributor
timestamp: August 09, 2024, 05:33:46 pm
content: Probably easiest to detect 2nd or 3rd harmonic content, both of which will jump in level on clipping (the 3rd particularly). Since this is a 15MHz amp you would couple a little of the output to a 45MHz tank and RF-detector, then have a slow comparator on its output.

author: bson
author_rank: Unknown
timestamp: August 09, 2024, 08:08:32 pm
content: If you detect on drive voltage you can also use it for a gain distortion warning (overdrive) that lights up well before clipping occurs.

author: Sensorcat
author_rank: Regular Contributor
timestamp: August 09, 2024, 09:55:20 pm
content: Put a diode in series with a capacitor at the output of the comparator. The capacitor gets charged whenever the condition detected by the comparator occurs. If the duration of the comparator signal is too short to charge the capacitor to the comparator output voltage (minus diode forward voltage) during one cycle or event, there will be a next one to finish charging. To reset (=discharge capacitor), use a switch in parallel to the capacitor, or a resistor. Measure the capacitor voltage to turn on the LED above a threshold. Capacitor should be small, input impedance of measuring circuit high. Manual switch: Reset by user. CMOS switch, or the like, operated by logic signal: Reset timer controlled or whatever logic signal suits your needs. Resistor: RC time constant determines reset.Two such circuits are required for positive and negative clipping detection.

author: Shay
author_rank: Regular Contributor
timestamp: August 10, 2024, 09:35:36 am
content: Thanks, everyone, for your great suggestions. @Sensorcat solution seems to be the most suitable for me.

author: tszaboo
author_rank: Super Contributor
timestamp: August 10, 2024, 11:11:03 am
content: Take the input, not amplified signal from the amplifier. Buffer it if necessary. Run it into a comparator, set to the voltage it starts clipping.

author: Shay
author_rank: Regular Contributor
timestamp: August 10, 2024, 07:54:18 pm
content: Quote from: tszaboo on August 10, 2024, 11:11:03 amGood idea. Thanks for your suggestion.

author: moffy
author_rank: Super Contributor
timestamp: August 10, 2024, 11:40:05 pm
content: If you get stuck a discreet alternative might help. A simple matched differential pair detects the threshold, and triggers a PNP which dumps charge into an capacitor. For the NPN matched pair the BFM520, 5GHz ft transistor pair would work well. The example uses a 50MHz sine wave just 200mV above the threshold.Don't exceed the reverse breakdown voltage of the NPN B-E junction.

author: magic
author_rank: Super Contributor
timestamp: August 11, 2024, 07:55:44 am
content: Quote from: moffy on August 10, 2024, 11:40:05 pmYeah, and this means protection diodes or something (or only use this at very low signal amplitude).In similar manner, I wonder if it would work to set up a common base amplifier whose emitter is connected to the signal through a reverse protection diode and the collector charges a capacitor like above. Some series resistance may be necessary to limit emitter current. Maybe less exotic parts could still give satisfactory performance in common base configuration.

author: moffy
author_rank: Super Contributor
timestamp: August 11, 2024, 08:38:51 am
content: Quote from: magic on August 11, 2024, 07:55:44 amLooking up Digikey the BFM520 is obsolete replaced by a slightly upgraded BFU520, the ft is 10GHz, they seem to have plenty in stock, 16k+ @ $1.49AUD per 1. You could probably use a less exotic matched pair like the BCM847 but then the base drive requirements become much higher and performance will suffer. Low impedance and keeping the signal within limits would be very important.

author: magic
author_rank: Super Contributor
timestamp: August 11, 2024, 05:49:00 pm
content: LTspice suggests that this can work.The only critical part is D2, which needs to have fast recovery because the waveform is negative for most of the time but we want positive DC to flow into Q2. Apparently 1N4148 may be good enough for 20MHz, but at 200MHz something better is called for like BAT83 (a fairly high voltage small signal Schottky).The output of D2 is AC grounded by C1 while DC is absorbed by Q2 and amplified by Q3.R1, D1, Q1 are an attempt at controlling thermal drift of detection threshold, not sure how effective they would be in practice. Due to aforementioned recovery effects, detection threshold varies somewhat with frequency - same simulation at 2MHz or 0.2MHz and 20Vpk amplitude shows ~50% duty cycle pulses at the LED, which disappear at 19.5Vpk amplitude.editI think R1, R2 could be removed because R3 limits base current of Q2 and hence emitter current too. R3 can apparently be increased even to 1MΩ without issues. The circuit's safety depends on the threshold not being much lower than the amp's supply rails.

author: moffy
author_rank: Super Contributor
timestamp: August 11, 2024, 07:13:20 pm
content: A clever use of components, uses a higher level signal and benefits from some component matching, but interesting.

author: Zoli
author_rank: Frequent Contributor
timestamp: August 11, 2024, 08:36:26 pm
content: Quote from: magic on August 11, 2024, 05:49:00 pmR1 can be removed, but R2 not so simple: when D2&Q2 is open, the current is freely flowing from the input thru (D2->Q2->Q3 EB diode) the ground. To solve the situation(and keep the components safe, R1&R2 removed), move R5 from the collector in the emitter of Q3.

author: magic
author_rank: Super Contributor
timestamp: August 12, 2024, 07:24:33 am
content: R1 is for symmetry, it's either both or none. Current can't flow entirely freely through Q2 because R3 limits base current.And by the way, R3 value should be higher than R4For minimum offset voltage and drift, it should be set to produce the same idle current in D1/Q1 as the current required in D2/Q2 to flow through R4 and turn on Q3, which is 70μA or so. Then offset voltage could be down maybe even to a few mV, but I dunno. No warranty, just a quick attempt to design something with jellybeansIn practice, such accuracy probably isn't required in a clipping indicator and the error at high frequency due to D2 reverse recovery is tens or hundreds mV anyway (with 1N4148 at 20MHz), so one could replace Q1 with another silicon diode or simply throw out all the thermal compensation stuff and connect Q2 base straight to a voltage divider or other reference potential.

author: moffy
author_rank: Super Contributor
timestamp: August 12, 2024, 07:56:57 am
content: To @magic, studying your circuit I realised I had the filtering cap in the wrong place, I moved it to the base of the PNP, which also removes the need for a fast PNP, I can go generic with a BC557. The threshold is pretty flat between 30Mhz to 300MHz @2.2V is on and 2.1V is off.