Hey, so I've probably beaten the preamp threads dead but I was curious about how the math checked out for the frequency response of this MPF102 (credit: James Hawes) take on the Tillman preamp. I wired it up and sounds great, but wanted to see how the response looked on paper.

Using the AC model for the JFET:

Where:

RL = R4 (in parallel with load resistance)
RD = R2
RS = R3
C = C1

Plotting this shows that it appears as a normal high pass filter, but I wanted to figure out the equation for the -3db frequency:

\(
\frac{V_{o}}{V_{i}} = [\frac{-g_{m}R_{D}}{1 + g_{m}R_{S}}][\frac{sR_{L}C}{1 + s(R_{L} + R_{D})C}]
\)

The conclusion was that while solving for the -3dB frequency, there was a range of load resistances 'RL' that would result in a negative/zero denominator under the square root which doesn't make sense to me.

Using the resistor values given in the original design I calculated that RL would have to be larger than around 3.5 Kohm for the calculation of f3dB to have a valid denominator under the root. With this in mind, I plotted the magnitude of the frequency response with a resistor within this 'unaccepted range' and it showed to be normal...

I am stumped. I don't know why the math doesn't check out... Any ideas?

Thanks,
JP

 

The circuit has a voltage gain of only about 1.4 times, slightly more than a piece of wire but it has a very high input impedance suitable for a guitar pickup preamp.

The high input impedance of the preamp allows the inductance of the pickup to resonate with the capacitance of its output cable at about 4kHz producing a peak of about +14dB (creating a crisp sounding TWANG) then the response drops off at higher frequencies, 6kHz is at the same level as low frequencies and 10kHz is at -12dB.

The datasheet for the MPF102 Jfet shows that it works very well at 400MHz.
Your drain resistor is 1.5k so stray capacitance to ground in its circuit of 10pF will cut frequencies above 10.7MHz (yes, a high radio frequency), but the guitar pickup has no output above about 15kHz.

 

Oh, you are calculating the low cutoff frequency of C4 and its load?
If the amplifier it feeds has an input impedance of 10k ohms then the cutoff frequency is 1 divided by (2 x pi x 10k x 4.7uF)= 3.4Hz, far lower than a guitar speaker can go.

 

Oh, you are calculating the low cutoff frequency of C4 and its load?
If the amplifier it feeds has an input impedance of 10k ohms then the cutoff frequency is 1 divided by (2 x pi x 10k x 4.7uF)= 3.4Hz, far lower than a guitar speaker can go.

Here is the frequency response using the small-signal model of the JFET that I found:

\(
\frac{V_{o}}{V_{i}} = [\frac{-g_{m}R_{D}}{1 + g_{m}R_{S}}][\frac{sR_{L}C}{1 + s(R_{L} + R_{D})C}]
\)

\(
\frac{V_{o}}{V_{i}}|_{f \rightarrow \infty} = [\frac{-g_{m}R_{D}}{1 + g_{m}R_{S}}][\frac{R_{L}}{R_{L} + R_{D}}]
\)

Why isn't the drain resistor taking part in the cutoff frequency for your calculation as it is in the response?

Breaking down the frequency response into its magnitude and solving for f-3dB I get the following equation:

\(
f_{3dB} = \frac{1}{2\pi C\sqrt{R_{L}^{2} - R_{D}^{2} - 2R_{L}R_{D}}}
\)

Subbing in:

C = 4.7 uF
RD = 1.5K
RL = 10K

\(
f_{3dB} = 4.11 Hz
\)

Maybe I am just thinking too much but based on the circuit model this makes sense to me...