Cmos transistor diagram6/22/2023 ![]() ![]() The upper transistors of both pairs (Q 1 and Q 2) have their source and drain terminals paralleled, while the lower transistors (Q 3 and Q 4) are series-connected. Notice also how transistors Q 2 and Q 4 are similarly controlled by the same input signal (input B), and how they will also exhibit the same on/off behavior for the same input logic levels. Both are controlled by the same input signal (input A), the upper transistor turning off and the lower transistor turning on when the input is “high” (1), and vice versa. Notice how transistors Q 1 and Q 3 resemble the series-connected complementary pair from the inverter circuit. This brings us to the next question: how do we design multiple-input CMOS gates such as AND, NAND, OR, and NOR? Not surprisingly, the answer(s) to this question reveal a simplicity of design much like that of the CMOS inverter over its TTL equivalent.įor example, here is the schematic diagram for a CMOS NAND gate: ![]() Of course, a separate pullup or pulldown resistor will be required for each gate input: ![]() Again, the value for a pulldown resistor is not critical:īecause open-collector TTL outputs always sink, never source, current, pull up resistors are necessary when interfacing such an output to a CMOS gate input:Īlthough the CMOS gates used in the preceding examples were all inverters (single-input), the same principle of pullup and pulldown resistors applies to multiple-input CMOS gates. When such a resistor is used to provide a “low” (0) logic level in the event of a floating signal source, it is known as a pulldown resistor. When used to provide a “high” (1) logic level in the event of a floating signal source, this resistor is known as a pullup resistor: This resistor’s value is not critical: 10 kΩ is usually sufficient. Whenever a single-throw switch (or any other sort of gate output incapable of both sourcing and sinking current) is being used to drive a CMOS input, a resistor connected to either V dd or ground may be used to provide a stable logic level for the state in which the driving device’s output is floating. Because such a TTL gate’s output floats when it goes “high” (1), the CMOS gate input will be left in an uncertain state:įortunately, there is an easy solution to this dilemma, one that is used frequently in CMOS logic circuitry. This may cause a problem if the input to a CMOS logic gate is driven by a single-throw switch, where one state has the input solidly connected to either V dd or ground and the other state has the input floating (not connected to anything):Īlso, this problem arises if a CMOS gate input is being driven by an open-collector TTL gate. Please note that this is very different from the behavior of a TTL gate where a floating input was safely interpreted as a “high” (1) logic level. For this reason, it is inadvisable to allow a CMOS logic gate input to float under any circumstances. Their inputs are, however, sensitive to high voltages generated by electrostatic (static electricity) sources, and may even be activated into “high” (1) or “low” (0) states by spurious voltage sources if left floating. Clearly, this circuit exhibits the behavior of an inverter, or NOT gate.ĬMOS circuits aren’t plagued by the inherent nonlinearities of the field-effect transistors, because as digital circuits their transistors always operate in either the saturated or cutoff modes and never in the active mode. Thus, the output of this gate circuit is now “low” (0). The upper transistor, having zero voltage applied between its gate and substrate, is in its normal mode: off. Now the lower transistor (N-channel) is saturated because it has sufficient voltage of the correct polarity applied between gate and substrate (channel) to turn it on (positive on gate, negative on the channel). Next, we’ll move the input switch to its other position and see what happens: This makes the output “high” (1) for the “low” (0) state of the input. Thus, the action of these two transistors are such that the output terminal of the gate circuit has a solid connection to V dd and a very high resistance connection to ground. The lower transistor, having zero voltage between gate and substrate (source), is in its normal mode: off. So, in the above illustration, the top transistor is turned on. When the channel (substrate) is made more positive than the gate (gate negative in reference to the substrate), the channel is enhanced and current is allowed between source and drain. The upper transistor is a P-channel IGFET. It takes an applied voltage between gate and drain (actually, between gate and substrate) of the correct polarity to bias them on. ![]() Please note that these IGFET transistors are E-type (Enhancement-mode), and so are normally-off devices. Let’s connect this gate circuit to a power source and input switch, and examine its operation. \)įield Effect Transistors in Gate Circuits ![]()
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