I need to insert a low-value series resistor in the power trace of a PCB to monitor power consumed by a microprocessor IC. I will be using a 50 ohm 1% resistor. At a current of 5 mA, the drop across the resistor will be 250 mV. The processor requires a supply voltage of 3.3V. Question is, is it ok to increase the power supply voltage to 3.55 V to compensate for the resistor drop in order to maintain the proper supply voltage to the processor? If so, is there a general rule of thumb for how much the supply can be increased (e.g. if a larger value resistor were used)? I'm curious if there are any detrimental effects to the processor circuit by doing so. For instance (an extreme case), could I use a 1k resistor (Vdrop = 5V) and increase the power supply to 8.8V?

 

A big problem when using a resistor as a voltage regulator is if the current is not constant. If it drops, all the sudden your micro is seeing the full voltage and will likely be destroyed. Resistors works for LEDs and lightbulbs, but not so well with smart devices.

 

How about a resistor and a zener diode to clamp anything over 3.3 volts? Problem is it may then go below 3.3 volts depending on how much curent the processor is using.

 

So if I understand correctly, the only concern is protecting the processor from over-voltage in case of resistor failure. Is this correct? If so, clamping the voltage using a 3.3V zener sounds like the way to go.

Problem is it may then go below 3.3 volts depending on how much curent the processor is using.

How is the voltage going to drop based on how much current the processor draws? The zener will be shunted to ground so that the voltage will never drop below 3.3V unless the main supply voltage drops below 3.3V.

 

So if I understand correctly, the only concern is protecting the processor from over-voltage in case of resistor failure. Is this correct? If so, clamping the voltage using a Zener sounds like the way to go.

How is the voltage going to drop based on how much current the processor draws? The zener will be shunted to ground so that the voltage will never drop below 3.3V unless the main supply voltage drops below 3.3V.

If you draw more than your assumed curent through the resistor, the resistor could drop more voltage than you expect. For example if you are using the micro outputs to drive leds the drop across the resistor will depend on how many leds you try to light.

 

Right. I understand. Thanks.

 

A 250mV drop to measure current seems a little high. A more typical shunt value of 50mV to 100mV is used, thus having less adverse effect on the circuit operating voltage.

How are you measuring the shunt voltage?

 

A 250mV drop to measure current seems a little high.

Sorry, I'm measuring current fluctuations so the greater the resistor the better.

 

Hi,

A shunt giving a 50mV drop can be used if you amplify this mV drop by means of an Op-amp circuit with pre-set gain to suit your readout requirement.

Just a thought!

 

Sorry, I'm measuring current fluctuations so the greater the resistor the better.

Not necessarily. What circuitry are you using to measure these "fluctuations".

 

5X differential probe with 100k impedance, sense resistor = 51 ohms, current through resistor = 5 mA.

 

What is the best sensitivity of the probe with your scope?

 

What is the best sensitivity of the probe with your scope?

Not sure.

The current setup I'm using isn't ideal. The probe is connected to an external power supply, which is connected via USB to a real time spectrum analyzer (the diff probe and power supply came with the RSA). So I'm basically using the RSA to power the probe and feeding the 50 ohm output of the power supply to an o-scope. It is working, but I'm wondering if my results could be better.