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www.play-hookey.com | Thu, 07-03-2008 |
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| DTL Inverter |
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One of the problems with RTL circuits is that the input resistor for each input has the effect of slowing down the rate at which the internal capacitance of the transistor can charge or discharge. This limits the frequency at which RTL gates can switch states. A second problem is that resistors take up a significant amount of space, or "real estate," on the silicon chip that comprises the integrated circuit. High-value resistors generally require more space than lower-value resistors, and in any case the shape of the resistor on the chip depends very much on the required resistance value. Real estate is very precious, and every effort is made to develop and use techniques that will minimize the amount of space required for any circuit.
Diodes take up far less room than resistors, and can be constructed to all be the same. In addition, the internal resistance of a diode is small when the diode is forward biased, thus allowing for faster switching action. As a result, gates built with diodes in place of most resistors can operate at higher frequencies. Because of this diode-transistor logic (DTL) rapidly replaced RTL in most digital applications.
The DTL inverter uses the same transistor as the RTL inverter, and the same 1K collector load resistor. However, in place of an input resistor in series with the base lead, we use a pair of diodes as shown. The diode connected directly to the transistor base serves to raise the input voltage required to turn the transistor on to about 1.3 to 1.4 volts. Any input voltage below this threshold will hold the transistor off. Thus, we can be sure that another circuit of this same type will have no trouble properly controlling the state of this circuit.
The base resistor is considerably smaller than 15K to turn the transistor on more quickly when the input rises to a logic 1. A 15K resistor would be enough to drive the transistor into saturation, but would limit the operating frequency of the gate just as it does with RTL. By reducing the value of this pull-up resistor, we allow the transistor to turn both on and off more rapidly, thus enabling higher switching speeds than are possible with RTL.
To construct and test the DTL inverter circuit on your breadboard, you will need the following experimental parts:
Select an area on your breadboard socket that is clear of other circuits. Our construction procedure places this circuit just to the right of the center divider of the breadboard socket, as shown in the construction image below. Refer to this image and the step-by step instructions as you install the experimental parts for this circuit.
Make sure all components from previous experimental circuits have been removed from the right hand side of your breadboard socket. You'll be re-using several of the parts left over from previous experiments, but you will be assembling them into a new configuration.
Click on the `Start' button below to begin construction of your experimental circuit.
You should have several 0.3" black jumpers left over from previous experiments. If not, use the guide above to create a new black jumper with a lead spacing of 0.3". Install the jumper as shown in the assembly figure to the right.
When the jumper is installed in the designated location, click on its image to the right to continue to the next step.
You should also have a 1K, ¼-watt resistor (brown-black-red) with its leads formed to a spacing of 0.5". If not, locate a new resistor of this value and form the leads to a spacing of 0.5". Install this resistor on your breadboard socket in the location shown to the right.
When the resistor is in place, click on its image to continue.
Locate a 4.7K, ¼-watt resistor (yellow-violet-red) and form the leads to a spacing of 0.5". Install this resistor on your breadboard socket in the location shown to the right.
As before, click on the resistor image to continue.
You should have several 1N914 diodes left over from your DL experiments, already formed to a spacing of 0.3". If not, locate a 1N914 silicon diode and form its leads to a spacing of 0.3". Remember to clip the leads to ½" length so the glass body of the diode will remain above the breadboard socket. Install this diode in the location shown in the assembly diagram to the right. Be sure to observe the correct orientation of the diode.
As usual, click on the image of the component you just installed to continue your construction of this circuit.
Locate a second 1N914 silicon diode and, if necessary, form its leads to a spacing of 0.3". Install this diode in the location shown in the assembly diagram to the right. As before, be sure to observe the correct orientation of the diode.
As usual, click on the image of this diode to continue.
Locate a 2N4124 NPN transistor and, if necessary, form the leads to spread them out to 0.1" spacing. If you have a transistor with formed leads left over from a previous experiment, by all means use it here. Install the transistor in the location shown to the right.
As always, click on the image of the component you just installed to continue.
You should have some orange jumpers left over from earlier experiments. If not, cut a 3" length of orange hookup wire and remove ¼" of insulation from each end. Connect this jumper from logic switch S0 to the point indicated in the assembly diagram.
As before, click on the image of this jumper to continue.
Locate the 10" white jumper you have used in earlier experiments. If you don't have one, cut a 10" length of white hookup wire and remove ¼" of insulation from each end. Connect one end to L0, the right hand LED indicator. Connect the other end to the inverter output, as shown to the right.
As before, click on the image of this jumper to continue.
This completes the construction of your experimental circuit. Check your assembly carefully against the figure to the right, and correct any errors you might find. Then, proceed with the experiment on the next part of this page.
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Turn on power to your experimental circuit, and move switch S0 back and forth between logic 1 input and logic 0 input. What is the output state of this circuit for each of the possible input states? Record your results in the table to the right. Does this circuit correctly invert the logic sense of the input signal? When you have made this determination, turn off the power to your experimental circuit and compare your results with the discussion below. |
You should have found that this circuit does indeed invert the input signal, just as the RTL inverter did. After all, it's the same transistor circuit, except that we have modified the input circuit to remove the resistor in series with the input signal. Thus, a logic 1 input produces a logic 0 output, while a logic 0 input produces a logic 1 output.
The main difference in behavior between RTL and DTL circuits is that with RTL, an open input connection provides no drive to the transistor, so it is read as a logic 0. This also makes the open RTL input susceptible to certain kinds of noise signals, so that an open RTL input can introduce spurious signals under some conditions. With DTL, the pull-up resistor causes the open input line to act as a logic 1. In addition, any noise signal will have to pull the input down nearly to ground in order to affect the transistor. As a result, the DTL circuit is much less susceptible to noise than the RTL circuit. This is one of many reasons why RTL circuits are not generally used any more.
When you have completed this experiment, make sure power is turned off and continue on to the next experiment when you are ready. Do not remove any components or jumpers from your breadboard socket. You will expand on this circuit in the next experiment.
| Your next experiment is: Three-Input DTL NAND Gate |
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