Uo=-KoUi (2) Substitute Eq.(2)into Eq.(1),we get: U-U。=- 1 I.= . U. R . R (3) Where Ko is the open loop gain(normally 105~10) Hence I,can be calculated if Uo is measured.We choose the feedback resistance R=1MO and digital voltmeter with the range of 200mV and resolution of 0.01mV.Then the minimum measurable current is: 0.01mV 1, =1×10-1A 1M2 Therefore it can be seen that this I-V transformer has a high sensitivity. 2.The I-V characteristics of the PN junction. It can be known from the solid state theories that the ideal forward I-V relationship of PN junction fulfills the following equation: (4) Where is the forward current passing through the PN junction,lo the inverse saturated current (proportional to the property of semiconductors and doping condition),U the forward voltage applied on the PN junction,T the absolute temperature,ka the Boltzmann constant,and e the elementary charge.At room temperature,e/ksT-38,exp(eU/ksT)>>1,so Eq.(1)can be approximately written like: eU I=1o exp (5) At room temperature,the forward current of PN junction changes exponentially with the forward voltage,so the current is very small with small voltage and thus need I-V transformer for the measurement.The above rule can be proved if the I-V characteristics can be measured. Furthermore,if the temperature T is measured,the Boltzmann constant can be concluded with the value of elementary charge already known.Take the logarithm of both sides of Eq.(5),we get: In /=In/o+ eU (6) kgT Taking In/and U as variables and making least square fitting,e/ksT and then ka can be calculated.In the experiment (see Fig.3),U is replaced by U and I=U2/Rs so Eq.(6)can be changed to: nU=(,+nR)小+e (7) pT The fitting can just be accomplished using U as abscissa and InU2 the ordinate. During the experiment,good results can normally not be obtained if using a diode as the sample.The reason is:(a)There is a depletion layer current,which is linearly proportional to exp(eU/2kaT);(b)there is a surface current that is linearly proportional to exp(eU/mkaT),m>2.Uo=-KoUi (2) Substitute Eq. (2) into Eq. (1), we get: 1 1 −   = = − +  −     i o o o s f f o f U U U U I R R K R (3) Where Ko is the open loop gain (normally 105~106 ) Hence Is can be calculated if Uo is measured. We choose the feedback resistance Rf =1MΩ and digital voltmeter with the range of 200mV and resolution of 0.01mV. Then the minimum measurable current is: 0.01 11 1 10 1 − = =   s mV I A M Therefore it can be seen that this I-V transformer has a high sensitivity. 2. The I-V characteristics of the PN junction. It can be known from the solid state theories that the ideal forward I-V relationship of PN junction fulfills the following equation: 0 exp 1     = −         B eU I I k T (4) Where I is the forward current passing through the PN junction, I0 the inverse saturated current (proportional to the property of semiconductors and doping condition), U the forward voltage applied on the PN junction, T the absolute temperature, kB the Boltzmann constant, and e the elementary charge. At room temperature, e/kBT~38, exp(eU/kBT)>>1, so Eq. (1) can be approximately written like: 0 exp   =     B eU I I k T (5) At room temperature, the forward current of PN junction changes exponentially with the forward voltage, so the current is very small with small voltage and thus need I-V transformer for the measurement. The above rule can be proved if the I-V characteristics can be measured. Furthermore, if the temperature T is measured, the Boltzmann constant can be concluded with the value of elementary charge already known. Take the logarithm of both sides of Eq. (5), we get: 0 ln ln = + B eU I I k T (6) Taking lnI and U as variables and making least square fitting, e/kBT and then kB can be calculated. In the experiment (see Fig.3), U is replaced by U1 and I=U2/Rf, so Eq. (6) can be changed to: ( ) 1 2 0 ln ln ln = + + f B eU U I R k T (7) The fitting can just be accomplished using U1 as abscissa and lnU2 the ordinate. During the experiment, good results can normally not be obtained if using a diode as the sample. The reason is: (a) There is a depletion layer current, which is linearly proportional to exp(eU/2kBT); (b) there is a surface current that is linearly proportional to exp(eU/mkBT), m>2