Diode failure

[chris kennedy]

I have a rectifier that has had diode failure twice in a year. I'm thinking TVSS on the AC side may mitigate this problem.

Any other ideas would be greatly appreciated.

Thanks

 

[hillbilly1]

Depot had that issue on their Peschel HID dimmers, the brushes that ran up and down the transformer face had diodes to limit current between windings, and when a storm came up, it would blow those diodes, and burn the surface contact. After a couple of fires, they took them all out of service. They even mounted a smoke detector above them as a stop gap measure. The surges are probably what’s taking yours out too. Florida, especially the Orlando area is pretty well known for lightning!

 

[Jraef]

Maybe. Diode failure are mostly the result of either a high DV/Dt (spike in voltage), or DI/Dt (spike in current) faster than what the device can handle. An SPD on the line side would take care of the DV/Dt issue if there is one, but does nothing for the DI/Dt.

“Dirty” power can cause the DI/Dt depending on what your rectifier is powering. If there are ringing transients on the line and in ringing, the voltage drops below the zero cross, a diode quits conducting. But if the diode is feeding a DC bus with capacitors, the load will pull energy out of the caps and when the next diode in the bridge conducts, the caps instantly pull ALL of their energy through that one diode and can damage it. This is the kind of issue that is best solved by putting a line reactor ahead of the rectifier. It slows down the reaction time for everything.

 

[drcampbell]

Make sure the diodes are the right ones for the application.
Sometimes a poor substitution is made, it fails, and then the poor substitution is repeated for eternity because that's what came out of there.

 

[chris kennedy]

It powers a Dings electromagnet and we got the replacement diode assemblies from Dings. (OEM)

 

[drcampbell]

Well, like I said, "sometimes". Apparently not this time.

 

[Besoeker3]

Maybe. Diode failure are mostly the result of either a high DV/Dt (spike in voltage), or DI/Dt (spike in current) faster than what the device can handle. An SPD on the line side would take care of the DV/Dt issue if there is one, but does nothing for the DI/Dt.

Cancel

 

[gar]

210904-1156 EDT

chris:

Is the diode in series with the inductor? From your description I assume that is likely.

If a diode is in series with an inductor fed from a low impedance DC source, then after the circuit has been closed and current reaches a steady state value, then there is a voltage drop across the inductor. Suppose a series circuit consisting of a battery with its positive terminal pointing to the right, an electromechanical switch, a diode pointing to the right, and a inductor that loops back to the battery negative terminal.

After the switch is closed and current reaches steady state, then positive current flows from left to right thru a forward biased diode and inductor, and the voltage drop across the inductor is positive on the left and negative on the right.

You can not instantaneously change the current thru an inductor. So when the switch is opened the instantaneous voltage across the inductor reverses to keep current flowing in the same direction and magnitude (the inductor has now become a voltage source). This voltage as large as necessary to keep the current at it present value. Thus, a very large reverse voltage is applied to the diode and it likely fails.

To prevent this problem you place a snubber across ( in parallel ) the inductor. This can be an RC network or another diode. A shunt diode across the inductor produces the longest drop out time of current thru the inductor, and the lowest reverse voltage across the inductor.

.

 

[Besoeker3]

210904-1156 EDT

chris:

Is the diode in series with the inductor? From your description I assume that is likely.

If a diode is in series with an inductor fed from a low impedance DC source, then after the circuit has been closed and current reaches a steady state value, then there is a voltage drop across the inductor. Suppose a series circuit consisting of a battery with its positive terminal pointing to the right, an electromechanical switch, a diode pointing to the right, and a inductor that loops back to the battery negative terminal.

After the switch is closed and current reaches steady state, then positive current flows from left to right thru a forward biased diode and inductor, and the voltage drop across the inductor is positive on the left and negative on the right.

You can not instantaneously change the current thru an inductor. So when the switch is opened the instantaneous voltage across the inductor reverses to keep current flowing in the same direction and magnitude (the inductor has now become a voltage source). This voltage as large as necessary to keep the current at it present value. Thus, a very large reverse voltage is applied to the diode and it likely fails.

To prevent this problem you place a snubber across ( in parallel ) the inductor. This can be an RC network or another diode. A shunt diode across the inductor produces the longest drop out time of current thru the inductor, and the lowest reverse voltage across the inductor.

.

Most of the diodes I have dealt with have been full-wave six pulse units. Usually or pretty much always these have had DC choke in series rectified input. And there was a snubber with each diode. I'm not sure if that's what you are referring to.

 

[gar]

210904-1307 EDT

Besoeker3:

A snubber is some sort of circuit that will absorber energy, or distribute the energy over time.

If you need fast response time of dropout of an inductive circuit, then you can use an appropriate series resistor and capacitor across the inductor to dissipate the stored energy in the inductor at a slower rate, and therefore lower peak voltage than without any snubber.

An old type automotive ignition system used the inductive kick back of a charged inductor at low voltage when the breaker points opened to produce a high voltage that was further increased by transformer action to produce the ignition spark. There was a capacitor across the breaker points. The combination of the breaker capacitor and various parameters of the ignition coil, and capacitance of the high voltage wiring all contributed to producing moderately rapid rising spark voltage. This rate of rise was not fast enough to overcome the effects lead fouled spark plugs. Thus, 10,000 miles was typical life of plugs with leaded gas.

.

 

[Besoeker3]

210904-1307 EDT

Besoeker3:

A snubber is some sort of circuit that will absorber energy, or distribute the energy over time.

If you need fast response time of dropout of an inductive circuit, then you can use an appropriate series resistor and capacitor across the inductor to dissipate the stored energy in the inductor at a slower rate, and therefore lower peak voltage than without any snubber.

