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Two, Two-Input DL AND-OR Gate

Introduction

Now that we have looked at both the Diode Logic (DL) OR gate and AND gate and found that they do operate within acceptable parameters, it is time to cascade them. That is, we will use an OR gate to combine the outputs of two AND gates. Then we will see how well this combination works.

The designation "2, 2-input AND-OR gate" means that we will set up a two-input AND gate with a second two-input AND gate, and combine their outputs with an OR gate. If we had specified a "2, 3-input AND-OR gate" it would have meant using a two-input AND gate and a three-input AND gate.



Diode Logic AND-OR Gate

Schematic Diagram

Combining Diode Logic ANDed signals with a DL OR gate is simple and straight-forward, as shown in the schematic diagram to the right. We have taken two basic 2-input AND gates and applied their output signals to the input diodes of an OR gate.

While such a circuit is easy to draw and equally easy to construct, there is some question as to how well it will work in practice. After all, the whole circuit consists of just diodes and resistors, which are both passive components. Some signal degradation must occur as the signal passes through multiple gates. We need to determine how much the signal will be degraded at each stage, and just how far we can cascade DL gates before losing control of the signal.

In this experiment, we will explore the possibilities of a two-level gating structure, AND followed by OR, and see how well it functions.



Parts List

To construct and test the 2, 2-input DL AND-OR gate on your breadboard socket, you will need the AND gate already in place plus the following components:



Constructing the Circuit

You should still have your Diode Logic AND gate in place from the previous experiment. If not, go back now and perform that experiment, leaving the components in place when you are done. You will expand on that circuit to prepare for this experiment.



Circuit Assembly

Start assembly procedure






Starting the Assembly

Your AND-OR gate will require more space on your breadboard socket, and more input jumpers. Therefore, we have left out the picture of the logic switches. Instead, we will simply label the unseen connection points of all jumpers, as illustrated to the right.

You should still have your DL AND gate in place from the previous experiment. Its layout should match the initial assembly diagram shown here. If not, or if you did not perform the preceding experiment, you should go back and perform it now. This will provide some background towards understanding this circuit, and leave you with the correct circuit in place as your starting point.

Click on the `Start' button below to begin assembling your experimental circuit.

Remove the White Jumper

Remove the 10" white jumper from its current connection and set it aside for now. Once you have constructed the OR portion of the gating structure, you will reconnect this jumper to monitor the final output signal.

Click on the image of the jumper you just removed to remove it from the assembly diagram and continue with your assembly.

10K, ¼-Watt Resistor

You should already have a 10K, ¼-watt resistor (brown-black-orange) formed to a lead spacing of 0.3", left over from your OR gate experiment. If not, form a 10K, ¼-watt resistor to a spacing of 0.3". Install this resistor in the location indicated in the assembly diagram.

Click on the resistor image to continue with your assembly.

10K, ¼-Watt Resistor

Locate another 10K, ¼-watt resistor (brown-black-orange) and form its leads to a spacing of 0.5". Install this resistor as shown to the right.

Again, click on the resistor image to continue with your assembly.

1N914 Silicon Diode

Locate a 1N914 diode and form its leads to a spacing of 0.3". Install this diode as shown in the assembly diagram. Be careful to observe the orientation of the diode.

When you have installed the diode correctly, click on its image to continue.

1N914 Silicon Diode

Locate another 1N914 diode and form its leads to a spacing of 0.3". Install this diode as shown in the assembly diagram. Be careful to observe the orientation of the diode.

As before, click on the image of the diode you just installed to continue.

1N914 Silicon Diode

Locate a third 1N914 diode and form its leads to a spacing of 0.3". Install this diode as shown in the assembly diagram. Again, be careful to observe the orientation of the diode.

Following the normal procedure, click on the image of the component you just installed to continue.

1N914 Silicon Diode

Form the leads of fourth 1N914 diode to a spacing of 0.4". Install the prepared diode as shown in the figure. Of course, be careful to observe the indicated orientation.

As usual, click on the image of the diode you just installed to continue.

10" White Jumper

Locate the 10" white jumper you removed and set aside at the beginning of this assembly procedure. Connect one end to the L0 input, and the other end to the point indicated on the assembly diagram.

Click on the image of this jumper to the right to move on to the next step.

6" Orange Jumper

Cut a 6" length of orange hookup wire and remove ¼" of insulation from each end. Install one end of this jumper as indicated to the right, and connect the other end to logic switch S6.

As usual, click on the image of this jumper to move on to the next step.

6" Orange Jumper

Cut a second 6" length of orange hookup wire and remove ¼" of insulation from each end. Connect one end of this jumper as indicated to the right. Connect the other end to S7.

As usual, click on the image of the jumper to move on to the next step.

