Nuts and Volts - December 2008

be 10 mV (5/5/100) and the current through the diode will be 1 mA. I designed the filter for a 100 μs period; you could make C1 larger and use a longer period but you have to wait five time constants before taking the Vtp measurement. With 1 μF, the time constant is: R1*C1 = 0.1 Meg * 1 μF = 0.1 second. You must wait at least 0.5 seconds for good accuracy.

I was going to run it from the five volts of the micro supply, but realized that there is not enough head room to measure a blue LED, so that is why VCC = nine volts. When measuring a red LED, Vtp is about three volts. That leaves six volts drop across Q1, and at 100 mA the power dissipation is 0.6 watts. That is a bit too much for a 2N3904, so I am recommending 2N2219. I don’t think you will need a heatsink.

SOLENOID DRIVER

QI need help making a circuit that activates a solenoid with a switch. The amount of time that it stays activated needs to be adjustable and it needs to activate as fast as possible. Can you help me?

— Dalton Jager

AThis circuit (Figure 7) will work with input from 5 to 15 VDC. I had to make some assumptions about the solenoid: operating voltage is 12 VDC; current draw is one amp. Most mechanical relays will pull in 15 milliseconds, so I used that number for the time the solenoid

■ FIGURE 7

■ FIGURE 8

will have to pull in.

Assuming that the solenoid will stay pulled in down to eight volts input, then the capacitor to energize the solenoid should be: C = I*dt/dV = 1 amp* 15 ms/4V = 4000 μF. You can use this same formula if you have a different solenoid.

The way this works is: The 555 turns on Q1, which turns on Q2. Q2 pulls C1 up to double the voltage on the solenoid. When the charge on C1 is depleted, current flows through D1 to keep the solenoid pulled in as long as the 555 output stays high. I would use a fast diode for D1, like a 1N4936. For Q1, use Si3442DV or similar; for Q2 use TIP42.

TELEPHONE LINE MONITOR AGC

QI need an audio amplifier that will provide an equal level of output for both sides of a telephone conversation. I searched the web for a schematic or a kit that will provide automatic gain control in this kind of application, but could not find one. We currently use a 600:600 ohm audio transformer (in series with capacitors) bridged on the POTS line on the input side and an LM386 on the other side of the transformer. This allows monitoring (and/or

QUESTIONS & ANSWERS

recording) of both sides of the conversation for quality control and order confirmation purposes. However, many times the caller can barely be heard while the agent’s voice is extremely loud. Can you suggest a circuit that will output both sides of the conversation at the same level?

— Stan Grupinski

AIn the circuit in Figure 8, the 2N3904 is acting as a variable resistor. It is very non-linear so it is necessary to keep the signal level small; in this case, about 10 millivolts. The amplifier gain is 101 so the output is about one volt p/p. R4 and the capacitors C1 and C2 are a low pass filter to provide some averaging and not respond to peaks. This slows the attack time but the attack time is about 200 ms, which should not be objectionable. R5 discharges the capacitors so that low level signals can be attenuated less. The decay time is about 300 ms. I assumed that you were using a single supply, so used IC1B to provide VCC/2 and prevent clipping of the signal. The circuit should work from 5 VDC to 12 VDC.