Decoupling high-frequency amplifiers

A radio receiver circuit contains several amplifiers and oscillators powered by a common DC voltage source. Problems can arise because the wire connecting the DC supply to the various amplifiers is not an ideal short circuit. Radio signals from one amplifier are communicated to another via the common DC line. Not only does this invalidate the standard theory of how the circuit works but also it may cause serious malfunction of the radio receiver because of consequent self-oscillation of amplifiers due to unwanted positive feedback.

A standard solution to the problem is to place L,C filters between each amplifier and the DC supply line (as shown in the diagram). The filter is called a decoupling mechanism because it inhibits the unwanted coupling of amplifiers via the DC supply.

The diagram on the right is drawn to facilitate analysis of the filter as a circuit which inhibits radio-frequency signals getting from the DC supply line in to the amplifier while having negligible effect on the supply of DC current to the amplifier. The real inductor can be modelled as an ideal inductor L together with a small series resistance r. The large resistance R is mainly due to the amplifier itself. In practice it would be so use that value. At a frequency of 10MHz it would be appropriate to choose and C=22nF. The value of r might come to . The analysis of the filter needs to be done at two frequencies only viz 0Hz (DC) and 10MHz. At DC, L can be treated as a short circuit and C as an open circuit (why?). At 10MHz the effects of r can be neglected compared to (show this).

Working to these approximations calculate :

• The fraction of the DC supply voltage that appears across the capacitor.
  

  SOLUTION:
  

  Refer to the diagram on the right. Remember that at DC ( ) the capacitor is an open circuit and the inductor is a short circuit. The amplifier is represented in the diagram on the right by the resistor R. The DC supply that the amplifier actually gets is what appears across R, say.
  

  

• The fraction of a 10MHz voltage on the DC supply that appears across the capacitor.
  

  SOLUTION:
  

  The magnitude of the impedance of the inductor at 10MHz is :

  

  The magnitude of the impedance of the capacitor at 10MHz is :

  

  Clearly the impedance of r is << than that of L and the impedance of C is << than that of R. So the 10MHz voltage that has managed to leak in to the DC line will come to the amplifier by an unwanted path. Its magnitude can be calculated by thinking of the voltage divider consisting of L and C alone.
  

  

  So this filter will be an effective decoupling mechanism.