I am following TI's pdf Non-inverting microphone pre-amplifier circuit with AOM-4544P-2-R microphone. I probe the circuit on the Vout after the op-amp, (not on the mic directly). I am playing around with the resistor and capacitor values to see practical changes. I noticed something about RL (or R6 in TI's notes).
When I use mic's load resistor RL=5k62, I get Vout+ = up to ~0.45V and Vout- = down to ~0.6V:
But when I use a lower value resistor of RL=2k2, I get a more "uniform" output where both Vmin and Vmax are +-0.6 V respectively:
A disclaimer, I tap my finger on the mic to generate noise, so its not 100% persistent, but I understand that there is this voltage difference and voltage uniformity when swapping RL.
So my schematic:
My probing:
Questions:
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The expected behavior would be, since I use a lower resistor as the mic's load, I would get a larger Vout, because more current flows through the mic (just an assumption of mine), but I guess the opposite happens? TI's pdf does not mention that RL is part of any equation, it just mentions these:
and that (but I cant tell if that is the reason behind the behavior):
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How can I read voltage below 0 V on the oscilloscope? I power the op-amp with 5/0 V. How can I see -0.6 V on its output?
Best Answer
To answer your second question first:
While the op-amp will indeed produce an output potential between 0V and +5V, with a mean DC value somewhere in that range, that DC component is removed by C16 and R9. They form a high pass filter that removes DC, and whose output will have a mean potential of 0V (no DC component present). A simple simulation will illustrate this:
simulate this circuit – Schematic created using CircuitLab
I use V2 to produce a 1V amplitude sinusoidal potential, and V1 simply adds a DC offset of 2.5V. Here's a plot of \$V_{IN}\$ (blue) and \$V_{OUT}\$ (orange) over time:
As you can see for \$V_{OUT}\$, the DC offset has gone, and the signal is now centered about zero, which means it becomes negative to retain the sinusoidal form.
You said:
The opposite happens, larger \$R_L\$ causes larger \$V_{OUT}\$. Here's why. You must first understand the microphone's internal construction, shown in the blue box here:
simulate this circuit
Voltage source V1 represents the microphone's capacitive diaphragm, which produces microvolts of fluctuations in response to incident sound waves.
The JFET here behaves like a voltage-controlled current sink, where the ratio of change in gate potential \$\Delta V_G\$ to the resulting change in drain current \$\Delta I_D\$ is called transconductance \$g_m\$:
$$ g_m = \frac{\Delta I_D}{\Delta V_G} $$
or
$$ \Delta I_D = g_m \times \Delta V_G \tag1 $$
In that second form, you can see than \$g_m\$ is a kind of gain, but instead of a varying output voltage, in response to a varying input voltage, the output is considered to be the current flowing through the JFET's channel.
Any change in that current will, by Ohm's law produce a change in the voltage \$V_L\$ across RL:
$$ \Delta V_L = \Delta I_D \times R_L $$
$$ \Delta I_D = \frac{\Delta V_L}{R_L} \tag2 $$
Substituting eq. 2 into eq. 1:
$$ \frac{\Delta V_L}{R_L} = g_m \times \Delta V_G $$
We can rearrange this to obtain voltage gain \$\frac{}{}\$:
$$ \frac{\Delta V_L}{\Delta V_G} = g_m R_L $$
In other words, whatever change in voltage occurs at gate G appears with \$g_mR_L\$ greater amplitude across \$R_L\$. Since the top of \$R_L\$ is fixed at +5V, it must be the bottom end D that is changing potential, and that's why we consider D to be the microphone's output.
Voltage gain, then, is proportional to resistance \$R_L\$. Output signal amplitude increases if \$R_L\$ increases, which can be seen in the plot below. Blue is the output OUT1 from the left circuit with \$R_L=10k\Omega\$, and orange is OUT2, on the right, where \$R_L=5k\Omega\$: