Design of a transistor flip flop data storage

To make a "data storage circuit" known as a flip flop, it begins with two inverters, coupled back to back, so the output of one feeds the input of the other, and vise versa.

This is a tutorial of how to design a very basic digital circuit for storing data known as a flip flop.

This tutorial will be put together from a hobbyist point of view, meaning the design of these circuits are from the very basic, NON professional way of designing circuits,

However still able to get very satisfactory results in the performance of these circuits.

Starting from scratch, step by step this will be for educational purposes of learning basic electronics, using discrete components.


Design:

First securing data (through measurement, or data sheets), find the forward voltage Vf, and current that will be needed to light up an LED.

I decided to use 10mA, with a Vf of 2V.
Also decided to design this with 5V. so a 9v. battery can run this.

VCC = 5v.
LED Vf = 2v. @ If = 10mA.

Now the series resitor (RS) needed to produce 10mA, across a 2v. drop of the LED is
RS =[ (VCC - Vf led) / If led] = [(5v. - 2v.) / 10mA] = 300 ohms

Now that is the establishment of the brightness of the LED in it's ON state.


1572204874773.png


The next part to design is how to turn the LED off.

A transistor (Q1) will be used to short out the LED so as to turn it off.

Assuming VCE sat. of 300mV, then the collector current will be
IC = [(VCC - VCEsat) / RS] = [(5v. - 0.3v.) / 300 ohms] ~= 16mA.

Therefore the base current (IB) needs to be 1/10th of this to put the transistor into saturation.
IB = 1.6mA.

The base resitor (RB1) needed to allow 1.6mA. to flow will be calculated on a first order basis, (ballpark calculation),also assuming base emitter voltage (Vbe) to be 0.7v. as RB = [(VCC - Vbe) / IB] = [(5v. - 0.7v.) / 1.6mA] ~= 2.7K ohms.


1572204881407.png


Now as can be seen, there is a lot of current being wasted to keep the LED in it's off condition.

So another circuit needs to be designed to lessen the current usage during this off cycle of the LED's.

SO try to reduce the idle current of 16mA, by a factor of 100, which makes this current (IC1) be now, 160uA.

To attain this new idle current (IC1) for the transistor (Q1), another transistor (Q2) in a emitter follower configuration needs to be employed, to reduce the amount of collector current from the first transistor.

First the RS value is recalculated, and now becomes the collector resistor (RC1) for Q1.
RC1 = [(VCC - VCEsat) / IC1] = [(5v. - 0.3v.) / 160uA] ~= 30K ohms.

1/10th of this is needed at the base to saturate Q1, which is 16uA. = (IB1)
And the base resistor (RB1) needed to saturate Q1 will be [(VCC - Vbe) / IB1] = [(5v. - 0.7v.) / 16uA.] ~= 270K ohms.

Now couple the output of Q1, to the base input of Q2, which is a emitter follower, so as to have the 10mA. of current flow through the LED, but when it is off, the only major current flowing will be around 170uA, through Q1 collector and around 16uA, through the base to draw around 180 to 200 uA. during idle time (LED in off condition).


1572204888750.png


Now by building a second one just like it, then both circuits can be crosscoupled together, to produce a simple flip flop. (RS flip flop)

In order for this to work, I need to take into consideration, that with the collector of Q1 directly connected to the base of Q2, will cause the base voltage of Q2 to rise due to the 30K ohm collector resistor of Q1, and vise versa.

So as to prevent that from happening, the two diodes are used, this arrangement allows the base of each transistor to be isolated from the voltage present at the bottom of the 30K ohm resistors.

But alows electron current flow through the diodes in only one direction, so as to bring the base of each transistor to a low voltage value, according to the voltage present at the collector of the opposing transistor.

These are used as steering diodes, to allow negative voltage to be felt at the base of the coupled transistor, while keeping the positive voltage isolated from it.


1572204894507.png


In this video
http://www.youtube.com/embed/FxBN3EjAUTc"

the jumper wire is connected to the negative supply, and the inputs are being made to the base terminal of each transistor, to cause the LED's to flip and flop back and forth.


"Next I'll work on designing this into a master-slave flip flop.

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