chris kennedy
Senior Member
- Location
- Miami Fla.
- Occupation
- 60 yr old tool twisting electrician
I reversed the leads and I'm getting 57V both ways.
Forum members never cease to amaze me
- Thread starter chris kennedy
- Start date
- Status
I reversed the leads and I'm getting 57V both ways.
Rick Christopherson
Senior Member
If I am seeing that paper correctly, you have several 1 meg resistors. Build a voltage divider without using the diodes, and see if the two meters are still in agreement then.
chris kennedy
Senior Member
- Location
- Miami Fla.
- Occupation
- 60 yr old tool twisting electrician
How do I do that?
mark32
Senior Member
- Location
- Currently in NJ
Hi Larry, the Greenlee model in question does not have such an option, I'm almost positive. Perhaps Chris can confirm this.
chris kennedy
Senior Member
- Location
- Miami Fla.
- Occupation
- 60 yr old tool twisting electrician
Ideal yes, Greenlee no.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090222-1230 EST
chris:
With rather high probability your Ideal meter is OK, and the Greenlee has a problem on AC, at least the range that includes the 115 V reading. Highly probable that the device or method of RMS to DC conversion in the Greenlee has failed. The Greenlee meter is the one to send for repair and calibration if you choose to do anything.
To explain the test circuit. It is a half wave rectifier with no explicit filtering. Total filter capacitance is maybe 100 to 200 pfd. Total load resistance is 2 megohms in parallel with 10 megohms or 5 megohms depending on whether one or two meters are across the 2 meg resistor (the 2 meg resistor is two 1 meg resistors in series), and assuming each meter input resistance is 10 megs. The heat shrink tubing simply covers the solder joint. You could put a 100,000 ohms resistor in place of the 2 meg and have very little effect on the reading. Maybe 0.1 V. The time constant at 200 pfd may be about 0.3 milliseconds, probably is less because the capacitance may not be this large.
The average DC voltage of a half wave rectified sine wave is (2/Pi)*(1/2) = 0.31831. The 1/2 comes from the half wave. 2/Pi is the average of a sine wave of unity peak from 0 to Pi radians. A 120 V RMS sine wave has a peak of 169.706 V. The average half wave rectified value of a 120 V RMS sine wave is 169.706 * 0.31831 = 54.02 V.
The ratio between the RMS value and the half wave rectified average is Pi / 2^0.5 = 2.22144146. Call it 2.222 for ease of memory.
Your DC readings were 57.1 and 57.2 very close and the implication is these are probably valid and that the DC function of both meters is probably OK. I will average these at 57.15 and this times 2.222 = 127.0 V . I suggested adding 0.3 V to compensate for diode drop. Thus, from the DC measurement the estimated RMS value is 127.3 V. Your Ideal reads 126.4 and the Greenlee reads 115.1. The Ideal differs by 0.7% and can for the moment be assumed to be within specification. The Greenlee appears to be incorrect by about 9.6%.
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chris:
With rather high probability your Ideal meter is OK, and the Greenlee has a problem on AC, at least the range that includes the 115 V reading. Highly probable that the device or method of RMS to DC conversion in the Greenlee has failed. The Greenlee meter is the one to send for repair and calibration if you choose to do anything.
To explain the test circuit. It is a half wave rectifier with no explicit filtering. Total filter capacitance is maybe 100 to 200 pfd. Total load resistance is 2 megohms in parallel with 10 megohms or 5 megohms depending on whether one or two meters are across the 2 meg resistor (the 2 meg resistor is two 1 meg resistors in series), and assuming each meter input resistance is 10 megs. The heat shrink tubing simply covers the solder joint. You could put a 100,000 ohms resistor in place of the 2 meg and have very little effect on the reading. Maybe 0.1 V. The time constant at 200 pfd may be about 0.3 milliseconds, probably is less because the capacitance may not be this large.
The average DC voltage of a half wave rectified sine wave is (2/Pi)*(1/2) = 0.31831. The 1/2 comes from the half wave. 2/Pi is the average of a sine wave of unity peak from 0 to Pi radians. A 120 V RMS sine wave has a peak of 169.706 V. The average half wave rectified value of a 120 V RMS sine wave is 169.706 * 0.31831 = 54.02 V.
