Dual-channel ten-segment level display circuit

It is a dual-channel ten-segment level display circuit composed of TA7666P and TA7667P. The circuit takes the input voltage
effective value 100mV as the 0∞ level. The low-power five-segment display is driven by TA7666, and the high-level five-segment display is
driven by TA7667. If the input stage gain of the two circuits is OdB (RF/ RS =1 at this time), then the fifth
(last-lit) LED emits light when the effective value of the input voltage is 912mV. Here, the fifth point LED of TA7666P is required to indicate a 2dB level, which corresponds
to the specified OdB voltage. The first point LED of TA7666P and TA7667P respectively displays -4cB and 106dB. This in
turn determines the gain of the input amplifier based on the 912mV input signal.
    First determine the gain required for the TA7666P. Assuming OdB=100mV, then -4dB-63.1mV, the required gain (voltage
amplification factor) Avl=912/63.1=14.5 times (23.2dB).
    For TA7667P, since +6dB-2×100mV, the gain Av2=912/200 -4. 45 times (13.2dB).
    Select the respective RF and Rs according to Avl, A, and tolerance. These are the 43k/12k (TA7667P)
and 160k/12k (TA7666P) marked in Figure 5-56.
    In the figure, there are two Rs connected to the input end of the same channel, so the input impedance is the parallel value of the two Rs
6kflo. This is the load of the preamplifier of the level display circuit , so the Rs value should not be too small. In addition, the output impedance of the preamplifier
should also be much smaller than Rs, otherwise the set level will not match the actual level.
    The value of the current-limiting resistor R connected in series with the light-emitting diodes needs to be determined according to the power supply voltage. For example, vcc - 9V,
the output conduction voltage of TA7666P7 7667P is about 1V, the forward voltage drop of LED is VF=2V, and the standard value of general LEI) luminous current is
lOmA, so R; (9V - 2V- 1V)/lOmA=600n ． In practice, take the nominal value 620flo