Diode failure
- Thread starter chris kennedy
- Start date
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gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210916-2053 EDT
Following is a simple experiment to see how "inductive kick"works.
I have a Stancor C1003 choke which measured 20.01 H, and 11 Q, on a General Radio bridge at 1 kHz, and DC resistance of 540 ohms.
To make a very simple test circuit I placed a shunt resistor across the choke. Could have added a diode in series with this resistor, but that simply adds a fix voltage drop in series with the resistance. But this does not have a major effect on the demonstration of the principle of what happens on turn off of current to the inductor. Prior to opening the switch from the battery this shunt is simply an added current load on the power source in addition to the inductor current.
Upon opening the power source switch the choke current from its stored energy goes thru the shunt resistor, but now that resistor is in series with the inductor.
My first experiment was with an 82 ohm resistor. Before switch opening 12 V DC was applied to the choke. This can be seen on the scope monitoring the circuit. Upon opening the power from the battery the voltage across the 82 ohm changed from a constant value of about +12 V to a negative spike of about -2 V. The exponential decay curve had a time constant of about 50 mS. Total R is 540 + 82 = 622 ohms. Calculated L/R is 32 mS. In the ball park of my rough time measurement.
Second experiment is with an external resistance of 5600 ohms. so L/R = 20/6140 = 4 mS. My measured peak reverse voltage was about 70 V, 10 mS time constant.
I might have expected a closer correlation on measured and calculated time constants, but the principle of what happens is illustrated.
.
Following is a simple experiment to see how "inductive kick"works.
I have a Stancor C1003 choke which measured 20.01 H, and 11 Q, on a General Radio bridge at 1 kHz, and DC resistance of 540 ohms.
To make a very simple test circuit I placed a shunt resistor across the choke. Could have added a diode in series with this resistor, but that simply adds a fix voltage drop in series with the resistance. But this does not have a major effect on the demonstration of the principle of what happens on turn off of current to the inductor. Prior to opening the switch from the battery this shunt is simply an added current load on the power source in addition to the inductor current.
Upon opening the power source switch the choke current from its stored energy goes thru the shunt resistor, but now that resistor is in series with the inductor.
My first experiment was with an 82 ohm resistor. Before switch opening 12 V DC was applied to the choke. This can be seen on the scope monitoring the circuit. Upon opening the power from the battery the voltage across the 82 ohm changed from a constant value of about +12 V to a negative spike of about -2 V. The exponential decay curve had a time constant of about 50 mS. Total R is 540 + 82 = 622 ohms. Calculated L/R is 32 mS. In the ball park of my rough time measurement.
Second experiment is with an external resistance of 5600 ohms. so L/R = 20/6140 = 4 mS. My measured peak reverse voltage was about 70 V, 10 mS time constant.
I might have expected a closer correlation on measured and calculated time constants, but the principle of what happens is illustrated.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210918-2103 EDT
Continuation of my previous experiment. The previous results illustrated what happens in the discharge of energy stored in an inductor into a resistance on removal of the excitation current to the inductor. But the calculated values did not agree closely with the measured value. Why?
The value of inductance previously used was from a General Radio LRC bridge at 1 kHz.
For this experiment I used a different means to measure inductance. In this case I placed an 0.47 ufd capacitor in parallel with the inductor, Then I put a 10 k resistor in parallel, and excited the inductor with 12 VDC. Upon removal of the excitation a damped oscillation from the parallel resonant circuit occurs. The frequency was measured via the zero crossing times. This was about 37 Hz, and knowing the capacitance value one can determine the associated inductance. This came out to about 37 H.
The results of the LR tests were:
Series R ... Vpeak ..... One Time ..... One Time
................................... Measured ..... Calculated
1 k ........... 10.2 V ..... 40 mS ........... 37 mS
3 k ........... 65 V ....... 10 ms ............ 12 mS
10 k ......... 210 V ...... 4 mS ............. 3.7 mS
This is an interesting experiment to run if you have the tools to do so. One Time Measured means One Time constant.
.
Continuation of my previous experiment. The previous results illustrated what happens in the discharge of energy stored in an inductor into a resistance on removal of the excitation current to the inductor. But the calculated values did not agree closely with the measured value. Why?
The value of inductance previously used was from a General Radio LRC bridge at 1 kHz.
For this experiment I used a different means to measure inductance. In this case I placed an 0.47 ufd capacitor in parallel with the inductor, Then I put a 10 k resistor in parallel, and excited the inductor with 12 VDC. Upon removal of the excitation a damped oscillation from the parallel resonant circuit occurs. The frequency was measured via the zero crossing times. This was about 37 Hz, and knowing the capacitance value one can determine the associated inductance. This came out to about 37 H.
