Keep setting fire to discharge load resistor - Wattage limit

[rwooduk]

We kept getting some voltage spikes back into our amplifier. One of the things we put in place is a 10kOhm resistor (discharge load) connected to the output of the amplifier. However, the resistor keeps burning and setting on fire.

The amp company offered a way on checking this...

Measure output voltage from the amplifier. Then calculate V² / R of the resistor and you may find you have dissipated more into R than it is designed for. The selection of resistor wattage for continuous operation should not dissipate more than 50% of it's capacity, lower if possible.

I'm really trying to understand what this guy is saying, I didn't even realize that resistors have a power limit. I can measure the voltage output (although it's a sinusoidal signal so not sure what to do here) and I have the resistor value. So I can calculate the power across the resistor.

My question is where do I check the Wattage limit of my resistor? It's just a general resistor from the electronics lab.

A suggestion was to use a CADDOCK power film resistor which are 15W, 20W, 25W or 30W non inductive resistors with cooling tab, would this stop the thing melting?

Thanks for any advice with this any suggestions would really be appreciated.

 

[BvU]

So I can calculate the power across the resistor.

So what did you measure and what did you calculate ?

it's a sinusoidal signal so not sure what to do here

In theory you take the effective voltage (the DC voltage that generates the same disspation in a resistance ). For a sine, $V_{\rm eff} = {1\over 2}\sqrt 2 V_{amplitude}$. However, many multimeters measure $V_{\rm eff}$.

Normal resistors are 1/4 W or so. Not meant to dissipate power.

This kind of feedback to an amplifier (guitar amplifier or something ?) doesn't help much against voltage spikes. At least not if you have to turn up the volume to compensate for the power loss.

Oh, and if you blow a 10 k$\Omega$ resistor, that means your voltages are considerable ! Be careful.

 

[rwooduk]

So what did you measure and what did you calculate ?
In theory you take the effective voltage (the DC voltage that generates the same disspation in a resistance ). For a sine, $V_{\rm eff} = {1\over 2}\sqrt 2 V_{amplitude}$. However, many multimeters measure $V_{\rm eff}$.

That's tomorrows job, I'm going to attempt measurement and thanks I see what you are saying, take the peak to peak voltage and calculate Vrms, I should have thought to do that, thanks!

Normal resistors are 1/4 W or so. Not meant to dissipate power.

hmm I was thinking about adding more in parallel, I'll see what the power deleivered is and go from there.

This kind of feedback to an amplifier (guitar amplifier or something ?) doesn't help much against voltage spikes. At least not if you have to turn up the volume to compensate for the power loss.

Oh, and if you blow a 10 k$\Omega$ resistor, that means your voltages are considerable ! Be careful.

For your interest, here is the amplifier...

http://www.ad-elektronik.de/AG1006revB.pdf

Yes, indeed, the voltage spikes are very high. We are using transducers, which kept destroying the MOSFET inside our amplifier which can take around 130V. The problem is that transducers act essentially as a capacitor, so when we disconnected the transducer it had charge stored. When we reconnected the transducer to the amplifier a voltage spike was sent through the unit killing the MOSFET inside. The solution for the amplifier was to install a TVS (transient voltage supressor), also we are discharging the transducer with a discharging plug after operation. The discharging load DURING operation is another safety precaution, however it seems the power delivered by the amplifier is too high.

It took us around 3-4 months to figure out that this was the problem and solve it, isn't research amazing lol. At least I'm not getting electricuted by the transducers anymore, the shock was quite painful!

Thanks again for your help and advice!

 

[BvU]

Oh boy, definitely not a guitar amplifier :) and if it can pump 300 W into 50 $\Omega$ your spikes will be hefty alright.
At least no worries about the power loss from this resistor. (130 V)²/10 k$\Omega$ gives 2 W so a 5 W resistor should last a bit longer than a small GP one from the lab. Wonder if you can find one. Else have two 22k $\Omega$ of 2 W in parallel.

@russ_watters : you any experience with bigger power ratings resistor availability ?

 

[Averagesupernova]

Your biggest challenge may be finding a higher wattage resistor that is not inductive.

 

[rbelli1]

Your biggest challenge may be finding a higher wattage resistor that is not inductive.

