##### 6.4 SWITCHING TIMES OF A TRANSISTOR

Instead of a step, if a pulse is applied to the transistor switch, how does the device respond? To understand this, we consider the switching times of a transistor (see Fig. 6.24).

Let the input to the transistor switch be a pulse of duration *T*. When a pulse is applied, because of stray capacitances, collector current will not reach the steady-state value instantaneously. To know exactly when the device switches into the ON state and also into the OFF state, we define the following switching times of the transistor.

**FIGURE 6.24** Switching times of the transistor

#### 6.4.1 The Turn-on Time of a Transistor

Turn-on time of the transistor is the time taken by the transistor from the instant the pulse is applied to the instant the transistor switches into the ON state and is the sum of the delay time and rise time. To find out the turn-on time of the transistor, the delay time and rise time have to be calculated.

**Delay Time ( t_{d}**). It is the time taken for the collector current to reach from its initial value to 10 per cent of its final value.

If the rise of the collector current is linear, the time required to rise to 10 per cent *I*_{C(sat)} is 1/8 the time required for the current to rise from 10 per cent to 90 per cent *I*_{C(sat)}.

It is given as:

where, *t _{r}* is the rise time.

**Rise Time.** Rise time, *t _{r}* is the time taken for the collector current to reach from 10 per cent of its final value to 90 per cent of its final value. From Fig. 6.24 it is seen that the moment input pulse is zero, the collector current is expected to fall to zero. However, because of the stored charges, the current remains unaltered for sometime interval

*t*

_{s1}and then begin to fall. The time taken for this current to fall from its initial value at

*t*

_{s1}to 90 per cent of its initial value is

*t*

_{s2}. The sum of these

*t*

_{s1}and

*t*

_{s2}is approximately

*t*

_{s1}=

*t*and is called the storage time. To calculate the rise time, consider the transistor switch as shown in Fig. 6.25.

_{s}Let *I _{B}* be the base current pulse that drives the transistor into saturation at

*I*

_{B1}and into OFF state at

*I*

_{B2}. If

*R*is large when compared to

_{S}*h*,

_{ie}The response of the transistor to the current step *I*_{B1} is given by:

If *R _{S}* is large,

where *f*_{2} = Upper 3-dB frequency of the low-pass *RC* circuit.

where, *τ* is the time constant.

Substituting Eq. (6.35) in Eq. (6.34),

**FIGURE 6.25** A transistor switch

If the device does not go into saturation, *i _{C}* would have increased as shown by the dashed line in Fig. 6.26 and would have reached a value

*h*

_{FE}I_{B1}as

*t*→ ∞. The gain bandwidth product of a transistor amplifier in which the transistor is replaced by an ideal current source

*h*(since (1

_{FE}I_{B}*/h*≈ ∞) is given as:

_{oe})where, *f _{T}* is the frequency at which the short circuit common emitter current gain has a value 1 and

*C*is the collector diode transition capacitance.

_{TC}From Eq. (6.37):

Therefore:

The variation of the collector current *i _{C}* is plotted in Fig. 6.26 using Eq. (6.36).

When in saturation Eq. (6.36) is written as:

As

If *N*_{1} > 1, the transistor is in saturation (over-driven transistor). Hence,

**FIGURE 6.26** Variation of *i _{C}* (i) when the device is in saturation (ii) when the device is not in saturation and (iii) when the device switches into OFF state

At *t* = *t*_{1}, 1*/N*_{1} is 0.1*/N*_{1}

Similarly *t* = *τ* ln

We know that ln (1 + *x*) =

For *N*_{1} >> 1, considering only the 1^{st} term

From Eq. (6.39),

The rise time *t _{r}*, is inversely proportional to

*I*

_{B1}. Therefore, if the turn-on time is to be small it is desirable to have a larger base current drive (over-driven transistor).

##### EXAMPLE

*Example 6.8:* Consider the switch in Fig. 6.27, *C _{TC}* = collector transition, capacitance = 7 pF,

*h*= 100 and

_{FE}*f*= 10 MHz. Calculate the turn-on time of the transistor.

_{T}**FIGURE 6.27** The given transistor switch

*Solution:*

To calculate the turn-on time, we have to calculate *t _{r}* and

*t*.

_{d}From Fig. 6.27, we have,

In the quiescent state the voltage at B is -4 V. However, when a pulse of 10 V appears at the input this voltage at B is 6 V.

We have from Eq. (6.39),

Hence, *N*_{1} ≥ 1. Equation (6.41) is valid when the overdrive factor *N*_{1} ≥ 1. *ω _{T}* is the radial frequency at which the current gain is unity. The rise time

*t*is calculated using Eq. (6.41) as,

_{r}and from Eq. (6.38),

where *τ* is the time constant.

Using Eq. (6.42)

From Eq. (6.41)

*t*_{turn−on} = *t _{r}* +

*t*= 184 + 23 = 207 ns.

_{d}##### EXAMPLE

*Example 6.9:* A transistor has *f _{T}* = 50 MHz,

*h*= 50,

_{FE}*C*= 5 pF, (

_{TC}*C*= collector transition capacitance) and the supply voltage

_{TC}*V*= 10 V,

_{CC}*R*= 0.5 kΩ. Initially the transistor is operating in the neighbourhood of the cut-in point. What base current must be applied to drive the transistor into saturation in 1

_{C}*µ*s?

*Solution:*

We have

Also

*ω _{T}* = 2

*πf*= 2 ×

_{T}*π*× 50 × 10

^{6}= 314 × 10

^{6}radians

*C _{TC}R_{C}* = 5 × 10

^{−12}× 0.5 × 10

^{3}= 2.5 × 10

^{−9}s

*t _{r}* = 1 × 10

^{−6}

A transistor is said to be switched from the OFF state into the ON state only when the collector current is 90 per cent of its final value. Thus we see that even though the transistor is expected to switch from OFF to ON at *t* = 0 when *v _{i}* =

*V*, the device switches actually into the ON state only when a finite time elapses and we call this time interval as the turn-on time of the transistor. From Example 6.8:

*t*_{turn-on} = *t _{d}* +

*t*= 23 + 184 = 207 ns

_{r}#### 6.4.2 The Turn-off Time of a Transistor

Again at *t* = *T* (at the end of the pulse), the transistor is required to switch into the OFF state instantaneously. But this is not going to happen. Once the device is driven hard into saturation, because of large number of stored charges on either side of the junction, the collector current is not going to fall to a smaller value instantaneously. To calculate the turn-off time we have to calculate the storage time and the fall time.

**Storage Time ( t_{s}**). Storage time,

*t*, is the time taken for the collector current to fall from its initial value to 90 per cent of its initial value.

_{s}*I*_{B1} is the base current, when the pulse amplitude is *V* = 10 V and *I*_{B2} is the base current, when the pulse amplitude is zero, (see Fig. 6.27).

Here

where, *α _{I}* and

*α*are the inverted mode and normal mode current gains in terms of Ebers–Moll parameters.

_{N}*α*

_{N0}is the normal direction current gain and its 3-dB frequency is

*ω*

_{N}. α_{I0}is the inverted mode current gain and the 3-dB frequency is

*ω*.

_{I}##### EXAMPLE

*Example 6.10:* From Fig. 6.27, calculate the storage time if *h _{FE}* = 100,

*α*

_{N0}= 0.99,

*α*

_{I0}= 0.5,

*f*= 1.2

_{N}*f*= 1.2 × 1 × 10

_{T}f_{T}^{6}= 1.2 MHz,

*f*= 1MHz

_{I}*Solution:*

From Eq. (6.44):

If *V* = 10 V

Using Eq. (6.43):

Further, the transistor is said to be switched from the ON state to OFF state only when the collector current falls to 10 per cent of its initial value. This is the fall time.

**Fall Time.** Fall time, *t _{f}*, is the time taken for the collector current to fall from 90 per cent of its initial value to 10 per cent of its initial value. To calculate the fall time, consider Fig. 6.28 when the base current is

*I*

_{B2}.

The fall time calculation is similar to the rise time calculation. Here, *I*_{B2} flows in the opposite direction. The device returns from saturation to active region where the collector current is *h _{FE}I*

_{B2}.

As *I*_{B2} flows in the opposite direction *N*_{2} is positive. We have,

From Fig.6.26 at *t* = *t*_{3}, 1*/N*_{2} is 0.9*/N*_{2}

**FIGURE 6.28** Base current is *I*_{B2}

Similarly, *t*_{4} = *τ* ln

Fall time, *t _{f}* again is inversely proportional to

*I*

_{B2}. If the turn-off time is to be small,

*I*

_{B2}should be relatively large.

It is given as:

Equation (6.47) is valid when *N*_{2} *>* 1.

##### EXAMPLE

*Example 6.11:* For the circuit in Fig. 6.27, calculate the fall time.

*Solution:*

From Eq. (6.47), we have,

Therefore, from the calculations made in Examples 6.10 and 6.11, we have the turn-off time of the transistor as,

*t*_{turn-off} = *t _{s}* +

*t*= 447 + 274 = 721ns.

_{f}The major concern now in switching applications is – how quickly can we drive a transistor from one state to the other. This switching speed obviously depends on the switching times of the transistor.