##### 14.2 MONOSTABLE BLOCKING OSCILLATORS

A pulse transformer is used as a coupling element in a blocking oscillator to provide positive feedback. If the blocking oscillator generates a single pulse, the circuit is called a monostable circuit. Monostable blocking oscillators can be of two types: Monostable circuit with base timing and monostable circuit with emitter timing. In a monostable blocking oscillator using base timing, the timing resistance *R* (a component that controls the gate width) is used in the base circuit of the transistor. This circuit, however, does not ensure stable pulse width. Alternatively, the timing resistance *R* is shifted to the emitter circuit of the transistor in the emitter-timing monostable blocking oscillator. This in turn generates a stable pulse, independent of the temperature-dependent parameters of the transistor.

#### 14.2.1 A Triggered Transistor Monostable Blocking Oscillator (Base Timing)

A monostable blocking oscillator circuit triggered by a pulse is shown in Fig. 14.1(a). The pulse transformer has a turns ratio of *n*:1 where the base circuit has *n* turns for every turn on the collector circuit. It is connected to provide polarity inversion (as indicated by the dots). A resistance *R* is included in series with the base of the transistor and this resistance controls the timing, i.e., the pulse duration (hence, the name base-timing). In the quiescent state (when the trigger is zero), the transistor is OFF. *V*_{BB} may be taken to be zero as the cut-in voltage is small. However, to ensure that the noise pulses do not trigger the circuit and drive it into free-running operation at higher temperatures, a small *V*_{BB} (few tenths of a volt) is provided. Since *V*_{BB} *V*_{CC}, *V*_{BB} can be neglected in the analysis.

The voltage at the collector initially is *V*_{CC} as, with *V*_{BE} ≈ 0, the device is OFF. A trigger (a negative pulse) is momentarily applied to the collector at *t* = 0. This pulse reduces the voltage at the collector. The winding polarities of the transformer (dot convention) are so chosen that when the voltage at the collector goes down, the voltage at the base rises. When *V*_{BE} > *V*_{γ}, the transistor draws a small base current. The collector current, therefore rises and the collector voltage decreases. This in turn raises the base voltage further. Hence, still some more base and collector currents flow, resulting in a further drop in the collector potential. If the ac loop gain exceeds unity, due to regenerative action, the transistor is driven into saturation quickly. To calculate the pulse width of the monostable multivibrator, we draw the equivalent circuit during pulse formation, as shown in Fig. 14.1(b). In the equivalent circuit, the pulse transformer is replaced by an ideal transformer. If the secondary winding of the transformer is disconnected then we only have the primary winding in the core, which behaves as an inductor.

**FIGURE 14.1(a)** A monostable blocking oscillator with base timing

This inductor is called the magnetizing inductance *L* and refers to the primary winding. Using this equivalent circuit, it is possible to find the amplitude and the width of the pulse; and to predict its shape. Since *V*_{BE} ≅ *V*_{CE} ≅ 0, the base, emitter and collector terminals are at the ground potential, as shown in Fig. 14.1(b). We know that, for an ideal transformer, the sum of the ampere turns is constant and the induced voltages are proportional to the turns. In the quiescent state, as all currents are zero, the algebraic sum of the ampere turns within the dashed box must be zero. Let *i* be the current in the ideal transformer winding in the collector circuit. The ampere turns in the collector winding is (*i* × 1) and that in the base winding is (*n* × *i*_{B}). Taking the winding polarities into account:

**FIGURE 14.1(b)** The equivalent circuit of Fig. 14.1(a) during pulse formation

If *V* is the amplitude of the negative pulse across the collector winding, then *nV* is the corresponding voltage across the base winding, where *n* is the number of turns in the base winding. From the collector circuit shown in Fig. 14.1(b):

From the base circuit,

From Eq. (14.1), *i* = *ni*_{B}. Using Eq. (14.3):

If *i*_{m} is the magnetizing current and since *V* is constant, then:

Or

But

*i*_{C} = *i* + *i*_{m}

That is,

From Eq. (14.6) it is seen that the collector current is trapezoidal [the sum of a step, *n ^{2}V_{CC}/R* and ramp, (

*V*)

_{CC}/L*t*] in nature. Though the collector current is trapezoidal, from Eq. (14.3), it is evident that the base current is constant, as shown in Fig. 14.1(c).