Assembly Complete

This completes the assembly portion of this experiment. Take the time now to check your work carefully against the assembly diagram to the right. Especially make sure that all diodes are installed with the correct orientation, since they do not all point the same way on the breadboard socket.

When you are ready, scroll down to the next part of the page and begin the actual experiment.

Restart assembly procedure


Performing the Experiment

Before applying power to your experimental circuit, determine the output voltage you would expect this circuit to produce if all input diodes are open-circuited (no input signal connected to any input diode). Assume a diode voltage drop of 0.65v and that resistance and voltage values in the circuit are precisely accurate.

When you have calculated this value, enter it in the top row of the table to the right.

Next, calculate the output voltage you would expect from this circuit if all four input diodes were connected to +5 volts (logic 1). Enter this value in the second row of the table to the right.

Then, calculate the output voltage you would expect to see if any one diode input is grounded (logic 0 input). Record this voltage in the third row of the table to the right.

Now turn on power to your experimental circuit. Set S0, S1, S6, and S7 to each of the 16 possible combinations of logic signals from four inputs. Note the resulting state of LED indicator L0, and record that state on the appropriate row of the table. Also measure the output voltage obtained for each input combination and record this value beside the observed state of L0. Continue until you have tested your circuit for all input combinations and recorded your observed results in each case.

Looking over your results, do you find any serious discrepancies between your calculated values and your measured values? Can you explain any such discrepancy? Do the measured output voltages represent valid logic 0 or logic 1 states for each input combination?

When you have completed your determinations, turn off the power to your experimental circuit and scroll down to the discussion below.

Expected
Open-Circuit
Output
Voltage
 volts
Expected
Output
Voltage
for All 1 Inputs
 volts
Expected
Output
Voltage
for One 0 Input
 volts

Inputs Outputs
S7 S6 S1 S0 L0 Voltage
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1


Discussion

In the first part of this experiment, you determined the output voltage of the AND-OR gating structure under various conditions. We note that if all inputs are at logic 1, the four input diodes will all be reverse biased and will therefore not conduct current. Thus, the output voltage for unconnected inputs will be the same as for all logic 1 inputs. Under these conditions, with an assumed diode voltage drop of 0.65v, we see 5 - 0.65 = 4.35v dropped across a combination of three 10K resistors. The two resistors connected to +5v are effectively in parallel, so their combined resistance is 5K. This gives us an output voltage of 4.35 × (10K/15K) = 2.9v.

As with the AND gate in the previous experiment, we didn't include the effects of the LED driver circuit that forms the input to L0. If you disconnect the white jumper to L0 temporarily, the output voltage will rise to a close approximation of 2.9 volts. With the jumper connected, the output is pulled down a little more.

If any one input is pulled down to ground (logic 0), it will pull the output of its AND gate down as well. This will cause the corresponding OR gate input diode to be reverse biased, effectively disconnecting that 10K resistor from the output circuit. Now we have only two 10K resistors dropping 4.35 volts, and the theoretical output voltage will be just half of the total voltage drop, or 4.35 × (10K/20K) = 2.175v.

Here we see the beginning of the practical problem of cascading Diode Logic gates. An output voltage of 2.9v is still above 2.5v, and is therefore clearly a logic 1. But an output voltage of 2.175v. is less than 2.5v, and should therefore really be a logic 0. It gets even worse with the L0 input circuit connected, due to the extra loading.

When the input combinations forced the output to logic 0, the output voltage was about 0.2 volt — clearly a valid logic 0. This changed only very slightly according to how many logic 0 inputs were present.

In spite of all this, L0 gave the correct logical indications for all input combinations. L0 turned on whenever both S0 and S1, or both S6 and S7, were set to logic 1. L0 remained on whenever you had any three or all four switches at logic 1. The logical expression for the state of L0 is L0 = (S0·S1) + (S6·S7).

The reason this circuit gave the correct visual outputs in all cases is that the LED indicator circuit will accept any input voltage above approximately 2.1v as a logic 1, since that will be sufficient to turn on the LED. A voltage of about 1.9v will leave the LED off. An input voltage within this range will leave the LED on, but not as brightly as a full logic 1. When both AND gates are pulled down to logic 0 outputs, the OR output voltage is very close to 0, and is therefore insufficient to turn L0 on. Thus, this circuit seems to work properly because as far as L0 is concerned, the threshold between logic 0 and logic 1 is not at 2.5 volts, but rather is much closer to 2.0 volts.

Only the nature of the LED driver circuit allows it to report the correct logic levels for this Diode Logic circuit. An LED driver that set its threshold at 2.5 volts would report a logic 1 output only if all four inputs were logic 1, thus claiming that this experimental circuit is simply a 4-input AND gate.

When you have completed this experiment, make sure power is off, and then remove the experimental components from your breadboard socket. Keep them handy, however; you'll be using almost all of them in your next experiment.


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