Ignition coil HV pulse on oscilloscope?

In summary, the conversation was about using a HV step up transformer and wanting to measure the HV pulses on an oscilloscope without damaging the equipment. It was suggested to use a differential probe, but since they are expensive, the idea of using a lower rated probe with a voltage divider was brought up. The calculations for the resistance values needed for the voltage divider were discussed, and some resources for high voltage resistors were provided. There were also suggestions for using a Zener diode for additional protection.
  • #1
BHY-BK
18
1
Hi all,
I've been using a HV step up transformer (10kV-20W on secondary) lately and I'm wanting to look at the HV pulses on an oscilloscope. The coil is isolated from ground. I know connecting a probe directly to it is dangerous and would ruin my scope. I want to use a differential probe but they are crazy expensive so this got me wondering...

Instead of using a differential probe rated at 10+kV could I use a lower rated probe (say 1kV) then a voltage divider on the circuit using some HV resistors across the components I want to take measurements across? Would this be safe to do as long as its done right. I should mention also the pulsing frequency is 25kHz so I shouldn't need to worry about parasitic elements on the divider, or do I?
 
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  • #2
BHY-BK said:
Instead of using a differential probe rated at 10+kV could I use a lower rated probe (say 1kV) then a voltage divider on the circuit using some HV resistors across the components I want to take measurements across?
Yes, but you need to isolate every bit of wire and then put the whole resistor chain inside an isolating sleeve. 10kV creates sparks at every sharp corner (or wire end).

Now for a bit of calculation: If you have a 10:1 probe connected to your scope, it will have an input impedance of 10MΩ (that is fairly standard). Assuming that your scope can handle a 10V input, the 10:1 probe expands that to 100V. In order to look at 10kV, you need a series resistance R given by [itex]\frac{100V}{10kV}=\frac{10M\Omega}{R+10M\Omega} [/itex], giving a value for R = 990MΩ. That kind of resistance is hard to come by! For an example of a commercial high voltage probe, see http://www.elexp.biz/tst_vp40.htm.
 
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  • #3
Svein said:
Yes, but you need to isolate every bit of wire and then put the whole resistor chain inside an isolating sleeve. 10kV creates sparks at every sharp corner (or wire end).

Now for a bit of calculation: If you have a 10:1 probe connected to your scope, it will have an input impedance of 10MΩ (that is fairly standard). Assuming that your scope can handle a 10V input, the 10:1 probe expands that to 100V. In order to look at 10kV, you need a series resistance R given by [itex]\frac{100V}{10kV}=\frac{10M\Omega}{R+10M\Omega} [/itex], giving a value for R = 990MΩ. That kind of resistance is hard to come by! For an example of a commercial high voltage probe, see http://www.elexp.biz/tst_vp40.htm.
Thanks for the help Svein,

So I'm looking at buying this diff probe
https://www.globalmediapro.com/dp/A01LT0/Pintek-DP-25-Differential-Probe-25MHz-1000V/#Compatibles

The specs say the probe has:
Input Impedance :
2M Omh, 2.3 PF between inputs and ground
4M Omh, 1.2 PF between inputs

and

Attenuation : x 20, x 50, x 200 (Into 1M Ohm scope) ; x 40, x 10, x 400 (Into 50 Ohm load)

So how do I calculate the resistance values I would need? Just to be on the safe side I would want to use the divider to reduce the voltage across the diff probe to 500V from 10kV. I know how to calculate the voltage drop across the resistances to get 500V but how do I know what resistance values to use?

I just don't understand the equation you showed...How did you get 990M ohms?Ok, I think I figured it out. I never thought to impedance match the scope. Glad I asked on this forum.
 
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  • #4
BHY-BK said:
Attenuation : x 20, x 50, x 200 (Into 1M Ohm scope) ; x 40, x 10, x 400 (Into 50 Ohm load)
So, if your scope has an input of 1MΩ and a max input level of 10V, using a 200x attenuation means a max input level of 200⋅10V = 2kV. If your scope can handle 50V input, you are just barely within range. Otherwise, look for a probe with a higher attenuation.
 
  • #5
It all depends on how much you want to load down the signal.

A series resistor on each leg, with a small resistor across them will get the job done. I wouldn't depend on the probe as a calibrated load.

For example two 1K resistors feeding a 1 ohm resistor gives 5V,(10KV/2001 ohms) but I'm sure that is too high a load (2000 ohms 10KV is 5 amps)
10K 10 ohms
100K 100 ohms
1M 1000 ohms will draw 5ma, which still may be too high.
etc
I found these 10M 10KV resistors for $113 each (lol) https://www.amazon.com/dp/B00Q7VQ9NI/?tag=pfamazon01-20

Here is a page of ohmite high voltage resistors
http://www.ohmite.com/search.php?appl=High Voltage&function=results

Here is a high voltage divider (I think you need two):
http://www.ohmite.com/contact.php?function=form&which=divider&set=slimmox
 
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  • #6
meBigGuy said:
It all depends on how much you want to load down the signal.
Exactly. 20W at 10kV is only 2mA of current so your Pot Chain needs to take significantly less than that if you don't want your measurement to affect the circuit operation. If you want better than 10% error then your chain must take less than 200μA. That would need a resistance higher than 10M. I looked at "high voltge probes" and they come at a couple of hundred GBP and have a resistance in the region of 10M (confirming my idea). I think that (safe) idea could be the best way to go. You could always try eBay . . . . In eBay UK, there are some for sale at around 50GBP.
 
  • #7
check this series of resistors

http://www.ohmite.com/cat/res_minimox.pdf

they're not very expensive

a string of 10 or 100 meg ersistors to make a divider sounds practical enough

pay attention to svein's advice about high voltage precautions
 
  • #8
It might be a good idea, if you are going for a DIY solution, to put a Zener diode across your scope input to short out any unwanted over volts due to breakdown. At least, you won't do a lot of damage with only 20mA,
 

1) What is an ignition coil HV pulse?

An ignition coil HV pulse is a high-voltage electrical pulse generated by an ignition coil in a vehicle. It is responsible for creating the spark that ignites the air-fuel mixture in the engine, allowing the vehicle to start and run.

2) How is an ignition coil HV pulse measured?

An ignition coil HV pulse can be measured using an oscilloscope, which is a device that displays and measures electrical signals. The oscilloscope is connected to the ignition coil and captures the voltage waveform of the HV pulse.

3) What is the purpose of measuring the ignition coil HV pulse?

Measuring the ignition coil HV pulse is important for diagnosing any issues with the ignition system. It can help identify if the ignition coil is producing the correct voltage and timing for the spark, which can affect the performance of the engine.

4) What should the ignition coil HV pulse waveform look like on an oscilloscope?

The ignition coil HV pulse waveform should have a sharp rise and fall, indicating a quick and powerful spark. The shape and height of the waveform can vary depending on the type of ignition system and the condition of the ignition coil.

5) How can I troubleshoot issues with the ignition coil HV pulse?

If the ignition coil HV pulse waveform is not within the expected range, it may indicate a problem with the ignition coil or other components in the ignition system. It is best to consult a professional mechanic or refer to the vehicle's manual for specific troubleshooting techniques.

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