The ratio between the RMS value and the half wave rectified average is Pi / 2^0.5 = 2.22144146. Call it 2.222 for ease of memory.
Your DC readings were 57.1 and 57.2 very close and the implication is these are probably valid and that the DC function of both meters is probably OK. I will average these at 57.15 and this times 2.222 = 127.0 V . I suggested adding 0.3 V to compensate for diode drop. Thus, from the DC measurement the estimated RMS value is 127.3 V. Your Ideal reads 126.4 and the Greenlee reads 115.1. The Ideal differs by 0.7% and can for the moment be assumed to be within specification. The Greenlee appears to be incorrect by about 9.6%.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090222-1233 EST
ohmhead:
The first picture has both meters in AC position and across the AC line. The diode and resistor have no effect on the readings.
What Chris did not do is simultaneously measure the DC voltage with both meters. If the AC voltage is not varying much, 0.1 V, then this is not important. But, if there is a slight variation in line voltage, then the 57.1 and 57.2 values might be closer or further apart.
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ohmhead:
The first picture has both meters in AC position and across the AC line. The diode and resistor have no effect on the readings.
What Chris did not do is simultaneously measure the DC voltage with both meters. If the AC voltage is not varying much, 0.1 V, then this is not important. But, if there is a slight variation in line voltage, then the 57.1 and 57.2 values might be closer or further apart.
.
electricmanscott
Senior Member
- Location
- Boston, MA
I didn't read the entire thread, too many words and numbers. :grin:
I'd say just be happy the bomb squad didn't get involved. Maybe not the best thing to send through the mail these days.
I'd say just be happy the bomb squad didn't get involved. Maybe not the best thing to send through the mail these days.
Rick Christopherson
Senior Member
The same as what you already had, except you are replacing the diode with a resistor. Put two resistors in series, with 120 volts across the pair, and measure the voltage across just one of the resistors.
Unfortunately, all this is going to do is tell you your meter is broken. I don't know enough about how meters are built, so locating a fried component would be by pure luck. If the rectifier section uses high power diodes or a full wave bridge, you might be able to spot them and replace them. If you pop the cover and take a picture, it is possible someone on the forum could spot them.
ohmhead
Senior Member
- Location
- ORLANDO FLA
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090222-1526 EST
ohmhead:
DC meters read a value of the input that is averaged over some time period. Consider a Simpson 260 in the DC 10 V range with a single diode in series. Assume this diode has zero forward voltage drop, and infinite reverse resistance.
First, I apply 10 V DC to this diode meter combination. What is the reading? 10 V.
Next I replace the DC source with variable frequency AC source. The peak value of the waveform is 10 V.
At 0.1 Hz this meter will track the instantaneous AC voltage from 0 to 10 V back 0 and remain at 0 for 5 seconds and then go back up.
Change to 60 Hz. Now the meter needle is steady at 3.18 V. At this frequency the meter is reading the average value of the waveform.
Somewhere between 0.1 and 60 Hz the needle will transition from tracking the signal to averaging it. In between these two frequencies the needle will do some fast wiggling as it converges toward 3.18 V.
Note: if you changed to a full wave bridge the reading would be 6.36 V .
An RMS meter would reaad 7.07 V .
I think you are wrong to characterize the diode as a voltage divider because it does not help you understand the physics of the problem, and provides no means for you to calculate the expected reading.
Digital meters on DC perform an averaging function but by a different mechanism. The Simpson mechanically averages based on the inertia of the mechanism, and the digital meters probably use an RC circuit.
To understand how you get the average value of a waveform consider this. Draw the waveform. At every 1/N point of the period to be averaged draw a vertical line to the curve. Add the lengths of each line and divide by N. As N approaches infinity this becomes equivalent to measuring the area under the curve and dividing by the length of the base of that area. By use of calculus it is easy to determine the area under a half sine wave. Similarly for the sine squared function to calculate RMS.
Digital true RMS reading meters have to have a device to convert AC to DC. This is not simply a bridge or half wave rectifier. In an earlier post on this thread I provided a reference to to an HP (Agilent) discussion of some methods of true RMS measurement.
