Working Voltage for Resistors in Series

Thanks. I thought this was the case, but most of the time this kinds of problems are not discussed in most electronic books.

I appreaciate the help.

Hector

Dear Electroman:,

If you put 10 0.25W 400Volt 1Meg resistors in series you will get a 10Meg 0.25W 2.5W 4000Volt resistor.

I think this is right. Think and reply.

That is incorrect. The current remains the same for a series circuit so the power rating for the package remains the same. The resistance and the voltage rating becomes the sum of the resistors. To increase the power rating they need to be in parallel.

Dear Mr.Jim_Wright,

P=VI

Here I is constant. But V is 4000V instead of 400V. So the total power also to be multiplied by 10. So 0.25*10=2.5W.

or

P=I²R

Here also I constant.

But R=10 M-ohm which is 10 times of individual resistor.

So power also to be 10 times higher.

Is it wrong?

not across the individual resistors. They form a voltage divider so while the current is constant the voltage is only the ratio that the resistor is to the entire stack. for example, if you have 10 100 ohm resistors in series only one tenth of the applied voltage is developed across each resistor.

not across the individual resistors.

We were not at all discussing about individual resistors. The volage , current, power ratings of individual resistors would not be affected.

See the answer of electroman,

Putting resistors in series will increase the overall voltage rating.

If you put 10 0.25W 400Volt 1Meg resistors in series you will get a 10Meg 0.25W 4000Volt resistor.

That 0.25W is told for total unit.

Thanks for the expanded explanation, this defenitely helps me and others that may be faced with the same or similar issues. I appreaciate the help.

Hector

Yet another great answer.

I feel the matter is a bit more complex than your liking .

Let us say there are Three resistors R₁, R₂, R₃ in series with voltages (break down) V₁, V₂, V₃

Let us now raise the voltage v_i, the current increases and we get voltages across resistors v1, v2 and v3

v_i = v1+ v2+v3

At a voltage V_i say the voltage v₂>= V₂ so the resistor R₂ breaks down .

Now the voltages re-distribute to v₁ and v₃ and now

v_i = v1+v3

Now the resitors R₁ and R₃ may break down or may not breakdown.

So safe the series connected resistors, the Breakdown voltage may be the least of

V (R_x/ΣR)

Since the voltage drop across any resistor = IR_j = V/ΣR . R_j

I understand what you're saying, but assuming that all components in the chain are operating as they should (no defects or out of specs) I'm not sure that in practice the failure would occur, should the sum of the Resistors in the chain be equal to or greater than the applied voltage. I say this because I know for a fact that this technique of series Resistors is used to make voltage deviders circuits to measure multi Kilivolt rated probes.

Can anyone shed any light on any facts that migh bridge the gap here if the there is a discrepency between theory and practice here?

Thanks for the example.

Hector

If all of the resistors break down* except one, then the final resistor must have a WV of at least the full applied voltage. As any of the resistors could be the final one in this scenario, then they all must have a WV of at least the full applied voltage. However, this situation may force the final resistor - whichever one it is - to fail anyway, as the full voltage is applied across it.

* High voltage resistors can fail by means of arc-over which effectively lowers the component's value. A high-voltage transient on the applied voltage (say, an inductive spike, for example) may far surpass the resistor's WV and initiate an arc. The resistance of an arc is very, very low, effectively turning the resistor into a short circuit. This failure may cause other resistors in the chain to fail as well - even after the transient has passed - either through arcing or by ohmic heating. Either way the system is hosed.

The lesson here is to design conservatively. Choose a WV that is at least twice the highest expected voltage and, if possible, a WV three or more times than that.

Another important guideline is to choose a WV that considers the gravity of the failure. For instance, if a failure would result in the loss of life or property, the WV should reflect this fact and be much higher than it might be for some other, less-critical application.

Yeah I had similar thoughts on arcing, so that's why I'm planning on putting the board in a pressurized SF-6 environment. I only have 40lbs of the stuff. LOL.

Thanks for the help.

Hector

SF₆ is a great insulator. We used pressurized SF₆ years ago to insulate a 250 kV power supply (@ 20 mA, no less) for one of our ion implanters.

What is your working voltage?

I operate my test devices at 41.5Kv with my small Spellman power supply at an average amperage of 2 micro amps. On my big Spellman operate at 100kv but can go to 130kv and at 4 milli amps.

Not a lot of power, but voltage is high.

Hector

sf6 use for so higher power generator? not bad,

can yoiu show us the set? or describle it size? are you sure you use this gas, instead of other stuff?

Are we talking about Break down (or the Flash over voltages) ? If yes then I stick to my Gun .

Because during my long ago Power Electronics Classes- one eminent prof form Canada - forgot his name has come- who was supposed to be at that time one of the topmost on HVDC- there we have learnt this- when the thyristors are put in the line, this concept has to be addressed for the redundancy purpose.

It is like any mechanical system- the system always fails on the wekest link, and that overstresses the other links.

In case of the voltage distribution - consider the chains in parallel and being pulled out, the individual loads on chains add up to the total load.

If a chain fails, what happens to the other chains ?

And this has got nothing to do with defects in the chain. The resistor (chain) is designed for that much voltage (stress)

You are absolutely right about resistive voltage dividers 0 but remember, the resistors are either rated for that much KV individually or else there is enough factor of safety eg the voltage drop across a resistor under the operating KV is much much less than the B/D rating of the resistor.

