John's Ramblings » Electronics

Why the diode in the circuit?

john — Sat, 11 Jul 2009 19:45:44 +0000

When the switch opens the falling edge will cause a negative pulse on the 4013 input. The diode removes that negative pulse.

Input to 4013 without diode – one pulse (on/off) – zero volts is the centre line. The signal goes negative to minus 1.44 volts. Positive pulse comes from direct connection to +3 volts, negative pulse is smaller because capacitor has to discharge through 1M resistor.

Input to 4013 with the diode – the negative signal is almost totally removed. The signal goes negative to minus 0.35 volts.

Final gas meter to current cost circuit

john — Fri, 10 Jul 2009 13:26:31 +0000

Here is the final circuit I will use, the resistor and capacitor values give a time constant of about 0.5 sec – I’m sure many other values would work but I have these ones in my box.
I have added a pull down resistor on the reset line just to make sure it is low unless deliberately raised high. I will build this over the weekend.

Energy tracking part 2

john — Wed, 08 Jul 2009 19:53:57 +0000

@MartinDix from CurrentCost kindly sent me a dev board so I could interface other feeds to the CurrentCost meter. After some experimentation this is the circuit I intend to use. I have not yet fully finalised values of components but it seems to work pretty well on the breadboard. Note my dev board is version 1.1, the positive connection to ch1 is connected to positive power line.
The idea of the circuit is that a rising edge (start of pulse) is latched until it has been transmitted to the meter.