.
ohmhead:
DC meters read a value of the input that is averaged over some time period. Consider a Simpson 260 in the DC 10 V range with a single diode in series. Assume this diode has zero forward voltage drop, and infinite reverse resistance.
First, I apply 10 V DC to this diode meter combination. What is the reading? 10 V.
Next I replace the DC source with variable frequency AC source. The peak value of the waveform is 10 V.
At 0.1 Hz this meter will track the instantaneous AC voltage from 0 to 10 V back 0 and remain at 0 for 5 seconds and then go back up.
Change to 60 Hz. Now the meter needle is steady at 3.18 V. At this frequency the meter is reading the average value of the waveform.
Somewhere between 0.1 and 60 Hz the needle will transition from tracking the signal to averaging it. In between these two frequencies the needle will do some fast wiggling as it converges toward 3.18 V.
Note: if you changed to a full wave bridge the reading would be 6.36 V .
An RMS meter would reaad 7.07 V .
I think you are wrong to characterize the diode as a voltage divider because it does not help you understand the physics of the problem, and provides no means for you to calculate the expected reading.
Digital meters on DC perform an averaging function but by a different mechanism. The Simpson mechanically averages based on the inertia of the mechanism, and the digital meters probably use an RC circuit.
To understand how you get the average value of a waveform consider this. Draw the waveform. At every 1/N point of the period to be averaged draw a vertical line to the curve. Add the lengths of each line and divide by N. As N approaches infinity this becomes equivalent to measuring the area under the curve and dividing by the length of the base of that area. By use of calculus it is easy to determine the area under a half sine wave. Similarly for the sine squared function to calculate RMS.
Digital true RMS reading meters have to have a device to convert AC to DC. This is not simply a bridge or half wave rectifier. In an earlier post on this thread I provided a reference to to an HP (Agilent) discussion of some methods of true RMS measurement.
.
ohmhead
Senior Member
- Location
- ORLANDO FLA
Well Gar lets look at the circuit the diode and then series resistors now the voltage is measured at ac lets say its 100 volts ac both meters. when you connect from neutral to the hot ac power 100 volts now we switch the meters to dc power and get half the voltage no i do not agree with what we are looking at. Your just going thur a diode it should be almost the full 100 volts dc minus the 1 volt drop lost across the diode do you agree with that . why do i see 50 volts dc half the volts on them meters there is a problem with that ? i see a divided circuit if not then explain the voltage cut in half at dc it should be 100 volts dc or ac . A half wave rect dc is no different then a fullwave rect dc its just more pure dc pulses of both neg and pos in one direction so why the voltage is cut in half ?comments best to ya
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090222-2022 EST
ohmhead:
In post #26 and #31 I described how the diode produced the DC voltage that we have discussed. If you do not believe this, then run the following experiments:
Components needed ---
1 1N4004 diode
1 Bridge rectifier made from 4 1N4004 diodes
1 2 megohm 1 W carbon composition resistor
1 1 mfd 400 V capacitor
The 2 meg resistor is the bleeder across the capacitor, Time constant 2 seconds.
Four experiments to run. Source voltage 120 V 60 Hz sine wave.
#1 One diode in series with a 10 megohm DC meter. Read voltage.
----- reading should be about 54 V.
#2 Same as #1 but put the capacitor across the meter leads.
----- reading should be about 170 V.
#3 Use the diode bridge in place of the diode.
----- reading should be about 108 V.
#4 Same as #3 with capacitor added.
----- reading should be about 170 V.
With the single diode case if the diode reverse leakage was equivalent to 1 megohm, then the DC reading in experiment #1 would be greatly reduced.
Experimentally at 122.2 V RMS input with one diode and only the Fluke 27 as a load I read 55.6 V DC. With a 1 megohm shunt across the diode I read 6.47 V DC.
Also see if you can go back and understand the discussion I presented on how to calculate the voltages.
Report back.
.
ohmhead:
In post #26 and #31 I described how the diode produced the DC voltage that we have discussed. If you do not believe this, then run the following experiments:
Components needed ---
1 1N4004 diode
1 Bridge rectifier made from 4 1N4004 diodes
1 2 megohm 1 W carbon composition resistor
1 1 mfd 400 V capacitor
The 2 meg resistor is the bleeder across the capacitor, Time constant 2 seconds.
