Fun with NOR Gates

See A Vintage Binary Electronic Adder Project for some background.

Another work-in-progress - this is something of a learning exercise for me! My own graphics & photos are to be added.

Non-Latching Switch/Invertor & NOR Gate

I have built and tested one of these successfully, based on the circuit above, but with:

• a +5V power supply

• an NPN transistor (BC548)

• an LED between point Q and ground (0 V) to indicate the output state.

Input resistors at left are 10k ohms, the resistor at upper right, between the collector/output and supply, is 1k ohm.

First consider the case where there is just one input, X, say.

When no voltage is applied to the base of the transistor (via a resistor), the LED is lit. In this scenario, I have measured the voltage at the collector as 1.84* V, indicating a drop of ~3 V across the 1k ohm resistor and a current of ~ 3 mA flowing through it.

(*This is about equal to the forward voltage of the LED, the minimum potential difference needed to light it, according to my multi-function tester which gives a value of ~1.86 V for this quantity).

Incidentally, if I temporarily remove the LED, with no base voltage applied to the transistor, the voltage at the collector becomes 4.95 V, which is the same as the measurement of the power supply, so there is no voltage drop across the 1k ohm resistor, no current flowing through it and hence none through the transistor either, showing that the transistor can be considered to be OFF here. In fact, the presence of the transistor makes no difference when the LED is on and could even be removed in this case! So, with the LED back in place, we can safely say that all the current flowing through the 1k ohm resistor flows through the LED.

If a voltage is now applied at X, current no longer flows through the LED and it is unlit. In this scenario, I have measured the voltage at the collector as 0.05 V,  as near as dammit to 0. This means a drop of ~5 V (actually measured as 4.89 V, as opposed to the 'full' 4.95V) across the 1k ohm resistor and a corresponding current of ~ 5 mA flowing through it. (In addition, there is a voltage drop, measured as 4.23 V across the 10k ohm input resistor, with a small current of around 0.4 mA). Where does the main current now go if not through the LED? It can only be flowing through the transistor via the collector and then to ground via the emitter. (The smaller current from the base also goes to ground via the emitter). So, in this case, we can consider the transistor to be ON (i.e. conducting, while the LED is OFF)  In fact, rather than using a transistor, we could just have connected the top of the LED to ground for the same outcome!

(Incidentally, with a 460 ohm resistor replacing the 1k ohm one, collector voltages are almost unchanged at 1.89 V and 0.07 V respectively. In the former case, the current through the resistor (and through the LED also) is now around ~6 mA. In the latter, the current through it to ground is now ~10 mA. When on, the LED appears brighter, even though the potential difference (PD) across it is about the same, but presumably because of increased current through it. So, a brighter LED here comes at a greater cost (higher current = higher power), even when it is off, as in this case the current flows to ground. The LED was still lit when using a 10k ohm resistor in place of the 1k ohm one. The voltage drop across it was 3.23V, indicating a collector voltage of 1.72 V and a current of only ~ 0.3 mA. If I do  get around to making a binary adder, then I will go with this higher value resistor to reduce power consumption.  Also, we can deduce that the voltage at the collector when the LED is on is largely determined by the LED itself).

We can consider this circuit with the one input, a non-latching switch, for if we remove the input X, the LED turns back on. At first glance it does not look very useful, given when the LED is off, it is consuming more power than when it is on! However:

• the circuit negates the logical truth of the input, so it can be seen as an inverting or NOT gate

• now consider the same circuit with two inputs, X and Y, say, to the base of the transistor via resistors. Let's also consider the output as being live/true if neither X nor Y is live/true. This is now a NOR gate. An online search for transistor-based NOR gates will come up with circuits having two transistors, one for each input. I'm focusing on the one-transistor design here, because that is the one that was used in the binary adder project.

A word or two of warning:

I've managed to blow 3 LEDs using a multimeter with this circuit! The first was when I was trying to measure the PD between transistor collector and ground. The circuit is quite crowded and perhaps I accidentally touched the wrong wire somewhere nearby. The other times, I was trying to measure current through the LED and then through the 1k/460 ohm resistor. There were no flashes, pops or smoke - the LEDs simply stopped working. I have plenty of others, so this was not a big deal, just a lesson learned.

Set-Reset Latch

This circuit consists of two NOR gates, the output of each being fed into the input of the other. It also has an input (not shown above) to each NOR gate. When running with no active inputs, the circuit will be in one of two states and hence can used to store a piece or one bit of binary information. As such, only one NOR gate needs to drive an LED to show the state of the entire circuit.  See here for example for more information. Note that the inputs cannot both be high at the same time.

Toggle Flip-Flop/Scale-of-Two circuit

A diode pulse routing network has been added to the S-R latch to ensure that each input pulse is sent to the required NOR gate.

I understand the diodes prevent current flowing between the bases of the transistors and hence interfering with the NOR gates and also prevent corruption of the input signal. Somehow the capacitors ensure the input goes to where it is needed.

Output is taken from one of the NOR gates, i.e. from one of the collectors and is at half the rate of the input. By connecting the output of one circuit to the input of another, we can chain these together to form a binary counter.

I'm using those tiny ceramic capacitors with thin leads and it has been a challenge to get good connections on a breadboard with these. I started with values of ~220 pF and with good connections they work fine, even though the Electronic Engineering article advises a minimum of 0.002uF, i.e. 2000 pF and the other material advises 0.01 uF, i.e. 10000 pF, although these are for a higher voltage supply. Bearing this in mind, I switched to ~2200 pF capacitors.

Testing with a momentary switched input, I found that with two LEDs (i.e. one per NOR gate): toggling mostly works as expected, but is not perfect, presumably because of switch bounce. However, with one LED, if the LED is lit when the momentary switch is pressed, it rarely stays lit when the switch is released. By contrast, the SR latch circuit above works as expected with one LED. The 'problem' is therefore with the diode pulse routing network, presumably because the circuit is no longer balanced, the LED setting the voltage at the collector, as noted earlier. To resolve the problem, I've moved the LED to a new sub-circuit consisting of the LED and a third transistor, to act as a switch, and a resistor (value tbd). The base of this transistor is connected to the output of the flip-flop (the NOR gate previously having the LED), the emitter is connected to ground via the resistor and the collector is connected to the cathode of the LED, the anode of which is connected to the 5V supply. 

This new set-up now works just fine, albeit still with occasional switch bounce. Hopefully, when driven by a pulsed input, this will not occur.

So, for just one element of the binary counter, we need:

• 3 transistors

• 2 diodes

• 1 LED

• 2 capacitors

• 7 resistors