In cut-off, a BJT switch behaves as:

An open switch with V_CE about VCC, so Vout is HIGH

A closed switch with V_CE about 0.2 V

In saturation, V_CE is approximately:

0.2 V, a closed switch with Vout LOW

0.7 V, a closed switch with Vout HIGH

What limits the BJT switching speed when leaving saturation?

Storage time to remove stored charge

What does the TTL NAND do when all inputs are HIGH?

Q2 and Q4 saturate, Q3 is OFF, Y = 0

2075 Asoj · 2+3 marks

TTL NAND gate; transistor switch

Short answer

The TTL NAND gate uses a multi-emitter input transistor; any low input keeps Q2 and Q4 off, Q3 on, giving high output. All inputs high saturate Q2 and Q4, turning Q3 off, output low. A BJT switch is in cut-off (V_BE < 0.7V, open switch, high output) or saturation (I_B ≥ I_C(sat)/β, V_CE ≈ 0.2V, low output).

Draw the schematic diagram of TTL NAND gate and explain about the transistor switch.

TTL NAND gate

              +5 V
   +-----------+----------+
   |           |          |
 R1 4k      R2 1.6k    R4 130
   |           |          |
   b           +-----+    c
A--e  Q1       |     +--b Q3
B--e     c--+  c          e
            +--b Q2       |
               e         D1
               |          |
               +-----+    +---- Y
               |     |    c
             R3 1k   +--b Q4
               |          e
   GND --------+----------+

(b, c, e = base, collector, emitter; Q1 is a multi-emitter transistor.)

Any input LOW → Q2, Q4 OFF, Q3 ON → Y = 1; all inputs HIGH → Q2, Q4 saturate, Q3 OFF → Y = 0. So $Y = \overline{AB}$ .

Transistor as a switch

In digital circuits a BJT is operated only in two regions:

        +VCC
         |
         RC
         |
         +----- Vout
         c
 Vin-RB--b
         e
         |
        GND

Region	Condition	$V_{CE}$	Acts as
Cut-off	$V_{BE} < 0.7$ V, $I_B = 0$	≈ $V_{CC}$	open switch, Vout HIGH
Saturation	$I_B \ge I_{C(sat)}/\beta$	≈ 0.2 V	closed switch, Vout LOW

So the transistor inverts: Vin LOW → OFF → Vout HIGH; Vin HIGH → saturated → Vout LOW.

Example: $V_{CC} = 5$ V, $R_C = 1\ \text{k}\Omega$ , $\beta = 100$ : $I_{C(sat)} = (5 - 0.2)/1\text{k} = 4.8$ mA, so the base current must be at least $4.8\ \text{mA}/100 = 48\ \mu$ A for saturation. Switching speed is limited by the time needed to remove stored charge when leaving saturation (storage time).

✓ Answer checked by Claude Opus 5.5 (high effort)

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