vinodquilon Member Aug 19, 2013 #1 For the below circuit, I have placed multimeter +ve probe at 'B' wrt GND and measures 4V against the expected value of 0V. Hence I have terminated the 'B' end with 10K to GND temporarily before taking the measurement and reads 0V. Why this happens? Attachments diode ORing.png 8.6 KB · Views: 296
For the below circuit, I have placed multimeter +ve probe at 'B' wrt GND and measures 4V against the expected value of 0V. Hence I have terminated the 'B' end with 10K to GND temporarily before taking the measurement and reads 0V. Why this happens?
ericgibbs Well-Known Member Most Helpful Member Aug 19, 2013 #2 hi, Check the reverse leakage current of the D2 diode and input impedance of your multimeter. E Attachments AAesp03.gif 32.3 KB · Views: 285
vinodquilon Member Aug 19, 2013 #3 ericgibbs said: hi, Check the reverse leakage current of the D2 diode and input impedance of your multimeter. E Click to expand... What happens to the anode of D2 while I connect a multimeter to 'B' end wrt GND ? I have replaced the 10K temporary resistor with 1M, then again 4V measured !
ericgibbs said: hi, Check the reverse leakage current of the D2 diode and input impedance of your multimeter. E Click to expand... What happens to the anode of D2 while I connect a multimeter to 'B' end wrt GND ? I have replaced the 10K temporary resistor with 1M, then again 4V measured !
ericgibbs Well-Known Member Most Helpful Member Aug 19, 2013 #4 vinodquilon said: What happens to the anode of D2 while I connect a multimeter to 'B' end wrt GND ? I have replaced the 10K temporary resistor with 1M, then again 4V measured ! Click to expand... hi, When you connect the multimeter to point 'B' you are providing a path for the reverse leakage current of the diode to flow to 0V. The leakage current will cause a voltage drop across the internal input resistance of your meter which will display as a voltage, its only nanoamps. E
vinodquilon said: What happens to the anode of D2 while I connect a multimeter to 'B' end wrt GND ? I have replaced the 10K temporary resistor with 1M, then again 4V measured ! Click to expand... hi, When you connect the multimeter to point 'B' you are providing a path for the reverse leakage current of the diode to flow to 0V. The leakage current will cause a voltage drop across the internal input resistance of your meter which will display as a voltage, its only nanoamps. E
vinodquilon Member Aug 19, 2013 #5 ericgibbs said: hi, When you connect the multimeter to point 'B' you are providing a path for the reverse leakage current of the diode to flow to 0V. The leakage current will cause a voltage drop across the internal input resistance of your meter which will display as a voltage, its only nanoamps. E Click to expand... That is when I connect 1M resistor, V=IR-- a considerable drop occurs and reads as 4V. In the case of 10K, measures very low as nA * KOhm and will be in MicroVolt range approximately 0V. Am I right ? Last edited: Aug 19, 2013
ericgibbs said: hi, When you connect the multimeter to point 'B' you are providing a path for the reverse leakage current of the diode to flow to 0V. The leakage current will cause a voltage drop across the internal input resistance of your meter which will display as a voltage, its only nanoamps. E Click to expand... That is when I connect 1M resistor, V=IR-- a considerable drop occurs and reads as 4V. In the case of 10K, measures very low as nA * KOhm and will be in MicroVolt range approximately 0V. Am I right ?
ericgibbs Well-Known Member Most Helpful Member Aug 19, 2013 #6 vinodquilon said: But what happens when I connect a 10K ? Click to expand... Consider 20nanoamps flowing thru a 10K resistor, thats Vdrop = 10000 * .00000002 = 200 microVolts! You may have a faulty diodes.
vinodquilon said: But what happens when I connect a 10K ? Click to expand... Consider 20nanoamps flowing thru a 10K resistor, thats Vdrop = 10000 * .00000002 = 200 microVolts! You may have a faulty diodes.