Precision OHM meter - Page 1
Forum section: electronics / general

author: behzad63
author_rank: Contributor
timestamp: October 10, 2016, 04:14:59 am
content: Hi dearhow dose a ohm meter works?whats its protection?Thanks

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: October 10, 2016, 06:36:14 am
content: Basicly you need a current source and you pass the current trough a DUT and measure its voltage across it, you know the current and measuring the voltage would help you determine the reistance

author: behzad63
author_rank: Contributor
timestamp: October 10, 2016, 11:40:37 am
content: Thanks my friendI want to know how do i protect ohm meter against high voltage?I saw many circuit ,do i must use TVS onlyfor protection?thanks

author: retrolefty
author_rank: Super Contributor
timestamp: October 10, 2016, 01:58:18 pm
content: Quote from: behzad63 on October 10, 2016, 11:40:37 amNot usually an important consideration as ohm meter should be used in only powered off equipment or components. What and how are you measuring?

author: zlymex
author_rank: Frequent Contributor
timestamp: October 10, 2016, 03:36:02 pm
content: The protection include the voltage sense part and current source part.The voltage sense part is simply a large value resistor because the input impedance is very high.The protection of current source can be done by a diode and a high voltage current limiter, as can be seen from the schematics of 34401A.If there is a positive voltage, CR202 is revered, no current.If there is a negative voltage, it will be evenly distributed among Q203 to Q210.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: October 10, 2016, 09:26:54 pm
content: Thanks zlymex for the info, But I wonder how this circuit works? where does the Current and voltage are measured?if q202 injects a know current where does the meter sense the DUT voltage? al

author: Dave
author_rank: Super Contributor
timestamp: October 10, 2016, 09:49:12 pm
content: The current from this circuit is injected into the positive input jack. The voltage is then measured on that same input. The voltage on that input will vary depending on the resistor connected to that input.If you were to use 4-wire measurement mode, the voltage would be measured on the sense inputs and the main input would just be used to source the current.Pictures tend to speak louder than words.2-wire measurement:4-wire meaurement:Download the full 34401A service manual if you want to see the whole schematic.Edit: Added the images.

author: behzad63
author_rank: Contributor
timestamp: October 13, 2016, 01:23:57 pm
content: Thanks for your time zlymexI understand how CR202 blocks the positive high voltages, But I do not understand how the Q203-Q210 blocks the negative voltages? or are they having another role?Also what's the purpose of U201,Q202 and Cr201? and finally what's Q211 is doing?

author: zlymex
author_rank: Frequent Contributor
timestamp: October 13, 2016, 02:56:04 pm
content: Quote from: behzad63 on October 13, 2016, 01:23:57 pmU201,Q202 and CR201 are part of the precision current source.Q211 is just used as a low leakage diode. It will switched off(none existance) when normal operation isolating the whole protection circuit. It will be on(short the base of Q204 to ground) if a negative voltage(say -1000V) appears and let the voltage evenly distributed among R203 to R206, each got 250V.Q204, Q206, Q208 and Q210 are voltage followers, so that those outputs(emitters) are almost the same as those resistors, therefore the voltage are also evenly distributed, each got 250V.Q203, Q205, Q207 and Q209 are another set of voltage followers and finally each got 250V too.Specifically, if we ignore the forward diode drops(about 0.6V each), if xref is -1000V, terminal 2 of R206 is also -1000V, terminal 1 of R206 is -750V, terminal 1 of R205 is -500V, terminal 1 of R204 is -250V, terminal 1 of R203 is 0V. Emitter of Q210 will be 750V because it follows the base, and emitter of Q209 will be 750V too because it follow its base. Likewise, emitter of Q207 will be 500V, emitter of Q205 will be 250V and emitter of Q203 will be 0V. Those Q203 to Q210 must be at least 250V rated, so there will be no problem for them to withstand 250V, protection is done for -1000V.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: October 13, 2016, 03:11:11 pm
content: Thanks zlymex,Then how the circuit works in normal conditions? I mean in the normal condition the current from the current source has to pass trough the Q203,Q204,R203 to R206 and CR202 and DUT? if so we would have lot's of voltage drop across them and we can not sense low resistances with enough headroom for resolution if I'm correct or Am'I missing something.

author: Kleinstein
author_rank: Super Contributor
timestamp: October 13, 2016, 03:24:28 pm
content: During normal operation the voltage drop across Q203-210 is somewhere is the 1.5-2V range. Its only 2 BE junctions and about 50-100 mV for the transistors (EC) in saturation and the resistors. In a mains operated DMM with a 15 V (or even higher) supply there is still plenty of headroom for the current source and the DUT. Usually the voltage one the DUT is only the 0.2 V range, maybe 2 V at high resistance.