Digikey to the rescue!

http://www.digikey.com/product-detail/en/bourns-inc/PWR221T-30-1002F/PWR221T-30-1002F-ND/2360938
[PLAIN]http://www.digikey.com/produ.../PWR221T-30-1002F/PWR221T-30-1002F-ND/2360938[/PLAIN]
Among many. This is just the cheapest. You'll want to bolt it to a heat sink to get any significant wattage out of it.

BoB

 

[rwooduk]

Oh boy, definitely not a guitar amplifier :) and if it can pump 300 W into 50 $\Omega$ your spikes will be hefty alright.
At least no worries about the power loss from this resistor. (130 V)²/10 k$\Omega$ gives 2 W so a 5 W resistor should last a bit longer than a small GP one from the lab. Wonder if you can find one. Else have two 22k $\Omega$ of 2 W in parallel.

@russ_watters : you any experience with bigger power ratings resistor availability ?

Yes that's a much better idea as I have to fit them into this box and soldering is not my strong point...

Thanks!

Your biggest challenge may be finding a higher wattage resistor that is not inductive.

Is this because when a resistor is inductive is will tend to block high frequency signals? I'm using between 22 kHz and 1 GHz from the amplifier. Is it this blocking that causes the heating and fire or just too much power running across it? Would I experience heating and fire with a resistor that has a higher power limit but also higher inductance?

Digikey to the rescue!

http://www.digikey.com/product-detail/en/bourns-inc/PWR221T-30-1002F/PWR221T-30-1002F-ND/2360938
Among many. This is just the cheapest. You'll want to bolt it to a heat sink to get any significant wattage out of it.

BoB

Many thanks for the link, just a couple of questions... "You'll want to bolt it to a heat sink to get any significant wattage out of it." I don't understand what this means, if you see the discharge box I made above, wouldn't I just solder it in there?

Also I am curious as to the difference between the resistor above and the one recommended below...

http://www.caddock.com/Online_catalog/Mrktg_Lit/MP9000_Series.pdf

I believe the caddock resistors have some sort of "cooling tab", would this be the same as a heat sink? I'm not familiar with electronics at all.

Thanks again!

 

[rbelli1]

Also I am curious as to the difference between the resistor above and the one recommended below...

Different manufacturers but otherwise the same.

Both need to be attached to a heat sink to have power handling capability. At your power level you could get a thermal silicone pad and bolt it to the case. Use the least amount of wire as possible.

I'm using between 22 kHz and 1 GHz

The impedance of the device will go up with frequency. Using film type resistors helps minimize this effect.

BoB

 

[davenn]

Digikey to the rescue!

http://www.digikey.com/product-detail/en/bourns-inc/PWR221T-30-1002F/PWR221T-30-1002F-ND/2360938

Many thanks for the link, just a couple of questions... "You'll want to bolt it to a heat sink to get any significant wattage out of it." I don't understand what this means, if you see the discharge box I made above, wouldn't I just solder it in there?

no, because unless you bolt it to a heatsink ... eg. your metal case then it won't be able to dissipate as much heat
and if you are really using significant power levels, > 50W, then the case will need a finned heatsink and probably even a cooling fan

if you don't do that, the resistor will fry very quickly

that 30W resistor will most likely be good up to 60 - 70 Watts when only short pulses of power are applied and it is well heatsunkDave

 

[davenn]

Also I am curious as to the difference between the resistor above and the one recommended below...

http://www.caddock.com/Online_catalog/Mrktg_Lit/MP9000_Series.pdf

I believe the caddock resistors have some sort of "cooling tab", would this be the same as a heat sink? I'm not familiar with electronics at all.

there is NO cooling tab. they are just like the other series
a 50 Ohm value in the MP9100 series would be nice 100W when heatsunk. and again when used for short pulses of test power
it would probably handle around 150 W when properly heatsunk

NOTE! ... even that 100W one is only rated at 3.5 W if it isn't heatsunk

 

[davenn]

I'm using between 22 kHz and 1 GHz from the amplifier.

are you ??
I wonder how you are getting 1 GHz out of an amplifier that only goes to 14 MHz ?