From the output characteristics of the transistor, it is seen that when the device switches suddenly from the OFF state into saturation, the path of the collector current is along the saturation line from *P* to *P*^{′}. The pulse ends at *P*^{′}, at which instant the transistor comes out of saturation and goes into the active region, as shown in Fig. 14.2(a).

At *t* = 0+, the operating point on the collector characteristics is at point *P* as shown in Fig. 14.2(a).

**FIGURE 14.1(c)** The base and collector current waveforms

From Eq. (14.6), at *t* = 0+:

and from Eq. (14.3):

However, the transistor is said to be in saturation only when:

*h _{FE}i_{B} > i_{C}*

i.e.,

Or

This condition in Eq. (14.7) can be easily satisfied as the turns ratio *n* is usually close to one and *h*_{FE} is reasonably large. For *t >* 0, the collector current increases and the operating point moves up the curve (point *P*^{′}) corresponding to the constant base current:

as shown in Fig. 14.2(a).

When the point *P*^{′} is reached, the voltage at the collector *V*_{C}(= *V*_{CE}) increases rapidly and, for the given winding polarities, the voltage at the base decreases, resulting in a decreased base current. At this point *P*^{′}, the transistor comes out of saturation into the active region. In the active region, as the loop gain is greater than unity, the transistor is quickly driven into cut-off by regenerative action and the pulse is terminated. The pulse width *t*_{p} is determined by the condition:

**FIGURE 14.2(a)** The output characteristics

From Eq. (14.3), we have:

From Eq. (14.6):

Substituting these values in Eq. (14.6), at *t* = *t*_{p}:

Therefore,

Therefore,

since *n* is very small when compared to *h*_{FE}. Eq. (14.11) indicates that the pulse width *t*_{p} varies linearly with *h*_{FE}. However, *h*_{FE} is a temperature-dependent parameter that may vary with transistor replacement also. Hence, stable pulse width can not be maintained with base timing. Thus, the major limitation with the monostable blocking oscillator with base timing is that a stable pulse width cannot be maintained, as elucidated in these discussions and exemplified by Example 14.1.

One way to ensure that the pulse width of the blocking oscillator is independent of *h*_{FE} (hence, stable) is to remove the timing resistance *R* from the base loop and include it in the emitter loop.

##### EXAMPLE

*Example 14.1:* The monostable blocking oscillator shown in Fig. 14.1 (a), has the following parameters: *V*_{CC} = 10 V, *R* = 0.5 kΩ, *L* = 5 mH, *h*_{FE} = 30 and *n* = 1.5. (a) Calculate and plot the base current and collector current waveforms and calculate the pulse width. (b) if due to temperature variation, *h*_{FE} increases to 40, calculate the pulse width.(c) Due to the replacement of the transistor, if the new transistor has *h*_{FE} = 25, what is the new pulse width?

*Solution:*

(a) From Eq. (14.3) we have:

From Eq. (14.6):

**FIGURE 14.2(b)** The waveforms for condition

Therefore, from Eq. (14.11):

*i _{C}* at

*t*= 0 is,

*i*= 45 mA

_{C}*i _{C}* at

*t*=

*t*is

_{p}*i*= 45 × 10

_{C}^{−3}+ 2 × 10

^{3}× 0.450 × 10

^{−3}= 945 mA

(b) Pulse width when *h*_{FE} = 40

(c) Pulse width when *h*_{FE} = 25

The waveforms for the condition (a) are plotted in Fig. 14.2(b).

#### 14.2.2 A Triggered Transistor Blocking Oscillator (Emitter Timing)

A triggered transistor blocking oscillator with emitter timing is shown in Fig. 14.3 (a). This blocking oscillator consists of a transistor with an emitter resistance and a three-winding pulse transformer. The turns ratio of the transformer in the base and collector windings is *n*: 1. Another winding, connected to load *R*_{L}(output winding), has *n*_{1} turns in it. The polarities of the base and collector windings are so chosen that the signal appearing at the collector undergoes phase inversion when it appears at the base, as indicated by the dots on the transformer windings. The polarity of the third winding (output winding) of the transformer is chosen, in this specific case, to be similar to that of the collector winding so as to derive a positive pulse. One can derive a pulse of opposite polarity by choosing an opposite polarity for the output winding. The equivalent circuit during pulse formation is shown in Fig. 14.3(b).

Let a negative pulse of magnitude *V* be applied at the collector. Then the voltage in the base winding is *nV* and that in the third winding (output winding) is *n*_{1}*V*. The base, emitter and collector terminals are at the same potential.

