Espelho de Corrente Desequilibrado

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Step of 4 13.027E Draw the imbalance current mirror circuit I I I Q₆ + + VBES VBE6 - - I R R₁ Figure 1 Apply voltage law at input loop of Figure 1. (1) Step of 4 Draw the equivalent circuit diagram with emitter resistors. - + + - - VBES VBE6 I Figure 2 Step of 4 Apply Kirchhoff's voltage law at input loop of Figure 2. (Since = The difference in emitter base voltage causes change in (2) Substitute equation (2) in equation (1). AR I The error is, AR = AR Hence, the error = is proved. Step of 4 Consider the value of resistor R is, The value of emitter resistor =2.63kQ = Determine the value of AR AR=0.02R = = Determine the value of error AR R + AR Substitute 2.63kQ for AR and for R 0.02kQ 0.02kQ = 3.65kQ 0.00547 Therefore, the value of error