Chapter 13
Q. 13.6
FET FREQUENCY DOUBLER DESIGN
A 12–24 GHz frequency doubler is designed using a GaAs MESFET with the following parameters: V_{t} = −2.0 V, R_{i} = 10 Ω, C_{gs} = 0.20 pF, C_{ds} = 0.15 pF, and R_{ds} = 40 Ω. Assume the operating point of the transistor is chosen so that V_{g\max} = 0.2 V,V_{g\min} = −6.0 V,V_{d\max} = 5.0 V,V_{d\min} = 1.0 V, and I_{\max} =80 mA. Find the conversion gain of the multiplier.
Step-by-Step
Verified Solution
We first use (13.74) and (13.75) to find the peak value of the AC input voltage.
The gate bias voltage is
and the peak AC input voltage is
V_{g} = V_{g\max} − V_{gg} = 0.2 + 2.9 = 3.1 V.Then the input power is given by (13.76):
P_{in}=\frac{\mid {V_{g}}\mid ^{2}R_{i}}{2\mid {R_{i}-j/\omega _{0}C_{gs}}\mid ^{2}} =\frac{(3.1)^{2}(10)}{2[(10)^{2}+(1/2\pi (12\times10^{9} )(0.2\times10^{-12}))^{2}]} =10.7 mW.The pulse width is found from (13.73) as
\cos \frac{\pi \tau }{T}=\frac{2V_{t}-V_{g\max}-V_{g\min}}{V_{g\max}-V_{g\min}}=\frac{2(-0.2)-0.2+6.0}{0.2+6.0}=0.29.for
\frac{\tau }{T}=0.406,Then the load current for the second harmonic is given by (13.72b):
I_{2}=I_{\max}\frac{4\tau }{\pi T } \frac{\cos (2\pi \tau /T)}{1-(4\tau /T)^{2}}=0.262 I_{\max}=21.0 mA,The load resistance required to match the transistor is found from (13.78):
R_{L}=\frac{V_{d\max}-V_{d\min}}{2I_{2}}=\frac{5-1}{2(0.021)}=95.2 \Omega ,The output power at 24 GHz is given by (13.79):
P_{2}=\frac{1}{2}\mid {I_{2}}\mid ^{2}R_{L}=\frac{1}{2}(0.021)^{2}(95.2)=21.0 mW.Finally, the conversion gain is, assuming the input is conjugately matched,
G_{c}=\frac{P_{2}}{P_{avail}}=\frac{21.0}{10.7} =2.9 dB.The load reactance required to resonate the second harmonic is X_{L} = 1/2ω_{0}C_{ds} =44.2 Ω, which corresponds to an inductance of 0.293 nH.