 

[rwooduk]

no, because unless you bolt it to a heatsink ... eg. your metal case then it won't be able to dissipate as much heat
and if you are really using significant power levels, > 50W, then the case will need a finned heatsink and probably even a cooling fan

if you don't do that, the resistor will fry very quickly

that 30W resistor will most likely be good up to 60 - 70 Watts when only short pulses of power are applied and it is well heatsunkDave

there is NO cooling tab. they are just like the other series
a 50 Ohm value in the MP9100 series would be nice 100W when heatsunk. and again when used for sort pulses of test power
it would probably handle around 150 W when properly heatsunk

NOTE! ... even that 100W one is only rated at 3.5 W if it isn't heatsunk

Thanks, decided to go with the suggested resistor above, although just for initial testing, so I'll have the following...

http://uk.farnell.com/bourns/pwr221t-30-1002f/resistor-thick-film-10kohm-1-to/dp/2328270
http://uk.rs-online.com/web/p/heatsinks/0263251/

are you ??
I wonder how you are getting 1 GHz out of an amplifier that only goes to 14 MHz ?

Apologies, sorry I meant 22 kHz to 1 MHz, good point!

Thanks for the help and advice!

 

[Nidum]

Why does the resistor have to be 10 k ?

Why not for instance 20 k - that would produce only 25% of the heat from the 10 k one .

In any case I suspect that there are other and better ways of dealing with this problem .

 

[BvU]

Old dutch saying that prevention is better than curing -- though it's attributed to Ben Franklin nowadays .

Any insight where the spikes come from ? Is your frequency characteristic important ? If not, damping the higher frequencies in the output with a parallel capacitor and/or serial inductor ?

[edit] Specs say input impedance is 50 $\Omega$. What does the source look like ?

getting some voltage spikes back into our amplifier

makes it look like the spikes aren't there in the input signal but are caused by some unpleasant positive feedback mechanism that you counter by adding a rather small resistive load.

 

[rwooduk]

Why does the resistor have to be 10 k ?

Why not for instance 20 k - that would produce only 25% of the heat from the 10 k one .

In any case I suspect that there are other and better ways of dealing with this problem .

I was just going to see if 10K was sufficient, no particular reason, but yes perhaps I should have gone with a higher value. Will see how it performs.

Old dutch saying that prevention is better than curing -- though it's attributed to Ben Franklin nowadays .

Any insight where the spikes come from ? Is your frequency characteristic important ? If not, damping the higher frequencies in the output with a parallel capacitor and/or serial inductor ?

[edit] Specs say input impedance is 50 $\Omega$. What does the source look like ?

makes it look like the spikes aren't there in the input signal but are caused by some unpleasant positive feedback mechanism that you counter by adding a rather small resistive load.

To both; yes exactly, the problem is the possibility of feedback of high voltage spikes to the amplifier from the transducer during operation due to the unpredictable nature of piezoelectric transducers (and RF!) which act as a capacitor. In particular the possibility when the amplifier signal is turned off and any stored charge in the transducer can shoot back down the line into the amp. That is the purpose of the resistor to allow it to discharge through it (see simple schematic below).The frequency of the output signal is important because we are looking at the frequency effects on our system.

We have solved the problem of discharge when connecting transducers to the amplifier by simply using a discharge plug. But it's during operation that concerns us and we are trying to put safety mechanisms in place. Hope that makes a little more sense, any further suggestions are more than welcome.

 

[davenn]

Why does the resistor have to be 10 k ?

Why not for instance 20 k - that would produce only 25% of the heat from the 10 k one .

In any case I suspect that there are other and better ways of dealing with this problem .

it doesn't and SHOULDNT be 10k, 20k etc
It SHOULD be 50 Ohms to suit the output impedance of the amplifier

I was just going to see if 10K was sufficient, no particular reason, but yes perhaps I should have gone with a higher value. Will see how it performs.

Again ... you should be using 50 Ohms

[edit] Specs say input impedance is 50 Ohms. What does the source look like ?

It's the output impedance and its load that is the concern and that is also stated as 50 Ohms

Dave

 

[BvU]

@davenn : it would pump up to 300 W into the 50 $\Omega$. Isn't that a bit steep ? And somewhat unnecessary ? From what I remember a piezo wants a voltage source and hardly any current.

