Probe and Attenuator - Page 1
Forum section: electronics / beginners

author: Nokia3310
author_rank: Contributor
timestamp: November 09, 2020, 04:10:05 pm
content: HiThis is a question about probes and attenuators. I'm new to electrics so please tell me if I'm wrong in my thinking. I bought one of those cheap DSO150 oscilloscopes. It came with a probe that has a x1 setting and x10 setting. I believe the x10 setting means a 10:1 reduction so if the probe touches 10V, 1V is displayed on the oscilloscope is that right?I also bought a x20 attenuator which I think causes a 20:1 voltage reduction?If I connect the x20 attenuator and the probe is on x10 is that 200:1 reduction? If 200V is on the probe 1V is displayed on the oscilloscope is that correct?What I'm doing is looking at the waveform of the primary side of an ignition coil. The peak voltage that the primary side gets to about 350V peak due to the self induction when the magnetic field collapses and the spark plug fires.Is it correct that with the probe and attenuator connected the oscilloscope would only have 1.75V peak across it? 350V/200=1.75V pk. Or am I totally wrong? The oscilloscope says it has a max of 50V peak.

author: Mechatrommer
author_rank: Super Contributor
timestamp: November 09, 2020, 04:32:56 pm
content: Quote from: Nokia3310 on November 09, 2020, 04:10:05 pmcorrect.Quote from: Nokia3310 on November 09, 2020, 04:10:05 pmnot quite correct... oscilloscope will see 1V at input, if you set DSO setting to 10X (or 1/10X depend on what dso) it will multiply by 10 and show 10V on the screen. but if you keep your dso setting at 1X, then you are correct, 1V will be displayed. same with any others probe attenuation.

author: Nokia3310
author_rank: Contributor
timestamp: November 09, 2020, 05:40:58 pm
content: Quote from: Mechatrommer on November 09, 2020, 04:32:56 pmThanks for the reply. I think this dso150 only has a x1 display option so it displays the voltage that's at the input and you have to multiply it yourself but I could be wrong.

author: tggzzz
author_rank: Super Contributor
timestamp: November 09, 2020, 07:58:33 pm
content: Quote from: Nokia3310 on November 09, 2020, 05:40:58 pmAh, the joys of complex instruments with important parameters buried in menu systemsSooner or later youwillhave the *1/*10 probe's switch in the "wrong" position. One way it is safe, the other it can be timewastingMake very sure you understand the limits of the probes and attenuators. Readallsections of the specifications and instruction manuals.

author: pwlps
author_rank: Frequent Contributor
timestamp: November 09, 2020, 08:24:08 pm
content: Quote from: tggzzz on November 09, 2020, 07:58:33 pmOne reason to have a good old Tek scope with auto probe detection

author: TimFox
author_rank: Super Contributor
timestamp: November 09, 2020, 08:50:56 pm
content: Quote from: Nokia3310 on November 09, 2020, 04:10:05 pmIs that 20:1 attenuator designed specifically for the probe and DSO? More often, attenuators labeled 20:1 are designed to be terminated with 50 ohms (and present 50 ohms to the source). The probe will not work properly when terminated with 50 ohms instead of the normal ‘scope input impedance of 1 megohm.When measuring high voltages, be careful to check the maximum voltage specification of the probe itself.

author: Nokia3310
author_rank: Contributor
timestamp: November 09, 2020, 11:21:23 pm
content: Quote from: TimFox on November 09, 2020, 08:50:56 pmNo the attenuator was a random one. I didn't realise they are designed specifically for a probe and DSOhttp://hantek.com/products/detail/67The probe came with the oscilloscope. It says max input voltage 300V RMS so 300V/0.707=424V Pk so that should be ok.Was I wrong to think that x10 probe and x20 attenuator gives x200 voltage reduction?

author: TimFox
author_rank: Super Contributor
timestamp: November 09, 2020, 11:47:50 pm
content: If the x20 attenuator were designed for 50 ohms, and you install it between the probe and a 1 megohm DSO input without 50 ohm terminator, the impedance seen by the probe output would still be close to 50 ohms, and the attenuation will jump from x10 (9 megohm in probe and 1 megohm at the DSO) to x180,000 (9 megohm in probe and 50 ohm load).The probe in your hantek.com link seems to be designed around 1 megohm and should increase the attenuation from x10 to x200. It is spec'ed for ignition primary side use. I assume that the 10 MHz bandwidth quoted there is accceptable for your automotive application.

author: tggzzz
author_rank: Super Contributor
timestamp: November 09, 2020, 11:59:52 pm
content: Quote from: Nokia3310 on November 09, 2020, 11:21:23 pmI doubt that is the entire specification.Work out the current flowing through the 15pF tip capacitance if the signal is 300V 100MHz !

author: vk6zgo
author_rank: Super Contributor
timestamp: November 10, 2020, 12:30:05 am
content: Quote from: Nokia3310 on November 09, 2020, 11:21:23 pmThe Hantek one in the link will, but it is pretty much an "outlier" among attenuators, apparently being specially designed for the application you have in mind.The standard convention for attenuators is that they are designed to be used in a 50, (or more rarely), 75 Ohm matched system, so if used with a 10x probe will deliver nonsense results.

author: Nokia3310
author_rank: Contributor
timestamp: November 10, 2020, 11:10:02 pm
content: Ok thanks for the info. I think I understand. Normally a 20:1 attenuator is matched with a specific probe in mind. The matched probe and attenuator together have 50 ohm resistance, or they need a 50 ohm terminator, and the probe has to be on x1. Otherwise the readings are distorted and don't make any sense? But the Hantek one I bought was designed for dummies who don't now anything about electricity so it doesn't need the 50 ohm terminator so it might actually work. I thought an attenuator was just resistors arranged as a voltage divider but there's obviously a lot more to it than that.

