Protection resistor at VDD viable?

Thread Starter

fieryfire

Joined Feb 14, 2017
150
Hello All,

I have an oscillator that consumes about 6mA from a 3.3V. I want to add a protection resistor of 500ohm in between 3.3V and VDD. Logically 3.3/500=6.6mA. Will the circuit still work? since it should theoretically be able to push a max of 6.6mA. and the requirement of the oscillator being 6mA? I need this resistor because otherwise it wouldnt pass a certification

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Ian0

Joined Aug 7, 2020
3,754
Hello All,

I have an oscillator that consumes about 6mA from a 3.3V. I want to add a protection resistor of 500ohm in between 3.3V and VDD. Logically 3.3/500=6.6mA. Will the circuit still work?
No. Because 6.6mA through 500Ω will cause a voltage drop of 3.3V. So the supply to the oscillator will be 3.3V less than Vdd, which is. . . . Nothing.
Why do you think it needs the resistor?
 

Thread Starter

fieryfire

Joined Feb 14, 2017
150
To limit the current. isnt that how its done? because the resistor is there to only pass a maximum of 6.6mA with 3.3V. im confused why it wouldnt work.
 

Thread Starter

fieryfire

Joined Feb 14, 2017
150
i mean i dont see why the 500ohm is unlike a current limiting resistor, in case of an event of a short. In order to limit excess current to drop on the oscillator. i have the 500ohm to only limit to a maximum of 6.6mA. When the requirement of the oscillator is only 6mA. Please help me to understand this
 

Ian0

Joined Aug 7, 2020
3,754
To limit the current. isnt that how its done? because the resistor is there to only pass a maximum of 6.6mA with 3.3V. im confused why it wouldnt work.
The supply is 3.3V.
The voltage across the resistor plus the voltage across the IC must add up to 3.3V.
if there is 3.3V across the resistor, how much is there across the IC?
 

Reloadron

Joined Jan 15, 2015
6,204
i mean i dont see why the 500ohm is unlike a current limiting resistor, in case of an event of a short. In order to limit excess current to drop on the oscillator. i have the 500ohm to only limit to a maximum of 6.6mA. When the requirement of the oscillator is only 6mA. Please help me to understand this
The 500 Ohms is a current limiting resistor just like we place a resistor in series with an LED. However, that 500 Ohms will have a voltage drop across it. Since your supply is 3.3 Volts with a current you mention as Ian pointed out you have nothing at the load.
No. Because 6.6mA through 500Ω will cause a voltage drop of 3.3V. So the supply to the oscillator will be 3.3V less than Vdd, which is. . . . Nothing.
Why do you think it needs the resistor?
<EDIT> I see Ian replied. </EDIT> :)

Ron
 

Thread Starter

fieryfire

Joined Feb 14, 2017
150
Hmm i see, so you found the resistance of the oscillator is about 550ohm, so if i had attachedthe 500 ohm, then the drop across the oscillator would be 1.72V. I see
 

Thread Starter

fieryfire

Joined Feb 14, 2017
150
What certification? I’ve read a lot of standards in my career, and none comes to mind.
well its the ATEX certification. the RTHJA of this component is about 400C/W. And the maximum allowable in my design is about 133C/W. But if i am able to limit the current i am able to increase the maximum allowable rthja ( the thermal resistance). so to do that i need to find a compensation here, which is quite hard. I realize that i am such a dummy
 

Thread Starter

fieryfire

Joined Feb 14, 2017
150
How much short-circuit current does this "certification" allow?
How much short-circuit current does this "certification" allow?
depends on the resistor, the limitation is on the thermal resistance of the oscillator. i can increase this limit by reducing the current. but i cannot reduce the current as much to balance between a good allowable RTHJA value and enough voltage and current consumption for this oscillator
 

crutschow

Joined Mar 14, 2008
28,170
If I understand the oscillator spec correctly, it will operate down to 2.5V, so you could put a resistor in series that would drop about a half volt or so (say 75Ω) at the typical operating current, and the circuit should still operate normally.
The short circuit power dissipated in the resistor would then be about 150mW.

Even a smaller 50Ω resistor, dropping about 330mV for normal operation, would only dissipate about 218mV for a short circuit.

Would either of those meet your requirements?

If not, you may have to go with a more elaborate current-limit circuit, which is likely not simple for such low voltage operation.
The simplest I know uses a matched, dual PNP transistor, plus a logic-level MOSFET.
 
Last edited:

BobTPH

Joined Jun 5, 2013
4,032
I have to think you are misinterpreting something in the requirements. Specifying a single thermal resistance regardless of the power being dissipated makes no sense.

Bob
 

Thread Starter

fieryfire

Joined Feb 14, 2017
150
If all if the power of your oscillator is dissipated in that one component, I calculate that the junction temperature rises by 8C. How is that a problem?

Bob
I know what you mean but let me explain. its to do with the ATEX Qualification from standard 60079-11. The Uo=8V, Io=3.11A. But i have a fuse of about 0.1A. And then the qualification throws in all kinds of safety rules if the PWA is placed in zone 0. So even if the component is fed 3.3V. The atex qualifiers would take 8V an 0.1A for the power measurement. And since the RTHja of the component is too high at about 421C/W or so. te only way is to have a sufficient enough resistor to with high enough power working at 2/3 its rating and again by 1/3 since its be qualified at 120C. that would mean i would need an unusually large resistor capable of very high power dissipation. either ill need to skip this part and use the internal oscillator of my mcu instead.
 
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