What do they mean with "Any load: up to 150 W continuous". Would that be an even smaller resistive load ?

high voltage spikes to the amplifier from the transducer during operation due to the unpredictable nature of piezoelectric transducers (and RF!)

Ah, much clearer ! The transducer whacks something and is whacked in return, which sends spikes back ?
And you want to study something frequency dependent, so of course frequency is important. By frequency characteristic I mean amplification (c.q. output amplitude) as a function of frequency. In other words: if you need the amplitude of the signal to the piezo and have independent means to measure it, then a simple LC filter would be much more effective in keeping the spikes away from the amplifier output.

That's tomorrows job, I'm going to attempt measurement

What came out ?

 

[davenn]

@davenn : it would pump up to 300 W into the 50 Ohms. Isn't that a bit steep ? And somewhat unnecessary ? From what I remember a piezo wants a voltage source and hardly any current.

it's an RF amplifier ... read the datasheet
50 Ohms is the standard impedance for everything other than TV

any and all correct tests should be done into the correct load. it's just a matter of finding resistive elements that can handle the power at that power out
... This is easy and typical comms and amateur radio stuff 300W is only moderate power. It's when you get up to a kilowatt or so that things get really interesting

I personally have never played with 300W or more, ~ 200W max, many of my ham mates have tho

for high power dummy loads multiple high wattage resistors in series/parallel configs to keep the 50 Ohms are used and they are usually immersed in cans of transformer cooling oil. Available from your local friendly power authority.

for the serious high power transmitters, there is often no choice but to use a BIG wire wound resistor. These are paralleled with the
appropriate capacitance value to negate the effects of the inductive reactance of the wire wound resistorDave

 

[Nidum]

The purpose of the 10 kΩ resistor is to provide a leak path for stored charge in the transducer isn't it ? Like the discharge resistor on a CRT ?

 

[rwooduk]

it doesn't and SHOULDNT be 10k, 20k etc
It SHOULD be 50 Ohms to suit the output impedance of the amplifier
Again ... you should be using 50 Ohms

It's the output impedance and its load that is the concern and that is also stated as 50 Ohms

Dave

Thanks for this, I'm still learning! Yes, I see what you mean about matching the impedance of the amplifier, indeed we were having trouble with feedback from the transducers because they were not matched correctly. Just one last question if I may, what is the best way to determine the wattage of the 50 Ohm resistor I need? I'm a little concerned because of the above statement that inductance rises with wattage.

From the amplifier I'll be using LP (power to the load) = 1 - ~50 W, no higher. But it's the high voltage spikes back that could potentially reach ~200V.

Ah, much clearer ! The transducer whacks something and is whacked in return, which sends spikes back ?

Exactly, this is the worry. Although it seems to only send spikes back when the amplifier is off becuase of the discharge from the transducer (capacitor). So the idea is to catch these spikes with the resistor rather than they go into the amplifier and kill it.

And you want to study something frequency dependent, so of course frequency is important. By frequency characteristic I mean amplification (c.q. output amplitude) as a function of frequency. In other words: if you need the amplitude of the signal to the piezo and have independent means to measure it, then a simple LC filter would be much more effective in keeping the spikes away from the amplifier output.

Thanks, yes I took a look at chokes but was concerned about the effect on the signal, but I'll take another look.

What came out ?

Ermm this didn't exactly go to plan, I decided to connect the oscilloscope directly to the amp (bad idea!) and the warning light came on as the amp needs a load present in order to function. So I'm going to have to do it this way...

So I've ordered the resistors and another T piece, just have to wait for them to arrive!

The purpose of the 10 kΩ resistor is to provide a leak path for stored charge in the transducer isn't it ? Like the discharge resistor on a CRT ?

Yes, exactly! The problem is the piezo can discharge high voltages (especially when the amplifier is switched off) that have destroyed the MOSFET in our amp.

 

[rwooduk]

Why does the resistor have to be 10 k ?

Why not for instance 20 k - that would produce only 25% of the heat from the 10 k one .

In any case I suspect that there are other and better ways of dealing with this problem .

Update, I understand this now. Basically the resistor is a high value so it doesn't drain anything from the amplifier and the signal goes straight to the transducer, the purpose is to discharge any static in the line. Therefore I need a bigger resistor, one that will not burn due to high powers from the amp.