The results of the LR tests were:
Series R ... Vpeak ..... One Time ..... One Time
................................... Measured ..... Calculated
1 k ........... 10.2 V ..... 40 mS ........... 37 mS
3 k ........... 65 V ....... 10 ms ............ 12 mS
10 k ......... 210 V ...... 4 mS ............. 3.7 mS
This is an interesting experiment to run if you have the tools to do so. One Time Measured means One Time constant.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210922-1006 EDT
I have run some further experiments with a little greater accuracy.
A review --- I believe posts #1 and #5 describe a single diode connected in series with an AC source, and an inductor with some internal resistance. It could be a diode bridge, doesn't matter. This was probably done to supply DC to an inductor. The purpose of the posts was to inquire as to why there were some failures of the diode.
If you know something about inductors, then you realize that the current thru the inductor can not instantaneously change value. Thus, in a series circuit of a battery, switch, resistor, and inductor the current in an inductor has to be exactly the same before and after switch actuation. From various posts of the past I have noticed that a number of responders do not understand, or know about this fundamental characteristic of an inductor.
To get at the basic problem brought up by chris in posts #1 and #5 I want to work with the simplest possible circuit. Thus, I selected the circuit of a 12 V battery plus terminal, switch, a shunt variable resistor from switch output back to said battery negative terminal. In parallel with the shunt resistor is the inductor and its internal series resistance.
I call this a shunt resistor by that name because it is in parallel with the inductor when said switch is closed, but it becomes a series resistor when the switch is open.
The inductor is a Stancher C1003 with 540 ohms measured internal resistance. The shunt resistor is a 0 to 10,000 ohm 10 turn Heliport.
The LRC bridge is a General Radio 1650-A operating at 1 kHz for inductance, and DC for resistance. The voltage across the inductor when at its measured condition was 0.74 V. Thus, core of the inductor was at a low flux level compared to my transient measurements. The measured values were 21 H, 9.8 Q @ 1 kHz, and 540 ohms DC.
The circuit source voltage for my transient test was a 12.2 V storage battery. This voltage did not change between the test load and no load. Thus, the inductor is charged to a current level of 12.2/540 = 22.6 mA. You will note in the following results that we get relatively close correlation between the measured results, and calculated results.
in all of the following tests the initial current into the circuit resistance at t = 0 is 0.0226 A. t = 0 is the time of opening the charging current from the battery. The total circuit resistance at this time is the sum of the internal resistance and the Heliport resistance. the measured voltage on the scope is only the voltage across the Helipot. The circuit resistance was setup by using a Fluke 27 ohmmeter to measure the sum of the internal resistance, and the external adjustable resistance. From the following results I assumed the inductance was 37 H at the operating flux level.
Loop R ..... Init V ....... V at Time Cnst ...... Time to 1 TC .... L/R
1 k ............ 10.4 ........ 3.85 ....................... 0.038 S ............. 0.037
2 k ............ 34 ........... 12.6 ....................... 0.018 S ............. 0.0175
3 k ............ 61 ........... 22.6 ....................... 0.011 S ............. 0.0123
5 k ............ 100 ......... 37 .......................... 0.0072 S ........... 0.0074
10 k .......... 210 ......... 78 .......................... 0.004 S ............. 0.0037
Loop R is the loop resistance measured with Fluke 27.
Init V is the initial peak voltage, across the external series resistance, measured on the scope.
V at Time Const is the calculated voltage of the waveform at 1 time constant using 0.37 times Int V.
Time to 1 TC is the time measured on the scope to the waveform point that is 0.37 of the Init V.
L/R is the calculated time constant from an assumed 37 H inductance, and the measured Loop R.
If you used only a back biased diode as the shunt across the inductor, then the time to 1 time constant is close to L/Rint.
Note the good correlation between Time to 1 TC and L/R
.
I have run some further experiments with a little greater accuracy.
A review --- I believe posts #1 and #5 describe a single diode connected in series with an AC source, and an inductor with some internal resistance. It could be a diode bridge, doesn't matter. This was probably done to supply DC to an inductor. The purpose of the posts was to inquire as to why there were some failures of the diode.
If you know something about inductors, then you realize that the current thru the inductor can not instantaneously change value. Thus, in a series circuit of a battery, switch, resistor, and inductor the current in an inductor has to be exactly the same before and after switch actuation. From various posts of the past I have noticed that a number of responders do not understand, or know about this fundamental characteristic of an inductor.