Say we are dividing 11KV, the voltage drop (theo across a particular resistor is 150 V, then may be I will select a resistor with B/D voltage say 1KV)

PS: The factor of safety should be taken like normal procedure keeping all the effects and after effects in mind

when the thyristors are put in the line, this concept has to be addressed for the redundancy purpose

Yes. Adding an additional thyristor (or SCR) to a string of thyristors is common and takes into account the potential failure of one or two thyristors in the string (ensuring that the voltage drop across the string and each individual thyristor is less than the safe operating voltage of each individual thyristor).

The key point is that multiple redundancy depends on the probability of a device, string of devices, etc failing. When designing a system the failure probability has to assessed and taken into account, which then gives you a good idea of just how much redundancy you really need (its nice to just stick an additional 100% redundancy in but is awful expensive when the system size gets big and expensive).

Another thing about resistor voltage (and current) ratings is that it really depends on the duration of the event (such as an over-voltage spike) and the duty cycle of the events (such as trying to measure a string of high voltage pulses). Some resistor data sheets include graphs which can give you an idea on just how stressed the resistors will get under the conditions you expect to use them (this information is also necessary to assess actual probability of failure and hence just how much redundancy is needed).

In the end choosing just how many resistors is needed can get rather confusing (especially if you are not able to obtain all the data, and even more so when circuit inductance due to the resistors is a potential issue). In a simple case where the application is not too critical it is just easier to perform a basic assessment and add a few extra resistors as a safety margin, this works in many applications but if the waveform is a long duration non-sinusoidal or a string of repeated pulses then further assessment is needed to prevent internal heating of the resistors causing unexpected premature string failure (which means adding more resistors).

This is very good experience based information, that you can't get out of books. Thanks for the good comments and suggestions.

Hector

IF your resistors are in separate packages, putting them in series will increase both the Voltage rating and the power rating - but you need to be careful about the physical construction of your assemblage, as Voltage ratings of high-Voltage resistors are often determined by the spacing of the terminals. Regarding the derating you need to apply for mismatch, this depends on the number of resistors in the chain.
For a long chain (which allows the worst-case situation), the design value for the total dissipation (based on the nominal resistor value) will be
P_single.N.(R_MIN/R_MAX)²; and if the resistors are nowhere near their dissipation limit, the minimum design design value for the supply Voltage will be V_single.N.R_MIN/R_MAX.

If the resistors are part of an array (i.e in a single package), you would need to know details of the array specification and/or construction to determine whether there is any advantage to be gained from placing them in series.

A final note - it's not a good idea to do this with resistors that fail to short-circuit - but fortunately these are not in the majority.

Good information to know.

Do you know if the breakdown point is an internal break down of the resistor material or is it just an arc over around the exterior from lead to lead?

I ask because from what I've seen, it seems like it's from arc over.

Hector

Most of the times (in case of mormal circumstances) will be an arc, since it is much easier to ionize the air sorrounding than pull out electrons from the valency band of the ceramics.

And this is the reason you have to be careful.

When the resistors arc over, the arc need not reach across resistore, it may as well jump and reach the casings in which these are housed. So you have to be very careful about the physical gap between the leads etc, supports and the housing.

SF6 will help, but in that case you have to very carefully seal it and keep it under pressure (SF6 is an ozone depleting as well as greenhouse gas). Though non toxic.

I agree with antimony on this - with most resistor types it's usually external.
But a couple of counter-examples: spiral-cut resistors in damp conditions can have long-term high-Voltage failure due to internal dendrite growth, and hermetic resistors can have internal spacing that is smaller than the external one. In neither case might you see or measure effects that could cause sudden failure after a while; so you should never exceed the specified working conditions of any individual resistor (which is what my calculations allow) - unless you know a great deal about the construction and physics of the resistor.

And I enlarge on his repetition - if you have multiple resistors in series, they almost certainly need to be arranged in a straight line, so a multi-resistor solution will take more space than a siongle one would.

Why dont youi take a photo to show us for more details?

Reading above discuss. Im sure most of guys havnt done it by themself, only talk about book.

your voltage is not so higher only tens kv. neednt insulated by sf6. exposing it in the air can do the trick. dont worry about it.

I know the Spellman. if you look at hteir products, you will find, even in 140kv situation, their sample resistors still were in the air.

I can guess you are calculating your sample resistor for hv sample. if yoiu hvent deat with hv, be careful. check out kinds of parameters to be sure theyare safe. hoever the calculation is very simple, most of above discuss are right at it.

most of arc is caused by technological, not by devices itself. if yoiu still have problem pm. practise is important.

what circuit mode are you using? flyback or direct boost?

Yes; but with some considerations:

1. Kept in straight Stick-like

2. If mounted on a PCB all the junctions be well isolated by removing copper around.

And on double-sided-PCB on the other side also. As normally component mounting side is having the 2nd side, keep resistors mounted by raising reasonable height from copper.

Note :

For increasing the WV of Diodes usually a small capacitor of WV more than the diode is connected in parallel to each diode.

PS In case anyone is tempted to try the capacitors in series: that works fine for AC. But for DC any mismatches between the leakage of the capacitors (absolutely never guaranteed to be equal - just less than some amount) will eventually cause the voltage distribution on the capacitors to become extremely uneven. If this is ever required, a parallel resistor chain is essential to even out the Voltages. (Capacitors are OK for Voltage multipliers because the diode charging keeps things even).