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Transcript
Self Bias and Source Bias for JFETs and DE-MOSFETs
Self bias and source bias FET circuits do not have a very simple formula for determining the
drain current. The development of their equations is a little more involved than a simple
KVL/Ohm’s Law solution. Instead, they will focus on the device transconductance equation.
Consider a basic self bias configuration:

Vgs 

I d  I dss 1 
 V

gs
(
off
)


2
Expanding this yields
2

2Vgs
Vgs 


I d  I dss 1 

2 
 Vgs( off ) V
gs ( off ) 

I d  I dss 
2 I dssVgs
Vgs ( off )

I dssVgs
2
Vgs( off )
2
Noting that
g m0  
2 I dss
Vgs( off )
The two can be combined to
I d  I dss  g m 0Vgs 
I dssVgs
2
Vgs( off )
2
Due to the fact that gate current is approximately zero
Vgs   I d Rs
Substituting this into the expanded transconductance equation yields
2
I d  I dss  g m 0 I d Rs 
I dss I d Rs
2
Vgs( off )
2
A modest rearrangement yields
2
0
I dss Rs
2
I  1  g m 0 Rs I d  I dss
2 d
Vgs( off )
It should be noticeable that this is a quadratic equation of the form ax2+bx+c and can be solved
using the quadratic formula:
y
 b  b 2  4ac
2a
The positive option in the numerator may be ignored as this occurs for Vgs less than Vgs(off). The
result is
 1  g m 0 Rs  1  2 g m 0 Rs
I d  2 I dss 

g m0 Rs 2





It is useful to graph this in terms of the current ratio using the parameter gm0Rs. To use the graphs
below simply find gm0Rs along the horizontal axis and trace up to the curve. Then move across to
find the ratio of Id to Idss. Multiply this by Idss to find the drain current.
Source Bias
For source bias, the original circuit is modified to include a negative power supply below the
source resistor.
The solution for this circuit is very similar to the preceding but with one twist, namely the voltage
across the source resistor is no longer equal in magnitude to the gate-source voltage but rather the
gate-source voltage plus the source power supply. Therefore,
Id  
Vgs  Vss
Rs
This leads to
Vgs   I d Rs  Vss
This is substituted into the expanded transconductance equation as before, simplified and solved
with the quadratic formula as before. To make the resulting equation more clear, it us useful to
define a “swamping factor”, ks=Vss/Vgs(off). This yields the final equation:
 1  g m 0 Rs 1  k s   1  2 g m 0 Rs 1  k s  

I d  2 I dss 
2




g
R
m0 s


Notice that if no source supply is used ks=0 and this equation reverts backs to the self bias
equation. Also, note that if ks is very large, meaning that the source supply is much greater than
the gate-source off voltage, the current may be approximated as
Id 
The plots were created using FooPlot at fooplot.com
Vss
Rs