author: TimFox
author_rank: Super Contributor
timestamp: November 11, 2020, 12:21:06 am
content: Your 10:1 probe is designed for a 1 megohm input impedance at the DSO input. That 20:1 attenuator should work between the probe and DSO.Most things sold as attenuators in that package are designed for 50 ohm systems, to terminate coaxial cables properly.Scope probes are not merely resistor dividers, since the input and cable capacitances would form a low-pass filter that ruins the frequency response. Note that scope probes must be adjusted (usually by screwdriver, sometimes by screwing the barrel) to obtain a flat top from a calibrator square wave.

author: Nokia3310
author_rank: Contributor
timestamp: November 11, 2020, 06:33:49 pm
content: I took the measurement today with the dso150 and the attenuator and it workedThe waveform looked like the waveform in this examplehttps://www.picoauto.com/library/automotive-guided-tests/moto-primary-voltage/I forgot to put the probe on x10 so it was on x1.Luckily the voltage didn't exceed the limit of the oscilloscope with the probe on x1.The peak voltage was just under 9 div high at 2V/div so 18V. Then multiply for the attenuator 18V x 20 = 360V pk. Which agrees with the example in the link.The voltage during burn was approx 1 div high at 2V/div so 2V x 20 = 40V. Which is the same as the exampleThe burn time was approx. 2 div wide so 2 x 0.5ms = 1ms. Also the same as the example.

author: tggzzz
author_rank: Super Contributor
timestamp: November 11, 2020, 07:51:36 pm
content: Quote from: Nokia3310 on November 11, 2020, 06:33:49 pmTold youI have never everevermade that mistake, oh no.You may find the references athttps://entertaininghacks.wordpress.com/library-2/scope-probe-reference-material/helpful for the future.

author: Nokia3310
author_rank: Contributor
timestamp: November 11, 2020, 09:06:10 pm
content: Quote from: tggzzz on November 11, 2020, 07:51:36 pmThanks I'll take a look at that.

author: TimFox
author_rank: Super Contributor
timestamp: November 11, 2020, 09:27:53 pm
content: That reference should include how to adjust the probe compensation: very important.

author: Nokia3310
author_rank: Contributor
timestamp: November 13, 2020, 06:30:49 pm
content: Something else I thought of with this. You can work out the speed of the engine when the voltage measurement was taken. This engine is an inline four cylinder but it has a waste spark ignition, so although it has four cylinders it only has two ignition coils. Each coil providing the spark for two different cylinders. That means if you take a measurement that has two successive voltage peaks in it you are seeing the spark event for two different cylinders. In this case it was the ignition coil for cylinders 2 and 3. It's a four stroke engine so there is 360 degrees of crank rotation between the spark event on cylinder 2 and cylinder 3. That means if you measure the time between the peak voltages you are measuring the time for 1 revolution of the engine.9 div x 5mS = 45mS for 1 crank revolution. Engine speed is usually shown in rev/min. Therefore in 1 minute 60,000mS/45mS = 1,333 Revolutions or 1,333 RPM. The bike idles at an indicated 1,250 RPM so not too far off. The engine computer must use those voltage peaks to calculate RPM. There's a RPM signal wire going from the engine computer and the dashboard display. I think I'll have a look at the signal on that wire next! See what shape it is.

author: Nokia3310
author_rank: Contributor
timestamp: November 22, 2020, 03:26:44 pm
content: Quote from: Nokia3310 on November 13, 2020, 06:30:49 pmJust to add a bit more to this. I had a look at the RPM signal going from the engine computer to the dashboard. It's turns out that it is a 12V square wave. From what I learned the first day I knew that because the engine is an inline 4 with waste spark, the two successive ignition coil peak voltages occurred in 1 engine revolution so I thought it would make sense if the computer used that fact to generate the RPM signal.To find out if that was true, when I took the RPM signal measurement I set the time/div scale to the same as it was on the ignition coil measurement, 5ms. If the computer was using the ignition coil peaks to generate the RPM signal the two wave forms should line up when they're on the same time/div and they doIt turns out that the computer is using voltage peaks from the right ignition coil (cylinder 2 and 3) to switch the RPM signal high and it's using the left ignition coil (cylinder 1 and 4) to switch the RPM signal low so that creates the square wave RPM signal.From the picture you can see how the RPM signal switches high when the right side ignition coil peak voltage occurs. In the middle of the RPM picture it shows the RPM signal switching low. This is happening because the left ignition coil peak voltage has occurred. Unfortunately I can't get the left and right ignition coil peak voltages on the screen at the same time to show the full engine cycle because the oscilloscope only has one channel. The full cycle would show a peak to the left for cylinder 1 and another peak in the middle for cylinder 4. And they would correspond with the RPM switching low.

author: Nokia3310
author_rank: Contributor
timestamp: November 22, 2020, 04:13:35 pm
content: The charging system on the bike generates 3 phase AC and that then that gets converted to DC by what I think is called a shunt regulator/rectifier. I would like to use the oscilloscope to look at the AC waveform coming from each phase of the stator. My question is about where the ground clamp on the probe should be attached? I think I saw somewhere that with an oscilloscope the ground clamp should always be connected to ground and never to any other part of the circuit. Is that right? In this case is ground the other side of the stator coil or is it battery negative? I think it should be the other side of the stator coil because the stator is insulated from the rest of the bike but I thought I'd ask before I connect it wrong and blow up the oscilloscope.