To get at the basic problem brought up by chris in posts #1 and #5 I want to work with the simplest possible circuit. Thus, I selected the circuit of a 12 V battery plus terminal, switch, a shunt variable resistor from switch output back to said battery negative terminal. In parallel with the shunt resistor is the inductor and its internal series resistance.
I call this a shunt resistor by that name because it is in parallel with the inductor when said switch is closed, but it becomes a series resistor when the switch is open.
The inductor is a Stancher C1003 with 540 ohms measured internal resistance. The shunt resistor is a 0 to 10,000 ohm 10 turn Heliport.
The LRC bridge is a General Radio 1650-A operating at 1 kHz for inductance, and DC for resistance. The voltage across the inductor when at its measured condition was 0.74 V. Thus, core of the inductor was at a low flux level compared to my transient measurements. The measured values were 21 H, 9.8 Q @ 1 kHz, and 540 ohms DC.
The circuit source voltage for my transient test was a 12.2 V storage battery. This voltage did not change between the test load and no load. Thus, the inductor is charged to a current level of 12.2/540 = 22.6 mA. You will note in the following results that we get relatively close correlation between the measured results, and calculated results.
in all of the following tests the initial current into the circuit resistance at t = 0 is 0.0226 A. t = 0 is the time of opening the charging current from the battery. The total circuit resistance at this time is the sum of the internal resistance and the Heliport resistance. the measured voltage on the scope is only the voltage across the Helipot. The circuit resistance was setup by using a Fluke 27 ohmmeter to measure the sum of the internal resistance, and the external adjustable resistance. From the following results I assumed the inductance was 37 H at the operating flux level.
Loop R ..... Init V ....... V at Time Cnst ...... Time to 1 TC .... L/R
1 k ............ 10.4 ........ 3.85 ....................... 0.038 S ............. 0.037
2 k ............ 34 ........... 12.6 ....................... 0.018 S ............. 0.0175
3 k ............ 61 ........... 22.6 ....................... 0.011 S ............. 0.0123
5 k ............ 100 ......... 37 .......................... 0.0072 S ........... 0.0074
10 k .......... 210 ......... 78 .......................... 0.004 S ............. 0.0037
Loop R is the loop resistance measured with Fluke 27.
Init V is the initial peak voltage, across the external series resistance, measured on the scope.
V at Time Const is the calculated voltage of the waveform at 1 time constant using 0.37 times Int V.
Time to 1 TC is the time measured on the scope to the waveform point that is 0.37 of the Init V.
L/R is the calculated time constant from an assumed 37 H inductance, and the measured Loop R.
If you used only a back biased diode as the shunt across the inductor, then the time to 1 time constant is close to L/Rint.
Note the good correlation between Time to 1 TC and L/R
.
I think every failed diode I’ve encountered was the victim of another failed component - usually a shorted electrolytic capacitor in a power supply.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210925-1252 EDT
Continuing my experiments with the same battery, switch, and inductor.
With infinite shunt resistance during charging the inductor, and thus no external series resistance during inductor discharge, I get electrical breakdown across the switch at a little over -300 V, then uniform decay of an exponential nature. This somewhat slow decay with infinite series resistance is not expected. It quite possibly results from core losses that I have not considered.
The high frequency voltage oscillation just following switch opening goes way positive relative to the battery positive voltage, and thus would apply a reverse voltage on a diode if I had one in the circuit. This is the voltage and energy that could damage a series diode.
In my experiments the high frequency oscillation might last to 2 mS.
Do my assumptions and experiments correspond with chris's problem I don't know.
Scope measurements may allow chris to see what is happening.
.
Continuing my experiments with the same battery, switch, and inductor.
With infinite shunt resistance during charging the inductor, and thus no external series resistance during inductor discharge, I get electrical breakdown across the switch at a little over -300 V, then uniform decay of an exponential nature. This somewhat slow decay with infinite series resistance is not expected. It quite possibly results from core losses that I have not considered.
The high frequency voltage oscillation just following switch opening goes way positive relative to the battery positive voltage, and thus would apply a reverse voltage on a diode if I had one in the circuit. This is the voltage and energy that could damage a series diode.
In my experiments the high frequency oscillation might last to 2 mS.
Do my assumptions and experiments correspond with chris's problem I don't know.
Scope measurements may allow chris to see what is happening.
.
Yes, that could be. I have had a a few electrolytic capacitor failures but not usually banks of them.