Four experiments to run. Source voltage 120 V 60 Hz sine wave.
#1 One diode in series with a 10 megohm DC meter. Read voltage.
----- reading should be about 54 V.
#2 Same as #1 but put the capacitor across the meter leads.
----- reading should be about 170 V.
#3 Use the diode bridge in place of the diode.
----- reading should be about 108 V.
#4 Same as #3 with capacitor added.
----- reading should be about 170 V.
With the single diode case if the diode reverse leakage was equivalent to 1 megohm, then the DC reading in experiment #1 would be greatly reduced.
Experimentally at 122.2 V RMS input with one diode and only the Fluke 27 as a load I read 55.6 V DC. With a 1 megohm shunt across the diode I read 6.47 V DC.
Also see if you can go back and understand the discussion I presented on how to calculate the voltages.
Report back.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090222-2232 EST
chris:
Some possible solutions:
See if the Greenlee is proportional to the Ideal over whatever AC range you typically work. May be 80 to 140 V is one range. A Variac (Powerstat) could be your test voltage source. If proportional, then you make a calibration curve for the Greenlee. No cost except time.
Send the Greenlee to a repair and calibration facility.
Junk the Greenlee and get something else suitable to your work.
Get a fancy one like the HP (Agilent). I still do not like their new name. HP was so easy and definitive. I am also unhappy that the computer side of HP gave up on calculators. The HP meter would provide new functions that might be useful, logging in particular, and a readable display.
.
chris:
Some possible solutions:
See if the Greenlee is proportional to the Ideal over whatever AC range you typically work. May be 80 to 140 V is one range. A Variac (Powerstat) could be your test voltage source. If proportional, then you make a calibration curve for the Greenlee. No cost except time.
Send the Greenlee to a repair and calibration facility.
Junk the Greenlee and get something else suitable to your work.
Get a fancy one like the HP (Agilent). I still do not like their new name. HP was so easy and definitive. I am also unhappy that the computer side of HP gave up on calculators. The HP meter would provide new functions that might be useful, logging in particular, and a readable display.
.
ohmhead
Senior Member
- Location
- ORLANDO FLA
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090223-0724 EST
ohmhead:
A conventional DC meter, moving coil (mechanical) or digital, has inertia (reacts to change slowly). This type of meter does not read the peak value or even wiggle in response to a fast changing waveform.
What this meter does is average the input.
Suppose you have a waveform that goes from 0 to 100 V, stays at 100 v for 1 millisecond, returns to 0, stays at 0 V for 9 milliseconds, and then continuously repeats this. What is the average voltage?
What do you think the DC meter will read when measuring this signal?
To format your posts so that a quote comes out quoted do the following:
Click on Wrap Quotes.
Between the inner two square brackets put the material to be quoted.
{QUOTE]here is where the quote goes[/QUOTE}
I have changed the outer square brackets here so this would not actually generate the quoted region so you can see the format.
.
ohmhead:
A conventional DC meter, moving coil (mechanical) or digital, has inertia (reacts to change slowly). This type of meter does not read the peak value or even wiggle in response to a fast changing waveform.
What this meter does is average the input.
Suppose you have a waveform that goes from 0 to 100 V, stays at 100 v for 1 millisecond, returns to 0, stays at 0 V for 9 milliseconds, and then continuously repeats this. What is the average voltage?
What do you think the DC meter will read when measuring this signal?
To format your posts so that a quote comes out quoted do the following:
Click on Wrap Quotes.
Between the inner two square brackets put the material to be quoted.
{QUOTE]here is where the quote goes[/QUOTE}
I have changed the outer square brackets here so this would not actually generate the quoted region so you can see the format.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090223-0807 EST
bth0mas20:
In post #1 Chris has the description of each photo above the photo. This is because it was written as a sequential story rather than as photos with captions below the pictures.
Even without the descriptions the photos are self explanatory.
.
bth0mas20:
In post #1 Chris has the description of each photo above the photo. This is because it was written as a sequential story rather than as photos with captions below the pictures.
Even without the descriptions the photos are self explanatory.
