My high-voltage differential probe design

[Dago]

There has been a couple of topics in here about DIY high-voltage differential probes. I finally managed to test mine and wrote a post about it: http://www.dgkelectronics.com/high-voltage-differential-probe/

I think it came out pretty nice. Seems to work comparably to a cheaper Agilent probe and better than Chinese probes. Hopefully helpful for someone!

[dom0]

Looks good basically, but I'd suggest connecting the caps and resistors in the input divider in parallel (at each node) so as to keep the LF potential on each cap about the same. Also ten resistors in series to withstand higher voltage doesn't make a lot of sense when you put just four caps in parallel with them ; the caps will fail before the resistors.

[Mechatrommer]

seems too easy...

1) did you measure cmrr using standard method? ie: test1) input1=signal, input2=gnd, result=output1, test2) input1=input2=signal, result = output2... cmr = 20 * log(output1/output2) because your cmrr measurement result seems optimistically straight line. i dont good in log math but it seems 40MHz and above signal only give -10dbm attenuation max which seems not alot. try to use the above formula for standard figure...

2) the challenge is not measuring large signal, but very small signal or common mode rejection. try connecting both input to either reallife fet drain or source and present the result...

[Dago]

Looks good basically, but I'd suggest connecting the caps and resistors in the input divider in parallel (at each node) so as to keep the LF potential on each cap about the same. Also ten resistors in series to withstand higher voltage doesn't make a lot of sense when you put just four caps in parallel with them ; the caps will fail before the resistors.

The resistors are specified for 100 V (maximum overload voltage of 200 V) and capacitors for 250 V each. Probe is designed for 500 V. I was thinking about the parallel connection but I haven't seen it done on commercial probes so I decided not to

[Dago]

seems too easy...

1) did you measure cmrr using standard method? ie: test1) input1=signal, input2=gnd, result=output1, test2) input1=input2=signal, result = output2... cmrr = 20 * log(output1/output2) because your cmrr measurement result seems optimistically straight line. i dont good in log math but it seems 40MHz and above signal only give -10dbm attenuation max which seems not alot. try to use the above formula for standard figure...

2) the challenge is not measuring large signal, but very small signal or common mode rejection. try connecting both input to either reallife fet drain or source and present the result...

CMRR was measured by shorting the inputs together, ground connected to probe ground. There is no offset or anything on the scale so it includes the probe attenuation. But thanks for the info on how to properly measure the CMRR, I'll try that method soon and see what it looks like. I'll also try to find a real example.

[Mechatrommer]

this is the best i can find, i remember reading better material not sure where...
http://www.analog.com/media/en/training-seminars/tutorials/MT-042.pdf

There is no offset or anything on the scale so it includes the probe attenuation.

yes thats how i read it. -35db probe attenuation. if my math is correct 57 - 35 = -22db rejection at 10MHz, say a 1Vpp signal, the output will be 0.08V ie 20log(CMRR), CMRR is 1/0.08 = 1:12.5 thats pretty low. at that freq, most off the shelf probe including clones are somewhere 1:100 - 1:1000

edit: i made an error and corrected the above formula, cmr is actually the 20log value, cmrr is the ratio Vcm/Voutput.

there better...  page 9, you should strive for that....
http://www.trs-rentelco.com/manual/at_1154a_manual.pdf

or if that is too much, here page 56...
http://www.trs-rentelco.com/Manual/AT_1141A_Manual.pdf

[bitwelder]

What do you estimate is the total cost of the project (i.e. including a suitable enclosure)?

[Dago]

this is the best i can find, i remember reading better material not sure where...
http://www.analog.com/media/en/training-seminars/tutorials/MT-042.pdf

There is no offset or anything on the scale so it includes the probe attenuation.

yes thats how i read it. -35db probe attenuation. if my math is correct 57 - 35 = -22db rejection at 10MHz, say a 1Vpp signal, the output will be 0.08V ie 20log(CMRR), CMRR is 1/0.08 = 1:12.5 thats pretty low. at that freq, most off the shelf probe including clones are somewhere 1:100 - 1:1000

edit: i made an error and corrected the above formula, cmr is actually the 20log value, cmrr is the ratio Vcm/Voutput.

there better...  page 9, you should strive for that....
http://www.trs-rentelco.com/manual/at_1154a_manual.pdf

or if that is too much, here page 56...
http://www.trs-rentelco.com/Manual/AT_1141A_Manual.pdf

I think the values are actually 10 dB better because the input power for the CMRR measurement was 10 dBm. So for 10 MHz the CMRR is (subtracting the probe attenuation of 43 dB) ~25 dB. Which is pretty much exactly same as Agilents N2791A 25 MHz low-end differential probe: http://literature.cdn.keysight.com/litweb/pdf/5990-3780EN.pdf (figure 9). I think I can make it maybe slightly better by tuning it better but otherwise I'm pretty happy with the performance.

What do you estimate is the total cost of the project (i.e. including a suitable enclosure)?

Not much. The most expensive part are the 0.1% resistors which are like 0.3€ a piece and you need 20 of them. And they are not a necessity, you could use just regular ones and trim the dividers by hand. The op-amps are 2€ and 3.5€ each from Digi-key. The enclosure is like 10€ or something. So maybe 40€ total all in all.

[dom0]

And they are not a necessity, you could use just regular ones and trim the dividers by hand. The op-amps are 2€ and 3.5€ each from Digi-key. The enclosure is like 10€ or something. So maybe 40€ total all in all.

Regular (1 % or worse) resistors will probably degrade temperature stability (esp. of CMRR) a lot due to their typically higher tempco.

I wouldn't use a metal case in this case (no pun intended) since a short from any of the inputs to the case will expose the user to quite a risk of electrical shock (and might cause some nice sparks, depending on what it is used on).

[DanielS]

I wouldn't use a metal case in this case (no pun intended) since a short from any of the inputs to the case will expose the user to quite a risk of electrical shock (and might cause some nice sparks, depending on what it is used on).

The box is presumably grounded through the scope's coax and possibly the power source. There may be sparks but the user should be safe as long as the ground connections from the box to actual ground are good enough. You would need shielding around the voltage divider network to minimize noise pick-up.

[Verdant]

Hey!
Why did you connect capacitors in parallel to R1 and R3?