Using a piezo-microphone

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Using a piezo-microphone

In the schematic above, the 100K potentiometer is adjusted until the transistor's collector current is approximately 100 microamps (10^-4amps). This is measured with a voltmeter across R_E until the voltage is approximately 0.33 volts. Please watch the video for a full explanation of the transistor model and how we build up to this circuit.

Gain Calculation

Following along with the assumptions made in the video, and ignoring R_M and the 10uF capacitor, the gain of the amplifier is approximately 22K/3.3K = 6.7. But at higher frequencies, the 10uF capacitor looks like a short, and the gain of the amplifier is roughly 22K / (3.3K // R_M). This last part in the denominator (R_A//R_B) means the parallel combination of two resistances, which equals (R_A*R_B)/(R_A + R_B).

If the resistance loading the emitter is small, then we can't make the same assumption that we made in the video that a change in voltage at the base will necessarily be followed by an almost identical change in voltage at the emitter. Previously, the change in V_BE was small enough compared to the change in voltage across R_E. But if R_E is small or zero, the change in V_BE will become important and limit the gain of the amplifier. Here's how this comes out of the equations:

As noted in the video, the two governing equations for this simple model of the NPN bipolar junction transistor (in forward active operation) are:

• I_C = β I_B
• I_B = I_S e^V_BE/V_th

The first of these says that the collector current is simply β times the base current, and the second says that there is a steep (exponential) relationship between V_BE and base current. (If you're curious about numbers here, then for our 2N3904 transistor, β is typically 100-200, I_Sis typically around 10^-14 amps, and V_th is about 25mV at room temperature.) Now we can also write, by definition, that the V_BE is simply the difference in voltages at the base and emitter:

• V_BE = V_B - V_E

and then, for a resistance R_E between the emitter and ground, we know that V_E = R_E * I_E, and since I_E is approximately equal to I_C, V_E = R_E * I_C. We plug this in, and move V_E to the other side of the equation, to get:

• V_B = R_E*I_C + V_BE

This just means that the voltage from the base to ground is a combination of the base to emitter voltage and the voltage drop across the emitter resistor R_E to ground. But this equation is interesting, because on the left side is V_B which is our input voltage, and on the right side, everything is a function of I_C, the collector current, which is related to our output voltage:

• ΔV_OUT = - R_C * ΔI_C

This is just Ohm's Law for the resistor R_C -- an increase in current through it (and also through the collector) increases the voltage dropacross it (which is why this is negative). We can rewrite this in a calculus way as a derivative:

• dV_OUT / dI_C = - R_C

And now that the above equation connects I_C to V_OUT, to complete the full gain chain and get from V_B (the input) to V_OUT, we really just need to connect from V_B to I_C, because:

• dV_OUT / dV_B = (dV_OUT / dI_C) * (dI_C / dV_B)

Fortunately, we can actually calculate (dI_C / dV_B) from the equations we wrote a few steps ago. Starting with V_B = R_E*I_C + V_BE, we can take the derivative of both sides with respect to I_C and find:

• dV_B / dI_C = R_E + dV_BE/dI_C

This last little part dV_BE/dI_C can be found right out of the two transistor equations we had earlier:

• I_C = β I_S e^V_BE/V_th
• dI_C / dV_BE = (1/V_th) * β I_S e^V_BE/V_th

where the (1/V_th) comes from taking the derivative of the exponential. Now, we can notice that the second part of the last equation is actually I_C itself, and we can simplify:

• dI_C / dV_BE = (1/V_th) * I_C
• dI_C / dV_BE = I_C / V_th

or taking one divided by both sides (inverting both fractions):

• dV_BE / dI_C = V_th / I_C

As we said before, V_th is about 25 millivolts at room temperature, and I_C was picked to be 100 microamps, so in this case,

• dV_BE / dI_C = (0.025 volts) / (0.0001 amps)
• dV_BE / dI_C = 250 ohms.

So now we can go back and substitute this in to figure out the amplifier gain:

• dV_B / dI_C = R_E + dV_BE/dI_C
• dV_B / dI_C = R_E + V_th / I_C
• dI_C / dV_B = 1 / (R_E + V_th / I_C)
• dV_OUT / dV_B = (dV_OUT / dI_C) * (dI_C / dV_B)
• dV_OUT / dV_B = (-R_C) / (R_E + V_th / I_C)

So if R_E is much greater than V_th / I_C (250 ohms in this case), then we get back to our original simplified gain equation that was derived in the video, dV_OUT / dV_B = - R_C / R_E. But now if, for example, R_E = 0, then while the earlier gain equation would have suggested infinite gain (dividing by zero), our new one would suggest that it's really - 22K / (R_E + 250), or a (negative) gain of about 88 for R_M = 0. With R_M = 1K, as in our actual potentiometer in our circuit, the (negative) gain is about 18. Having a potentiometer or just swapping out resistors lets you change the gain over a wide range.