The power supplies I’m referring to are relatively small, < 1KW. Most have a single capacitor for a filter.
Ah, OK. A single failure would do it.
Yes. My area was 100kW to 10MW.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210928-2021EDT
I have run some additional experiments that are interesting. The assumed circuit is interesting to study. But without feedback from chris on precise details on his circuit I am going to quit thinking about this problem.
.
I have run some additional experiments that are interesting. The assumed circuit is interesting to study. But without feedback from chris on precise details on his circuit I am going to quit thinking about this problem.
.
chris kennedy
Senior Member
- Location
- Miami Fla.
- Occupation
- 60 yr old tool twisting electrician
Very simple circuit.
480V 3 phase primary, 96V 3 phase secondary transformer
Bank of 6 diodes producing 110V DC
MOV in parallel with DC output
Electro magnet.
480V 3 phase primary, 96V 3 phase secondary transformer
Bank of 6 diodes producing 110V DC
MOV in parallel with DC output
Electro magnet.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210929-1009 EDT
chris:
Your information helps some.
A summary of some of my experiments. The first experiments were with a 12 V DC storage battery, a mechanical switch, an inductor with internal series resistance, and a variable resistance shunt in parallel with the inductor and its series resistance. As long as my shunt resistance kept the "inductive kick" voltage below about -340 V I had very clean perfect LR exponential decay curves.
When the shunt resistance was greater or none at all, then I got a high frequency oscillation, in the MHz range, for about 0.7 mS following switch opening, and then followed by the nice exponential decay.
Next I added a series diode in the circuit. The high frequency oscillation remained about the same. During this time the diode was slow enough that it was still in a forward biased condition. At the end of the oscillation the diode turned off, and a large damped oscillation occurred. Peak positive voltage was in the range of 400 V, and oscillation frequency about 1.3 kH. Thus, the diode possibly saw a peak reverse voltage of 400 V.
.
.
chris:
Your information helps some.
A summary of some of my experiments. The first experiments were with a 12 V DC storage battery, a mechanical switch, an inductor with internal series resistance, and a variable resistance shunt in parallel with the inductor and its series resistance. As long as my shunt resistance kept the "inductive kick" voltage below about -340 V I had very clean perfect LR exponential decay curves.
When the shunt resistance was greater or none at all, then I got a high frequency oscillation, in the MHz range, for about 0.7 mS following switch opening, and then followed by the nice exponential decay.
Next I added a series diode in the circuit. The high frequency oscillation remained about the same. During this time the diode was slow enough that it was still in a forward biased condition. At the end of the oscillation the diode turned off, and a large damped oscillation occurred. Peak positive voltage was in the range of 400 V, and oscillation frequency about 1.3 kH. Thus, the diode possibly saw a peak reverse voltage of 400 V.
.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
210929-2018 EDT
chris:
If you can tolerate slow turn off of the magnetic field, then I would suggest a reverse biased diode shunt across the magnetic terminals.
If you just use a diode with no series resistance, then you get the longest dropout time, and the lowest voltage across the inductor terminals. This is simply the diode forward diode drop, and peak current equals the current before turn off. Power dissipation in the diode is based on discharge current and duration. You must properly size the diode and its heat sink.
If you add external series resistance in the diode path, then peak terminal voltage across the inductor increases, less energy is dissipated in the diode, and discharge time, and dropout time decreases. You can make the series resistance as large as you want that does not produce excessive voltage across the inductor.
Often times an RC snubber circuit is used instead of the diode and resistance,
.
chris:
If you can tolerate slow turn off of the magnetic field, then I would suggest a reverse biased diode shunt across the magnetic terminals.
If you just use a diode with no series resistance, then you get the longest dropout time, and the lowest voltage across the inductor terminals. This is simply the diode forward diode drop, and peak current equals the current before turn off. Power dissipation in the diode is based on discharge current and duration. You must properly size the diode and its heat sink.
If you add external series resistance in the diode path, then peak terminal voltage across the inductor increases, less energy is dissipated in the diode, and discharge time, and dropout time decreases. You can make the series resistance as large as you want that does not produce excessive voltage across the inductor.
Often times an RC snubber circuit is used instead of the diode and resistance,
.
chris kennedy
Senior Member
- Location
- Miami Fla.
- Occupation
- 60 yr old tool twisting electrician
Thank you Gordon
I am meeting next week with one of the engineers on a new build which also has the identical electro magnet.
I will share this thread with him and report back.
I am meeting next week with one of the engineers on a new build which also has the identical electro magnet.
I will share this thread with him and report back.
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