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I am interested in starting an intro electronics project using a IR sensor and am having a little trouble understand the application circuit given on the datasheet of the circuit. The value of resistor they recommend using is far less then the one I calculate I should need.

circuit

datasheet

The supply current given on the datasheet is 0.27mA - 0.45mA and the supply voltage is 2.5V - 5.5V. I intend to run this sensor at 0.35mA with 3.3V and this yields a needed resistance of around 10K ohms; far greater than the 33 - 1K ohms suggested.

As this is my first venture into electronics, could someone help me to understand how using such a low resistor as specified would work when the math says otherwise?


On a slightly related note, from what I understand, the capacitor in the circuit is called a decoupling capacitor and is used to provide the extra current during short periods of high current draw so the voltage supply does not. Because this is such a low current circuit (or is 0.35mA low?), is it fine if I leave the capacitor out or should I include it because the datasheet lists the absolute maximum current as 3mA and the circuit may try to draw that much?

Also, I am not familiar with the component on the far right of the diagram (marked "micro" C). Is that a generic representation of a microchip?

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To start with your last question: yes, it's a microcontroller (not a microchip, that's a confusing word since it's the name of a manufacturer of a.o. microcontrollers).

Unfortunately the supply current is not for you to choose. At 3.3 V it will be typically 0.35 mA, but it can be anything between 0.27 mA and 0.45 mA. And the 10 kΩ you calculated is the equivalent resistor of the receiver's power consumption.

The resistor in the schematic forms a low-pass filter with the capacitor, letting DC and low frequencies pass and filter out the higher frequency noise. That resistor will cause a small voltage drop from the power supply; the higher the current, the higher the voltage drop, due to Ohm's Law. So worst case the current will be 0.45 mA, and then the 1kΩ resistor will cause a 450 mV drop. If you supply 3.3 V then the receiver will have 2.85 V left. This is enough to operate, but you have to take into account that it will only output 2.85 V as a high level, and you have to check that this is enough for the microcontroller.

You're right about the working of a decoupling capacitor, and that's part of its job. Like I said the capacitor is also part of a filter. If you choose 1 kΩ for the resistor, and 100 nF for the capacitor it will have a cutoff frequency of 1600 Hz.

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  • \$\begingroup\$ How do I go about choosing what cuttoff frequency I need for the low-pass filter? I'm assuming after I choose the resistor for the voltage drop acceptable, I can choose the capacitor to give me the frequency I require. \$\endgroup\$ – OSaucey Aug 10 '12 at 0:55
  • \$\begingroup\$ @OSaucey - that's right. The higher the resistor and capacitor values the lower the cutoff frequency: \$F_C =\frac{1}{2 \pi R C}\$. And the lower the cutoff frequency the more noise will be attenuated, up to a point. A 10000 µF should give you a 100000 x lower frequency than the 100 nF, but your power supply will take half a minute to rise to the required level. I wouldn't go much higher than about 10 micro;F, and then you'll need the 100 nF in parallel too. \$\endgroup\$ – stevenvh Aug 10 '12 at 5:48
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\$ \mu C\$ stands for microcontroller, which is what you'll be most likely using to process the data you receive.

As for the current, you're misunderstanding the way the sensor works. It has its own internal impedance and the external resistor is there just as a part of power supply filter. You simply provide the needed voltage to the circuit and the current will be set to appropriate level using the components marked as "Circuit" in the picture.

In some cases when filter is not used such receivers can provide false output signals which will produce extra load for the microcontroller, so it's best to include it unless you have a good specific reason why not to have it.

As for decoupling capacitor, yes you need it. You always need it! Why: Well the device is digital. The 0.35 mA you expect the device to consume is the average current over some time. The moment the device switches its internal state (or to say it a bit differently, produces a digital signal), it will consume huge amounts of current for a very short time. That current has to come from somewhere and the decoupling capacitor is the source.

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