Brandeis University |
Physics 29a |

Fall 2016 |
Kevan Hashemi |

# Lecture 8: Transistor Radio

## Diode Output by Analytical Approximation

## Diode Output by Numerical Integration

The following Pascal program integrates the sinusoidal exponential numerically, and so produces a plot of detector output versus input amplitude for many orders of magnitude. The units of amplitude are volts, and the units of the detector output are volts.

{
Crystal Diode Response 03-MAY-12. We integrate exp(a*sin(x)) numerically so as
to obtain the rectification response of a crystal diode. This is a Pascal source
code file.
}
program p;
const
dt=0.001;
pi=3.141592654;
a_scale=1.1;
a_min=0.0001;
a_max=10.000000;
vT=0.0271; {this is kT/q for T=310K}
fsd=10;
fsr=1;
var
i:integer;
integral:real;
a,y,t:real;
begin
a:=a_min;
while a<=a_max do begin
integral:=0;
t:=0;
while t<=1.0 do begin
integral:=integral+exp(a*sin(2*pi*t)/vT)*dt;
t:=t+dt;
end;
y:=vT*ln(integral);
writeln(a:fsr:fsd,' ',y:fsr:fsd);
a:=a_scale*a;
end;
end.

## Measured versus Calculated Response

In this circuit, we use a P-type Schottky diode with *I*_{S} = 3 μA to detect incoming 146-MHz radio frequency power. We connected a 146-MHz signal of known amplitude to our demodulator and measured the voltage on output of the diode detector. The plot below shows the detector output we calculate using our numerical integration, as well as our measurements.

Because the saturation current of the diode is 3 μA, its equivalent resistance for zero bias is around 10 kΩ. Compare this to several megaohms for a silicon PN diode. A PN diode made out of germanium can have saturation current as high as 10 μA, which gives an equivalent resistance at zero bias of around 3 kΩ.