An old type automotive ignition system used the inductive kick back of a charged inductor at low voltage when the breaker points opened to produce a high voltage that was further increased by transformer action to produce the ignition spark. There was a capacitor across the breaker points. The combination of the breaker capacitor and various parameters of the ignition coil, and capacitance of the high voltage wiring all contributed to producing moderately rapid rising spark voltage. This rate of rise was not fast enough to overcome the effects lead fouled spark plugs. Thus, 10,000 miles was typical life of plugs with leaded gas.

.

A few things.
The diodes in my case* have been in power circuits. High power rectifiers and variable speed drives. Fast they are not. The fundamental frequency is either 50Hz or 60Hz. The choke is to smooth the six-pulse unit.
*We have used some fast recovery diodes for a few applications.

 

[gar]

210914-1947 EDT

Besoekwe3:

Fundamentals --- the current thru an inductor can not change instantaneously, the voltage across a capacitor can not change instantaneously.

So consider an inductor with some steady current thru the inductor, and a series switch that can open the circuit. Open the switch and that exact same current will continue to flow, and in the same direction. This means the inductance goes from being a load to a voltage generator, and the voltage across the switch goes from positive to negative, and of a sufficiently high voltage to maintain the previous current. An arc occurs across the switch to maintain current flow.

Put a series diode in the circuit, and that diode probably will be destroyed. Same is true of a bridge rectifier source. So something must be done to prevent a large reverse voltage across the diode. A reverse biased diode across the load will solve the problem. Some other ty[pe of snubber can be adequately effective.

.

 

[Besoeker3]

210914-1947 EDT

Besoekwe3:

Fundamentals --- the current thru an inductor can not change instantaneously, the voltage across a capacitor can not change instantaneously.

So consider an inductor with some steady current thru the inductor, and a series switch that can open the circuit. Open the switch and that exact same current will continue to flow, and in the same direction. This means the inductance goes from being a load to a voltage generator, and the voltage across the switch goes from positive to negative, and of a sufficiently high voltage to maintain the previous current. An arc occurs across the switch to maintain current flow.

Put a series diode in the circuit, and that diode probably will be destroyed. Same is true of a bridge rectifier source. So something must be done to prevent a large reverse voltage across the diode. A reverse biased diode across the load will solve the problem. Some other ty[pe of snubber can be adequately effective.

.

It's not quite what happens. The choke is smoothing. Quite often we have six, twelve, or twenty-four devices. I'll see if I can dig out a diagram if you wish.

 

[gar]

210915-1348 EDT

Besoeker3:

A choke is an inductor. You can not instantaneously change the current thru any inductor. Basic electrical theory. When you try to change that current, then the inductor will produce whatever voltage that is needed to maintain that current flow. As time passes after the circuit is opened the current will gradually decrease as energy is dissipated somewhere in the circuit. This will be in circuit resistance, and dominantly in the opened switching element.

When you place a reverse biased diode across the inductor then there is a low resistance for this current to flow and thus a long discharge time constant. You can add resistance to the path to shorten the discharge time constant. How much can be added is determined by the amount of arcing you want in the switch.

.

.

 

[gar]

210915-1412 EDT

Besoeker3:

I suggest that you study some books on differential equations, and electrical transients. I would believe such books and discussions are found in any undergraduate electrical engineering program. Probably at a sophomore or junior level. May be taught in physics classes as well.

.

 

[Besoeker3]

210915-1348 EDT

Besoeker3:

A chole is an inductor. You can not instantaneously change the current thru any inductor. Basic electrical theory. When you try to change that current, then the inductor will produce whatever voltage that is needed to maintain that current flow. As time passes after the circuit is opened the current will gradually decrease as energy is dissipated somewhere in the circuit. This will be in circuit resistance, and dominantly in the switching element.

When you place a reverse biased diode across the inductor then there is a low resistance for this current to flow and thus a long discharge time constant. You can add resistance to the path to shorten the discharge time constant.

.

.

The devices we used are regular semiconductors, not fast devices.
I did use fast devices too for variable frequency applications. This one was a Marconi unit.

 

[Besoeker3]

210915-1412 EDT

Besoeker3:

I suggest that you study some books on differential equations, and electrical transients. I would believe such books and discussions are found in any undergraduate electrical engineering program. Probably at a sophomore or junior level. May be taught in physics classes as well.

.

I'll let that pass...........

 

[gar]

210915-1958 EDT

Besoeker3:

Look at post #5. The diode appears to be in series with an inductor to provide DC to the coil. Thus, the subject of this thread appears to be a diode in series with an inductor, resistance, some series switch, and why does the diode fail.

This is a classic series circuit of a DC or AC power source, a switch, a resistance, and inductor. A typical circuit analyzed in a differential equations course.

Why do you not address what appears to be a simple circuit. You list yourself as an electrical engineer. Is that really true?

.

 

[Besoeker3]

210915-1958 EDT

Besoeker3:

Look at post #5. The diode appears to be in series with an inductor to provide DC to the coil. Thus, the subject of this thread appears to be a diode in series with an inductor, resistance, some series switch, and why does the diode fail.

This is a classic series circuit of a DC or AC power source, a switch, a resistance, and inductor. A typical circuit analyzed in a differential equations course.

Why do you not address what appears to be a simple circuit. You list yourself as an electrical engineer. Is that really true?

.

I have been designing and manufacturing power electronics for fifty years, Usually from 30kW to 10MW. Fiddling about with differential equations, Fourier analysis, Bode plots, and Nyquist diagrams. Oh, and I do PAT training - that's pets as therapy.

 

[gar]

210916-0732

Besoeker3:

Then why don't you address what happens in the probable circuit described in posts #1 and #5?

.