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Commercial circuit, 1210 resistor dissipating 1 Watt

After replacing a defective 9V regulator, an output Mosfet a transistor, and a 100 resistor that looked like it had been hot, I have a working FM transmitter. But then I noticed the 100 ohm 1210 smd resistor running hot. The same 100 resistor that I had previously replaced. The resistor has 8.7 Volts across 100 ohms or 0.76 Watts. (a 1210 smd is a 1/2 watt device) Before I even started the repair I bought another FM transmitter of the same model. (The model does have same updates to the circuitry. But I'll leave that for later discussion.) I got out the new FM transmitter and measured the voltage across the 100 resistor (btw, it also feels very hot) the voltage is 9.8V.

9.8V across 100 ohms is 0.95 Watts. Voltage measured with 10Meg input DVM.

The original unit worked for about 4 years averaging 4 hrs a day. So the questions! Do Leave it as is? Replace the 1210 smd with a 1 watt part? 2 Watt part?

The 100 ohm is a voltage dropping resistor to the output Mosfet when the low power mode is selected. I always run in low power mode, cause the resistor to dissipate high power. Here's the updated circuit.

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Thanks, Mikek

Reply to
amdx
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The average buyer of a cheap FM transmitter probably uses only high power :-)

Reply to
Rob

Hi Rob, For some reason, I don't see anything in your post and I can't respond to it. Could you post again? Thanks, Mikek

Reply to
amdx

It's bypassed with a cap, so only sets a DC current. I don't see any reason you couldn't replace it with some higher power R.

George H.

Reply to
George Herold

A 1206, or even an 0805, can safely dissipate a watt if the end caps are heat sunk to decent copper pours. If there's just small pads and skinny traces, the heat can't get out.

What do the pads look like?

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

Gosh, that's a bad design.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

What in particular sticks out as bad to you> The whole thing is not a good answer.

Reply to
amdx

I don't see much copper there, but the resistor lasted 4 years, and still measured within spec when I changed it.

Resistor in question is labeled 1000.

Mikek

Reply to
amdx

Model.jpg?dl=0

Resistors need to be derated for temperature.

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So you need to measure the temperature and check the datasheet. One old fas hion test for semiconductors was the spit test. You wet the part and if it doesn't boil away immediately it was below 100C and you were probably okay. This test was okay for debugging and deciding that a part was not the prob lem, but for design you really need to measure the temperature.

A higher wattage resistor won't run cooler if it uses the same package, but it might tolerate the higher temperatures better. To lower the temperature you need more heatsinking from a bigger part or bigger pads or an actual h eatsink.

Reply to
Wanderer

If I replace the part, I would use a 1 Watt through hole part. Mikek

Reply to
amdx

The biasing for the P-channel MOSFET doesn't make a lick of sense, and the biasing for Q3 doesn't make a lick of sense, either!

If the circuit is drawn correctly it looks like the designer is using the signal voltage itself to set both the quiescent current through Q3 and the bias on the gate of the output MOSFET. Unless "OSC_INPUT" is pulled down whenever the device isn't transmitting Q3 will short the power supply straight to ground through the 100 ohm emitter resistor.

The reason the 100 ohm resistor is getting hot and Q3 isn't during "normal" operation is probably that the collector of Q3 is swinging above the PS rail as "OSC_INPUT" is shutting Q3 down, and shoving gobs of current straight to ground thru the 100 ohms and the capacitor next to the 2.2k resistor. Which seems like madness, they've basically built a shitty switching converter with a 100 ohm load

Reply to
bitrex

Beta dependencies, for one. And open-loop biasing of the output fet.

Looks like a fiddled-until-one-worked circuit.

--

John Larkin         Highland Technology, Inc 
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jlarkin att highlandtechnology dott com 
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Reply to
John Larkin

It would make slightly more sense if the 100 ohm were in series with the power supply rail and the 8.2k/2.2k were a voltage divider from the supply rail to the MOSFET gate and then ground - is OP sure he got the layout correct?

Reply to
bitrex

I don't think it can possibly be drawn correctly.

Reply to
bitrex

If the correct layout is as stated above then the reason the 100 ohm is getting hot is a shorted bypass cap across the 2.2k.

Reply to
bitrex

Replace with a pair of 200R 1/2W 1210s....?

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Reply to
TTman

Unit is always in transmit mode.

Sorry, I should have made it clear which 100 ohm resistor is getting hot. Refering back to the schematic in my OP, the hot 100 ohm resistor is the one at the top of the page connected to 12.2V and 2.6V on the other end. Thanks, Mikek

Reply to
amdx

Finally can read your post.

Yes, I agree most people probably use high power, I'm trying to stay within FCC rules. I run a couple of attenuators and a low pass filter.

Mikek

Reply to
amdx

On Friday, August 18, 2017 at 11:12:05 AM UTC-7, bitrex wrote: ...

There is a potential divider of the 8.2k and 2.2k to provide the gate voltage for the PA (I'm skeptical that it is really an P-channel device - I expect it really is N-channel).

Q3 is capacitatively coupled to the gate - so the bias conditions of Q3 don't affect it's DC conditions.

It is not a good arrangement as it depends upon the Hfe of Q3 but as the emitter voltage rises the base current falls so it will find an equilibrium.

Huh - Q3 is just a badly biased class A amplifier. (assuming it is not being overdriven).

The 100 ohm resistor getting hot is the one at the top of the schematic that is used to drop the voltage to the PA in low power mode.

kevin

Reply to
kevin93

It seems very bizarre to bias a P channel output device from a common emitter NPN collector like that; guess that's why I was assuming the lower 100 was the problem. I didn't really see how it could be dissipating that much power unless it were actually in the supply line if that one was the issue

Reply to
bitrex

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