One-bit memory circuit using transistors

Question

I'm trying to redesign this flip-flop circuit on the picture so I can use it to store one bit of memory. This circuit on the picture is operated with pushbuttons, but I want to operate it via a ESP8266 digital pis.

I will use the circuit to save the state of a PIR sensor which I will read later with an ESP8266 (ESP is in deepsleep to conserve energy).

Circuit initially has a low state (0) and should change to a high state if it recives a 3.3 V signal from the PIR.

There should also be an option to set it to a low state (erase data) via a signal from an ESP8266 pin.

I know this can be done with a 555 timer, but I'm looking for a NPN transistor version since I have a supply of BC547 transistors.

The circuit will allways be powered with 5 V or 3.3 V so memory loss is not a problem.

Updates from comments:

• ESP is in deep-sleep and turns on every hour just to check if PIR was activated.

• ESP deep sleep power consumption is acceptable for me.

• Waking up ESP on hardware interrupt will happen too often since it will turn on every time PIR is triggered.

• I'm using HC SR501 PIR sensor.

Answer 1

Have a go with this:

^{simulate this circuit – Schematic created using CircuitLab}

The "data" output is Q. Its state will be indeterminate upon power on, due to the circuit's symmetry. Use the microcontroller to explicitly clear Q upon power-on as I will explain below.

S and R are "set" and "reset" inputs. A short positive pulse on S will cause Q to go high and stay there, even after the 'S' signal has returned low. Similarly, a short positive pulse on R will cause Q to go low and stay there.

Here's a plot of this happening:

I have replaced all resistors and LEDs with values which will be much less power-hungry, and still work. The only other thing I've done is replace the two switches with transistors that perform the exact same function in response to digital signals at their bases.

If I have understood you correctly, you should connect output Q to a microcontroller input, and input R to a microcontroller output. Then you may program the microcontroller to provide a momentary pulse on R in order to bring Q low.

Connect S to whatever source you like; a short pulse there will bring Q high, where it will stay until R input receives a reset pulse from the microcontroller.

The whole circuit draws about 500μA (0.5mA) from the 5V supply.

Alternative with fewer parts

It's safe to short circuit Q to ground (or NQ), since there's nothing but a weak pull-up resistor holding it high. That means this circuit lends itself well to a sort of diode-AND arrangement:

^{simulate this circuit}

_{Note: I've offset two of the plots by 6V and 12V, so they all appear stacked one above the other. In reality all signals are digital 0V or 5V}

The active level of the set and reset inputs is now low, and they are called NS and NR ("N" for "not"). When either NS (or NR) is brought momentarily low, its respective diode becomes forward biased, dragging Q (or NQ) with it, with sheer brute force. This causes the flip-flop to flip, or flop, if necessary, to adopt the new state. When the NS (or NR) input is high, its diode is reverse biased, isolating that high potential from Q (or NQ).

Answer 2

Since your main issue is to conserve energy, and you only want to see if PIR was going high during your sleep, you can use a edge detection coupled with sample and hold.

You can see that the capacitor goes high when the PIR is going high. Then as the GPIO is in INPUT mode, you can read the value.

Then to reset the edge detector, you put briefly put the GPIO into OUTPUT mode and pull LOW.

If you revert back to INPUT mode, and go to sleep again, you're ready to wait for another detection.

The whole setup doesn't use much current. Only to charge/discharge the condensator and then it takes a couple of nA of leak.

The 3.6V Zener is used to clamp the voltage to safe levels for the MCU, adjust to your needs. You can put it eventually after the 1k resistor if the input is too high.

Additional info not directly relevant to your question:

You can even trigger an interrupt on that pin to wake the MCU up upon PIR going high to be able not to miss edges. If you want to know "how much time", you can use a discrete edge counter with several CMOS IC to be able to avoid waking up to MCU.