Attenuators are passive devices. It is convenient to discuss them along with decibels. Attenuators weaken or attenuate the high level output of a signal generator, for example, to provide a lower level signal for something like the antenna input of a sensitive radio receiver. (Figure The attenuator could be built into the signal generator, or be a stand-alone device. It could provide a fixed or adjustable amount of attenuation. An attenuator section can also provide isolation between a source and a troublesome load.
Constant impedance attenuator is matched to source impedance ZI and load impedance ZO. For radio frequency equipment Z is 50 Ω.
In the case of a stand-alone attenuator, it must be placed in series between the signal source and the load by breaking open the signal path as shown above. In addition, it must match both the source impedance ZI and the load impedance ZO, while providing a specified amount of attenuation. In this section we will only consider the special, and most common, case where the source and load impedances are equal. Not considered in this section, unequal source and load impedances may be matched by an attenuator section. However, the formulation is more complex.
T section and Π section attenuators are common forms.
Common configurations are the T and Π networks shown in Figure
Multiple attenuator sections may be cascaded when even weaker signals are needed as in Figure
Voltage ratios, as used in the design of attenuators are often expressed in terms of decibels. The voltage ratio (K below) must be derived from the attenuation in decibels. Power ratios expressed as decibels are additive. For example, a 10 dB attenuator followed by a 6 dB attenuator provides 16dB of attenuation overall.
10 dB + 6 db = 16 dB
Changing sound levels are perceptible roughly proportional to the logarithm of the power ratio (PI / PO).
sound level = log10(PI / PO)
A change of 1 dB in sound level is barely perceptible to a listener, while 2 db is readily perceptible. An attenuation of 3 dB corresponds to cutting power in half, while a gain of 3 db corresponds to a doubling of the power level. A gain of -3 dB is the same as an attenuation of +3 dB, corresponding to half the original power level.
The power change in decibels in terms of power ratio is:
dB = 10 log10(PI / PO)
Assuming that the load RI at PI is the same as the load resistor RO at PO (RI = RO), the decibels may be derived from the voltage ratio (VI / VO) or current ratio (II / IO):
PO = V O IO = VO2 / R = IO2 R
PI = V I II = VI2 / R = II2 R
dB = 10 log10(PI / PO) = 10 log10(VI2 / VO2) = 20 log10(VI/VO)
dB = 10 log10(PI / PO) = 10 log10(II2 / IO2) = 20 log10(II/IO)
The two most often used forms of the decibel equation are:
dB = 10 log10(PI / PO) or dB = 20 log10(VI / VO)
We will use the latter form, since we need the voltage ratio. Once again, the voltage ratio form of equation is only applicable where the two corresponding resistors are equal. That is, the source and load resistance need to be equal.
Power into an attenuator is 10 watts, the power out is 1 watt. Find the attenuation in dB.
dB = 10 log10(PI / PO) = 10 log10 (10 /1) = 1 log10 (10) = 10 (1) = 10 dB
Find the voltage attenuation ratio (K= (VI / VO)) for a 10 dB attenuator.
dB = 10= 20 log10(VI / VO)
10/20 = log10(VI / VO)
1010/20 = 10log10(VI / VO)
3.16 = (VI / VO) = AP(ratio)
Power into an attenuator is 100 milliwatts, the power out is 1 milliwatt. Find the attenuation in dB.
dB = 10 log10(PI / PO) = 10 log10 (100 /1) = 10 log10 (100) = 10 (2) = 20 dB
Find the voltage attenuation ratio (K= (VI / VO)) for a 20 dB attenuator.
dB = 20= 20 log10(VI / VO )
1020/20 = 10 log10(VI / VO )
10 = (VI / VO ) = K
The T and Π attenuators must be connected to a Z source and Z load impedance. The Z-(arrows) pointing away from the attenuator in the figure below indicate this. The Z-(arrows) pointing toward the attenuator indicates that the impedance seen looking into the attenuator with a load Z on the opposite end is Z, Z=50 Ω for our case. This impedance is a constant (50 Ω) with respect to attenuation– impedance does not change when attenuation is changed.
The table in Figure lists resistor values for the T and Π attenuators to match a 50 Ω source/ load, as is the usual requirement in radio frequency work.
Telephone utility and other audio work often requires matching to 600 Ω. Multiply all R values by the ratio (600/50) to correct for 600 Ω matching. Multiplying by 75/50 would convert table values to match a 75 Ω source and load.
Formulas for T-section attenuator resistors, given K, the voltage attenuation ratio, and ZI = ZO = 50 Ω.
The amount of attenuation is customarily specified in dB (decibels). Though, we need the voltage (or current) ratio K to find the resistor values from equations. See the dB/20 term in the power of 10 term for computing the voltage ratio K from dB, above.
The T (and below Π) configurations are most commonly used as they provide bidirectional matching. That is, the attenuator input and output may be swapped end for end and still match the source and load impedances while supplying the same attenuation.
Disconnecting the source and looking in to the right at VI, we need to see a series parallel combination of RO, RI, RO, and Z looking like an equivalent resistance of Z, the source/load impedance: (Z is connected to the output.)
Z=RO + (R2 ||(R1 + Z))
For example, substitute the 10 dB values from the 50 Ω attenuator table for R1 and R2 as shown in Figure .
Z = 25.97 + (35.14 ||(25.97 + 50))
Z = 25.97 + (35.14 || 75.97 )
Z = 25.97 + 24.03 = 50
This shows us that we see 50 Ω looking right into the example attenuator (Figure ) with a 50 Ω load.
Replacing the source generator, disconnecting load Z at VO, and looking in to the left, should give us the same equation as above for the impedance at VO, due to symmetry. Moreover, the three resistors must be values which supply the required attenuation from input to output. This is accomplished by the equations for R1 and R2 above as applied to the T-attenuator below.