SEND AND RECEIVE CIRCUIT ON THE SAME SPLIT POWER SUPPLY

[Oxford]

Gentlemen,

The subject I am on is Pg. 120-121 of Science and communiction circuits and projects (Engineers's MiniNotebook) Forrest Mims III
This circuit has two sides . The sending side uses a microphone to modulate an I.R. beam which is focused onto an I.R. phototransitor on the other circuit. Both circuits are drawn with battery power.
I have assembled the parts on a breadboard using a split +- 12 volt power supply. It appears that the microphone is feeding the receiver circuit directly.
Is there any neat way to prevent the microphone input from influencing the receiver circuit?? Remember that a + and - supply voltage is called for.

Oxford

[Chris Smith]

I don’t have the circuit in front of me, but often placing two diodes in the power path will block a given signal, especially if one is a positive Dc signal, and the other is a negative Dc signal.

IF the signal is AC it wont work as well. The diode has to be held open in order for a AC signal to pass.

[philba]

what makes you think the mike is directly feeding the receiver?

Have you completely shielded the transmitter LED and/or the receiver unit to see if your suspicion is correct.

[Oxford]

I can block the light path (I.R.) or simply disable the sending cicuit and the microphone is still audible on the amplification side. Some how the Mic is feeding thru the common power supply.
I will try to feed one side thru one diode and the other thru another diode from the +12 post. I guess the -12 portion of the circuit should be feed with the diode band toward the -12??

Oxford

[Oxford]

The problem was solved with a diode on the + hookup to the microphone. The sound quality is terrible but the sound stops when the I.R. light path is interrupted.

Oxford

[Chris Smith]

If it worked but its terrible, possibly a isolation transformer is next?

Also a cap?

There are many ways to skin a cat.

[rshayes]

A likely cause of your problem is coupling through the power supplies, as you suggested. The output impedance of the power supplies is probably in the ohm range. The current change when the LED is modulated may be in the 5 milliamp range. Under these conditions, it is possible to have a millivolt or two of voltage change in the power supply when the LED is modulated.

On the receiving side, the light is probably detected with a phototransistor with a load connected to one of these supplies. The signal level at this point is probably also in the millivolt range. The voltage change on the power supply is mixed with the received signal and both are amplified.

This can be reduced by decoupling networks. The first thing to try would be large capacitors, such as a 100 uF electrolytic, directly across the power supplies.

The next thing to try would be adding a resistor between the power supply and the load resistor of the phototransistor. The load resistor is probably fairly high, probably above 10K, so an additional 1K resistor will not have much effect. Add another large capacitor, in the 100 uF range, between the junction of the added resistor and the load resistor to ground. This acts as a low pass filter for AC currents coming from the power supply.

If this is still inadequate, a similar network can be added between the power source and the transmitter. In this case, the resistor will have to be lower, since the transmitter draws much more current. A resistor value in the 10 to 100 ohm range may be the highest that is practical. This forms a low pass filter between the transmitter and the power source.