Electronics > RF, Microwave, Ham Radio

Frequence response of through hole resistor vs power rating? (attenuators)

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coppercone2:
I have a attenuator that is busted and it calls for a 1/8 watt resistor. It looks like it was repaired before and there was not a 1/8 watt resistor in there in the first place.

I am waiting for other parts to fix this one (HP461), but I am wondering if anyone has any information on how big a difference there is for a 1/8th watt and a 1/4 watt resistor.

The resistor is 499 Ohms, so its probobly not the best thing to try to test on a VNA, and I would also need to buy it.

Are these small parts critical? The amplifier I have Fmax is 160MHz

I thought there might be a table some where that exists with this information.

I had to bend the leads out quite a bit to make it fit, it has a notch for a real 1/8th watt resistor that the previous repair man did not bother using.

FYI the attenuator had a drifted resistor, a cracked resistor that measured ok (possibly my noise source that was ruining the circuit) and a 1pF capacitor that read 500ohms.

coppercone2:
Also does this have to do with power level or just physical size?

For instance I have a 0.4W resistors, that are some how in a 1/6th or 1/8th watt size package. Not sure how that works (way smaller then a 1/4 watt), from a real manufacturer (they advertise as reduced size package).

Are these 0.4W resistors going to be the same for RF as the 1/8th watt resistors? I have the 0.4 reduced size ones and I also have a 1/4 watt normal size ones (but the reduced size ones with the high power level are not the correct value).

Do you think I can test the reduced size 0.4W resistors vs the 1/4 watt regular resistors to make VNA plots and basically know what will happen when I get a 1/8th watt resistor in traditional size? (since if I buy a 1/8th watt resistor I will likely buy a 1/8 watt resistor rather then try to find a 'increased rating' resistor, which I just happen to have).


Is there some kind of other magic going on with the reduced size higher power ones? I don't know how they make it like 3x smaller then a 1/4 watt but then it has 40% more power rating).

mag_therm:
Hi Cc,
Do you have enough  info about , or by tracing circuit of, the attenuator  to be able to calculate the power loss in that resistor at max output?

coppercone2:
the power loss is well handled by the 1/8th resistor, its a LNA with 50 ohm input impedance, the input voltage limit is low, the maximum output power is ~7dBm, so the input should not really get above -30dBm since it has 40dB gain (and the attenuator is front end and cuts input in half, it does not change gain)

its a LNA input attenuator not output attenuator

so the non attenuated maximum input is -30dbm to get the 7dbm output power maximum, if you swap in the 20dbm attenuator it should be -10dbm input maximum, since -10dbm goes to -30dbm and that is gained to ~10dbm, rounding the numbers hard. that should be 100uW input for -10dbm I think.

the manual specifies that maximum allowed input is 1Vrms or 2V pulse but the output max is 0.5Vrms so yeah

https://xdevs.com/doc/HP_Agilent_Keysight/HP%20461A%20462A%20Operating%20&%20Service.pdf

At 1Vrms input the dissipation in a 600 ohm resistor is like 1mW, 3.5mW for the DC signal of +2V if you figure that the attenuator resistor and the load resistor are in series to divide 2/550

mag_therm:
Then rating isn't critical,  you could just try for a physical size close to the original, so leads are approx same.
That is about all you can do, isn't it?

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