# Armstrong Oscillator and Feedback Mechanisms Resulting

TMA 3 question Help

Q5 (e) The given L1:L2 ratio is not necessarily the optimum value to give a good sinusoidal output. [A Fourier probe on the output will give a spectral response]. Try to devise an experiment to find the optimum ratio*. Express the ratio as a turns ratio.

The report should include copies of any graphical responses produced in the investigation.

*This can be done by performing an AC sweep on the inductance parameter L2. However a sweep cannot be performed without a voltage source in the circuit. For purposes of this analysis a small voltage source can be inserted into the feedback loop as shown in the second version of the circuit on Blackboard.

The simulation software used is SIMATRIX SPICE which can be downloaded for free From: https://www.simetrix.co.uk/downloads/. This is an urgent request SPICE files and question paper uploaded. Any help an advantage. No description or support material in distance learning course work. Question not detailed enough. No support yet from tutor. I think you some how find the turns ratio from the graphs produced in the SIMATRIX SPICE files rather than just use equations for transformers.

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#### Solution Preview

Please find attached. Note I haven't had time to check the arithmetic,

Analysing the above schematic we can write the output equation as the input to the gain block

multiplied by the gain, thus

s_0=G.(s_i-Hs_o)

Collecting like terms we can write

s_o+GHs_o=Gs_i

s_o (1+GH)=ã€–Gsã€—_i

Re-arranging we can therefore write the transfer equation (ratio of output signal to input

signal) as

s_o/s_i =G/(1+GH)

Examining the above we consider the case when the denominator goes to zero. This results

in unstable oscillatory action and will occur when the product of gain and feedback portion

equals -1, GH=-1

Thus the conditions for oscillation (known as the Barkhausen criteria) are that

The magnitude of the gain/feedback product is unity, |GH|=1

The phase change of the feedback portion is anti-phase with the input, 180^0phase difference

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We need to convert the dB value to a linear value so

H=ã€–10ã€—^((H(dB))/10)=ã€–10ã€—^(-10/10)=ã€–10ã€—^(-1)=0.1

Condition for oscillation when GH=1 so we calculate the required value of G from

G=1/H=1/0.1=10

This is equivalent to a gain in dB determined as

G(dB)=10.Log(10)=10 dB

Note in addition the loop feedback fraction would also need to introduce the 180^0 phase

change for oscillation to occur

As the microphone is brought closer to the load speaker (considered as the output of the feedback system) a proportionally larger signal is feedback to the microphone (considered the input of the feedback system). This has the effect of increasing the feedback portion or H in the feedback equation. As H rises so the gain feedback product GH (for a set amplifier gain G) increases eventually becoming unity

(|GH|=1 condition). This produces the positive feedback and oscillation in the system, resulting in howling.

The howling effect can by mitigated by

Reducing the gain of the amplifier to such an extent as to reduce the gain feedback product below unity again

Make the microphone more directional so in effect reducing the feedback portion and making the gain feedback product come in below unity again

1% of the output power represents a feedback fraction H=0.01

The point at which howling occurs is when |GH|=1 thus

G=1/H=1/0.01=100

This is equivalent to an amplifier dB gain of

G(dB)=10.Log(100)=20 dB

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(a) The transformer dot notation represents ends of the transformer coils that are in phase with each other. For controlled oscillation to occur three things need to be provided

We provide a means of amplification in the circuit which is achieved using the BJT transistor

We provide unity loop gain with of loop phase in the feedback loop (ie we provide positive feedback)

We provide a frequency selective network

Looking at the ac equivalent circuit (figure b) we can see that the BJT common emitter arrangement provides of phase shift. This output is then fed to the grounded end of the primary winding. The secondary transformer winding is configured in the reverse polarity to the primary so with reference to the dot notation, which will represent points where voltages are of the same polarity, the secondary transformer coil will produce a voltage waveform in anti-phase to that ...

#### Solution Summary

Analysing positive and negative feedback, the Armstrong Oscillator and determining the optimal turns ratio of transformer to provide optimal oscillation from the Armstrong oscillator. Also a description of the workings of the Armstrong Oscillators operation are described in detail