Question

INDICATE HOW AFFECTS OR IS AFFECTED THE +5 VDC (INFORMATION IS PROVIDED IN THE LINK)

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3.2. DESIGN OF THE ELECTRONICS

According to the electrical characteristics of the PT202C, a maximum collector current (Ic) of \(9 \mathrm{~mA}\) and a bias of 5 Vdc will be used. The value of the resistor is

calculated according to the ohm's law \((R=550 \Omega\) ). This resistance value is made of a resistor of \(220 \Omega\) and a potentiometer set to \(300 \Omega\). The configuration of the photo transistor can be seen in Fig. 3 .

The \(220 \Omega\) resistor controls the maximum \(\mathrm{I}_{\mathrm{C}}\) which flows through the photo transistor. The \(5 \mathrm{k} \Omega\) potentiometer is to set this current.

This configuration works as a current-voltage converter where \(\mathrm{l}_{c}\) is proportional to voltage variations between the emitter and ground. Thus, when incident radiation modifies le, voltage will vary corresponding to the radiation received by the photo transistor.

The output of the configuration provides an analogue value in the range of \(0-5 \mathrm{Vdc}\) which can be send to an analogue-digital conversion stage. This conversion is to generate an irradiance scale in \(\mathrm{W} / \mathrm{m}^{2}\).

However, the response of the PT202C photo transistor to irradiance is limited by its intrinsic characteristics; it covers from \(0.5\) to \(5 \mathrm{~mW} / \mathrm{cm}^{2}\). This means that the maximum irradiance level obtained by this photo transistor will be \(50 \mathrm{~W} / \mathrm{m}^{2}\) at \(5 \mathrm{Vdc}\). This value is dramatically lower than the maximum level of irradiance measured in the Northern territory of Chile, which is about \(1350 \mathrm{~W} / \mathrm{m}^{2}\). In this way, for a maximum irradiance value of \(40 \mathrm{~W} / \mathrm{m}^{2}\), on the photo transistor's sensitive surface, it is necessary an attenuation from \(1350 \mathrm{~W} / \mathrm{m}^{2}\) to the desired value. This attenuation is calculated according to Eq. (1).

$$ x=\frac{40 * 100}{1350}=2.96=97.03 \% $$

Hence, in order to avoid saturation in the photo transistor, it is necessary to place an attenuator element over the sensor.

Answer 1

No, the +5V is not affected.

Given The irradiance is limited by its characteristics, it covers 0.5 to 5mW/cm². Also mentioned that maximum irradiance obtained at 50W/m² at 5Volts. The maximum irradiance measured in Northern territory of Chile, is about 1350W/m² which is almost 27 times of previous value. We cannot apply a 27times of 5Volts (i.e., 135 volts) to an Photo transistor, which damages the transistor. Instead we have to design an additional circuits or Design a suitable resistances.

They are using an attenuator circuit in Northern territory of Chile to change 1350W/m² to 40W/m², here also there is no affect on 5Volts. Just design a resistor and potentiometer values to get a desired 40W/m².