.
ohmhead
Senior Member
- Location
- ORLANDO FLA
[QUOTE=gar;
ohmhead:
A conventional DC meter, moving coil (mechanical) or digital, has inertia (reacts to change slowly). This type of meter does not read the peak value or even wiggle in response to a fast changing waveform.
What this meter does is average the input.
Suppose you have a waveform that goes from 0 to 100 V, stays at 100 v for 1 millisecond, returns to 0, stays at 0 V for 9 milliseconds, and then continuously repeats this. What is the average voltage?
What do you think the DC meter will read when measuring this signal?
Well i know how a dc meter works so lets get to the point why are we measuring 57 volts dc at the output of that diode ? Its not the dc meter time lapse to compute or to calculate the result in dc voltage it should read 119 volts + or less in micro seconds most meters work at well above 400 cycles today ? Is it a series circuit were working with yes or no ? input around 120 volts to output on that diode is not a 50 volt drop in voltage no meter takes that long to register we just would like to know why its half the voltage of the 120 volts + or minus to the original input to the circuit. No one has yet to give me a answer ? I see a series circuit current is the same thur it less than one milliamp i see a resistance than a diode the voltage out or in the center of circuit is half why ? Tell me in the simplest way i really like to know why ? please explain why its not a voltage divider circuit ? Best to ya Gar
ohmhead:
A conventional DC meter, moving coil (mechanical) or digital, has inertia (reacts to change slowly). This type of meter does not read the peak value or even wiggle in response to a fast changing waveform.
What this meter does is average the input.
Suppose you have a waveform that goes from 0 to 100 V, stays at 100 v for 1 millisecond, returns to 0, stays at 0 V for 9 milliseconds, and then continuously repeats this. What is the average voltage?
What do you think the DC meter will read when measuring this signal?
Well i know how a dc meter works so lets get to the point why are we measuring 57 volts dc at the output of that diode ? Its not the dc meter time lapse to compute or to calculate the result in dc voltage it should read 119 volts + or less in micro seconds most meters work at well above 400 cycles today ? Is it a series circuit were working with yes or no ? input around 120 volts to output on that diode is not a 50 volt drop in voltage no meter takes that long to register we just would like to know why its half the voltage of the 120 volts + or minus to the original input to the circuit. No one has yet to give me a answer ? I see a series circuit current is the same thur it less than one milliamp i see a resistance than a diode the voltage out or in the center of circuit is half why ? Tell me in the simplest way i really like to know why ? please explain why its not a voltage divider circuit ? Best to ya Gar
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090223-1501 EST
ohmhead:
I am sorry but you do not know how a conventional DC voltmeter works. You need to erase some of your misconceptions.
I indicated experiments for you to run. When you run these experiments you will find the results contradict your current concept.
I am now adding to the experiments.
The experiment now consists of two load resistors in addition to whatever small load the meters add in parallel to the load resistors. I used a 1 megohm 1/2 watt and a 150,000 ohm 1/2 W in the data that follows. The values are not too critical. 0.47 to 2.2 megs 1/2 W could be used in place of the 1 meg. 47,000 ohm to 150,000 1/2 W can serve as the second resistor. Keep the relationship greater than 10 to 1 if convenient. If you go below 47,000 ohms you need to consider wattage. 14,400/47,000 = 0.31 W.
Use two meters. Use a Simpson 260, or equivalent, and a DVM. These will be in parallel simultaneously. I used a Simpson 270 and a Fluke 27.
The capacitor I used was 2.25 mfd.
The circuit consists of a diode with its anode connected to the hot side of the 120 V supply. The two resistors are connected to the cathode of the diode. The other end of the higher resistance resistor is connected to the neutral line. When required, the other end of the lower resistance is connected to neutral in addition to the high resistance unit to create a lower load resistance as part of the experiment.
After the first measurements are made a capacitor is added in parallel with the load resistor.
#1. Both meters in AC position read the AC line voltage.
----- 270 read 124 V and 27 read 121.9 V
#2. Next change the meters to DC volts, and connect COM to neutral and + to the diode cathode.
#3. Read the DC voltage with only the high resistance load.
----- 270 read 55 V and 27 read 54.7 V
#4. Parallel the low resistance with the high resistance and read the DC voltage.
