## Emitter current is 2.5 mA less

 With this circuit, i expected emitter current around 9.7-9.8 mA but i am getting only 7.2 mA. I am new to electronics, can anyone explain why i am getting less current. by manikm78 August 22, 2020 VBE = 0.7 volt (difference of voltage between the base and the emitter of the NPN polarized in forward active mode but not fully saturated, to be check further down). The 0.7 is given in the datasheet as VBE ON. Can be as high as 0.77 though, and as low as 0.55. VR1 = 9 - 0.7 = 8.3 volt IR1 = I{sub]B[/sub] = 8.3 / 10 000 hfe = 90, from the datasheet IC = 90 x 8.3 / 10 000 = 0.74 mA Checking polarization: VC = 9 - 0.74E-3 x 1000 = 8.26, so VCE > VBE, ok. If the transistor is closing the loop, you cannot have more than 9 - 0.25 = 8.75 volt (under full saturation, V[sub}CEsat[/sub] from the datasheet), so you cannot have more than 8.75 mA. So, you cannot have 9.7 to 9.8 mA, unless you reduce R2 (and R1 to reach full saturation with the conditions: 0.7 < VBE >= VCE). by vanderghast August 22, 2020 There is a typo, It should be 7.4mA, not 0.74mA. by vanderghast August 22, 2020 Add comment... Please sign in or create an account to comment.

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