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Electronic Gain in Amplifiers

  • Thread starter cjm181
  • Start date
  • #1
69
1

Homework Statement


FIGURE 1 shows the block diagram of a superhetrodyne radio receiver.
In a test a 20 μV signal was fed from the aerial into the first stage of the
receiver, a radio frequency amplifier. This signal is the passed through
several stages of the receiver to eventually appear at the input to the AM
detector. For the AM detector to work satisfactory it requires a minimum
signal level of –3dB(mW). Calculate the required minimum gain of the
fifth intermediate amplifier given the data in TABLE 1.

Fig 1 looks like the below

Ariel (20microV, 75ohms) > Amplifier > Mixer . Band Pass Filter > Intermetiate Amp 1 > Inter Amp 2 > Inter Amp 3 > Inter Amp 4 > Inter Amp 5 > AM Detector (needs -3dB (mW).



Table 1 is below

RF amp = +10dB
Mixer = -7dB
Band pass filter = -1dB
Intermediate amp = +15dB
IA2 = +15dB
IA3 = +20dB
IA4 = 10dB

Homework Equations




The Attempt at a Solution


Find input power level coming into the system:

P = V^2 / R = 0.00002^2 / 75 = 5.3333e-12W (or 5.3333e-9 mW)

Convert mW to dB(mW)

10 x Log (5.3333e-9) = -82.73 dB(mW)

So, the total system gain is the sum of all gains, so if we have -82.73 in, and want -3 out, then:

-3 = (-82.73) + 10 + (-7) + (-1) + 15 + 15 + 10 + IA5

-3 = -20.73 + IA5

IA5 = -3 + 20.73 = 17.73.

Therefore the gain at intermediate amplifier 5 is 17.73dB (or 18dB)

Is this going in the right direction? or correct?

Thanks
 
Last edited:

Answers and Replies

  • #2
69
1
Fig 1 and Table 1
 

Attachments

  • #3
Baluncore
Science Advisor
2019 Award
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Input signal is 20uV in 75 ohms. Convert to power, to mW then to dBm.
W = V^2 / R = 20e-6 * 20e-6 / 75 = 5.333e-12W = 5.333e-9 mW
dBm = 10 * Log( 5.333e-9 ) = -82.73 dBm
Advice; Use a spreadsheet to balance the energy account.
Code:
Description   Debit   Credit dB    Balance dBm
Input signal   –      -82.73 dBm   -82.73
RF amp         –       10 dB       -72.73
Mixer         7 dB      –          -79.73
BPF           1 dB      –          -80.73
IF Amp 1       –       15 dB       -65.73
IA2            –       15 dB       -50.73
IA3            –       20 dB       -30.73
IA4            –       10 dB       -20.73
Detector       –        3 dBm      -17.73
So; IF amp 5 will need gain a gain of +17.73 dB to balance the energy account.

You are correct.
 
  • #4
69
1
Thankyou Baluncore.

I didnt think of using a spread sheet. Thanks for the tip...

Kr
Craig
 

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