----- 270 read 55 V and 27 read 54.7 V
#5. Parallel the capacitor across the load resistor. Note on removal of the capacitor it will have a high voltage and this needs to be bled off.
----- 270 read 169 V and 27 read 167.8 V
After you run these experiments you will have first hand experimental evidence of what happens. How to proceed from this point to have you understand why the results are what they are I have no clue. I have already explained what happens.
One remaining thought. Connect a 5,000 ohm 5 W resistor as the load on the diode. Connect a 100,000 resistor from the diode cathode to one terminal of the capacitor and capacitor's other end to neutral. Setup an oscilloscope with two channels, both scaled the same. Assume the scope is grounded thru the EGC and your neutral also is tied to the EGC at the main panel, then the scope chassis is close to the potential of the neutral end of the load resistor and the filter capacitor. DO NOT connect probe ground leads or the scope chassis to neutral. Use only the probe center point for measurement. With scope probes shorted to scope chassis adjust both traces to overlap and be 1 CM from the bottom. Adjust sensitivity so 8 CM is 200 V. Connect one probe to the diode cathode, and the other probe to hot terminal of the capacitor.
You will see the half sine wave pulses on one trace, and the other trace will be a straight line at 31.8 % of the half sine wave peak.
.
ohmhead:
I am sorry but you do not know how a conventional DC voltmeter works. You need to erase some of your misconceptions.
I indicated experiments for you to run. When you run these experiments you will find the results contradict your current concept.
I am now adding to the experiments.
The experiment now consists of two load resistors in addition to whatever small load the meters add in parallel to the load resistors. I used a 1 megohm 1/2 watt and a 150,000 ohm 1/2 W in the data that follows. The values are not too critical. 0.47 to 2.2 megs 1/2 W could be used in place of the 1 meg. 47,000 ohm to 150,000 1/2 W can serve as the second resistor. Keep the relationship greater than 10 to 1 if convenient. If you go below 47,000 ohms you need to consider wattage. 14,400/47,000 = 0.31 W.
Use two meters. Use a Simpson 260, or equivalent, and a DVM. These will be in parallel simultaneously. I used a Simpson 270 and a Fluke 27.
The capacitor I used was 2.25 mfd.
The circuit consists of a diode with its anode connected to the hot side of the 120 V supply. The two resistors are connected to the cathode of the diode. The other end of the higher resistance resistor is connected to the neutral line. When required, the other end of the lower resistance is connected to neutral in addition to the high resistance unit to create a lower load resistance as part of the experiment.
After the first measurements are made a capacitor is added in parallel with the load resistor.
#1. Both meters in AC position read the AC line voltage.
----- 270 read 124 V and 27 read 121.9 V
#2. Next change the meters to DC volts, and connect COM to neutral and + to the diode cathode.
#3. Read the DC voltage with only the high resistance load.
----- 270 read 55 V and 27 read 54.7 V
#4. Parallel the low resistance with the high resistance and read the DC voltage.
----- 270 read 55 V and 27 read 54.7 V
#5. Parallel the capacitor across the load resistor. Note on removal of the capacitor it will have a high voltage and this needs to be bled off.
----- 270 read 169 V and 27 read 167.8 V
After you run these experiments you will have first hand experimental evidence of what happens. How to proceed from this point to have you understand why the results are what they are I have no clue. I have already explained what happens.
One remaining thought. Connect a 5,000 ohm 5 W resistor as the load on the diode. Connect a 100,000 resistor from the diode cathode to one terminal of the capacitor and capacitor's other end to neutral. Setup an oscilloscope with two channels, both scaled the same. Assume the scope is grounded thru the EGC and your neutral also is tied to the EGC at the main panel, then the scope chassis is close to the potential of the neutral end of the load resistor and the filter capacitor. DO NOT connect probe ground leads or the scope chassis to neutral. Use only the probe center point for measurement. With scope probes shorted to scope chassis adjust both traces to overlap and be 1 CM from the bottom. Adjust sensitivity so 8 CM is 200 V. Connect one probe to the diode cathode, and the other probe to hot terminal of the capacitor.
You will see the half sine wave pulses on one trace, and the other trace will be a straight line at 31.8 % of the half sine wave peak.
.
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