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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.

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

$V_{gg} = (V_{g\max} − V_{g\min})/2 = (0.2 − 6.0)/2 = −2.9 V,$

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.