**FIGURE 14.3(a)** A monostable blocking oscillator with emitter timing

**FIGURE 14.3(b)** The equivalent circuit of Fig. 14.3(a) during pulse formation

Applying KVL to the outside loop, as shown in Fig. 14.3(c):

Therefore,

*V*_{CC} = *V* (*n* + 1)

Hence,

From the base circuit shown in Fig. 14.3(d), the drop across *R* is:

**FIGURE 14.3(c)** The outer loop of the oscillator circuit

**FIGURE 14.3(d)** The base circuit of the transistor

From the given current direction:

For the given current directions, from Eq. (14.14) and (14.15):

However, from Eq. (14.13):

Therefore,

The emitter current as given by Eq. (14.17) is constant. To find *i*_{C} and *i*_{B}, we need to have another relation between *i*_{C} and *i*_{B}. The sum of the ampere turns in an ideal transformer is zero, as shown in Fig. 14.3(b):

where, (*i* × 1) is the ampere turns in the collector winding, (*n* × *i*_{B}) is the ampere turns in the base winding and (*n*_{1} × *i*_{1}) is the ampere turns in the output winding.

From the load circuit:

From the equivalent circuit:

Therefore,

Substituting these values in Eq. (14.18):

Equation 14.22 is the required second relation involving *i*_{C} and *i*_{B}. From Eq. (14.17):

Substituting Eq. (14.23) in Eq. (14.22):

From Eq. (14.13):

Therefore,

From Eq. (14.23):

And from Eq. (14.24):

The emitter current:

Again, using Eq. (14.13):

Therefore,

The waveforms of these three currents, *i _{C}, i_{E}* and

*i*, are plotted in Fig. 14.3(e).

_{B}**FIGURE 14.3(e)** The waveforms of the currents

From the waveforms, it is seen that the collector current is trapezoidal (step followed by a positive ramp), as is the base current (step followed by a negative ramp). However the emitter current is constant during the pulse formation. The waveforms of the voltages are shown in Fig. 14.3(f).

**FIGURE 14.3(f)** The voltage waveforms

From the waveforms shown in Fig. 14.3(f), it is seen that when a negative trigger of magnitude *V* is applied, the voltage at the collector, *V*_{CN} falls by *V*. At the end of the pulse, this voltage is once again required to return to *V*_{CC}. However, a spike appears and eventually decays as the inductor current does not immediately become zero when the transistor is switched into the OFF state. Because of the polarity inversion, the voltage at the base, *V*_{BN} is of positive polarity. The voltage in the output winding is typically the inverted version of *V*_{CN}, but is referenced to the zero level.

At *t* = *t*_{p}, due to regenerative action, the transistor turns OFF and terminates the pulse when *i*_{C} = *h*_{FE}*i*_{B}.

Substituting Eq. (14.24) and Eq. (14.25), at *t* = *t*_{p} in *i*_{C} = *h _{FE}i_{B}*

Therefore,

Usually *n* is small and lies typically between 0.2 and 1. As *h _{FE} >> n* in Eq. (14.27a),

Therefore,

For *t*_{p} to be positive:

Or

This is the condition to be satisfied for pulse formation. Thus, the pulse width is independent of *h*_{FE} and this circuit generates a pulse of stable duration.

##### EXAMPLE

*Example 14.2:* The blocking oscillator with emitter timing shown in Fig. 14.3(a) has the following parameters: *n* = 1, *n*_{1} = 1, *L* = 5 mH, *R* = 0.5*k*Ω, *h*_{FE} = 25 and *V*_{CC} = 10 V. Calculate (a) the amplitude of the trigger; (b) the value of *R*_{L} that allows the pulse formation; (c) the pulse width with *h*_{FE} = 25 and also calculate *t*_{p} using the approximate relation; (d) the base, collector and emitter currents and (e) plot the current waveforms.

*Solution:*

(a) The Amplitude of the trigger pulse is:

(b) The value of *R*_{L} that allows the pulse formation is given by the relation:

As *R*_{L} should be greater than 500 Ω, choose, *R*_{L} = 1000 Ω

(c) The pulse width is calculated using the relation:

The pulse width using the approximate relation is given by:

(d) The base current is given by the relation:

Collector current *i*_{C} is given by:

The emitter current *i*_{E} is given as:

Therefore,

(e) The current waveforms are shown in Fig. 14.3(g).

**FIGURE 14.3(g)** The waveforms of the currents