I believe circuit and PCB design is an important aspect of realizing prototypes and communication with other engineers, and therefore it is a good skill to have (at least the basic) knowledge of and experience in. Actually ordering and soldering the PCB was of great value. Next time, for instance, a ground-plane is necessary when working with RFID technology as it is easily influenced and easily influences the rest of the circuit due to peak current.

Initially I opted for integrating a clap sensor into a lamp. More specifically into a socket for a lamp. This would require me to ‘play around’ with the 230V power supply in the PCB design. As she was strongly advised not to, by experts, she decided to consider other options; finding an even better one: controlling multiple existing sockets with a sensor.

I designed and ordered the PCB and made calculations regarding its use of current. After soldering the ordered PCB’s I needed to add to components to make the connection to the RFID chip less sensitive for in circuit interference. The RFID technology requires peak current and, with this current, influences the LEDs in the current circuit. The breadboard already filtered out these peak currents due to its own standard resistance; therefore this was not noted before. The second addition, to solve this same problem, was to make the sensor even less influenceable by non-clapping sounds: another capacitor which influences the ‘cut off’ frequency.

It was calculated that a 10000mAh powerbank could power the sensor, taking into account heat loss and quality loss of the powerbank, for ±14 days which makes the sensor very energy effective. Nevertheless it was not taken into account that a powerbank requires a minimum current use for it to stay on. Thus the sensor cannot make use of a every powerbank as it requires too little power.

Calculations

Led light (red) 4.7 mA (because I aim to achieve little current usage, 20mA is normal)

Forward voltage red led = 1.8 V at 5V (looked up in datasheet)

I = V/R with a 680 Ω resistor and a 5V power source this results in) I = (5-1.8) / 680 = 0,0047

Microphone circuit, in a microphone I is negligible as it has extremely small changes. R= 100R + 10K + 10K = 100 + 10000 + 100000 = 20100Ω

I = 5 / 20100 = 0,0002488 = 0,25mA

ATTINY 85 power consumption at 5V, 20MHz (in the most unfavorable situation) taken from datasheet = ± 11,5mA

Power consumption LM358 from datasheet: ±28nA, this is such a small amount that it is negligible.

The RF module, FS1000A operates on a 5V current. The output power is 22mA. In theory thee is no current when in rest but earlier people experienced a rest current of max. 4mA. The same as for the receiver. (Taken from Arduino forum) To be certain 4mA is used in the total calculation.

The capacitors only draw a little current once, to charge. This is negligible in the total amount.

In rest the sensor draws: I = 4mA + 11,5mA + 0,25mA = 15,75mA.

When all sockets are turned on there has 3 times been a peek of 33,75mA after which a led turns on and a varied current between 15,75 and 29,85 mA starts. (15,75+(3*4,7))

Let’s say the sensor is on half of the time and 1 or more devices are on (not always all are on). 1/2(3*4.7) = 7.05mA.

Half of the time all devices are off (during the night for instance) so 15,75mA + . 1/2(7.05)mA = 19,275mA as typical consumption.

A 10000mAh powerbank is an often occurring powerbank. With such powerbank the system works approximately 10000/19,275 = 518,8 = ±518 hours = 21 days.

Therefore it is not necessary to continuously plug the sensor into a socket. Even when 20% of the power is lost to heath and the powerbank drops ±10% in quality the sensor still can be employed using the powerbank for ±14 days.