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对于典型应用(输入电压(VI)低于10V),您无需考虑最大允许基极电压,但是当VI高于10V时,则需格外小心。
最大输入电压(VI)主要取决于以下因素:
1.串联基极电阻(R1)的值(即R1的功耗)
2.最大集电极电流(IC)
最大输入电压因BRT(例如,hFE)使用条件的不同而有所变化,但是如果R1值较小,则该电压通常取决于集电极电流;如果R1值较大,则取决于R1的值。
1. 串联基极电阻(R1)的值与输入电压(VI)之间的关系
BRT的内置电阻的允许功耗为1/8W。假设处于“导通”状态的内部晶体管的基极-发射极电压为Vbe。然后,流过R1(IB)的电流用以下等式表示。为简单起见,设Vbe =0.7V。
IB=(VI – Vbe )/R1 =(VI – 0.7)/R1
由于IB而引起的R1的功耗不得超过1/8W。因此:
1/8 W > R1 * { ( VI – 0.7 ) / R1 }^2
> ( VI – 0.7 ) ^2 / R1
VI < √( R1 / 8 ) + 0.7
2. 最大集电极电流(IC(最大值))和输入电压(VI)之间的关系The 将最大集电极电流(IC(最大值))规定为绝对最大额定值,即使瞬时也不得超过该最大额定值。
由于BRT在饱和区工作,因此其hFE在10-20的范围内。可通过从上面计算的IB值减去流至R2(IR2)的电流来计算流至内部晶体管(Ib)的基极(b)电流:
Ib = IB – IR2 = ( VI – Vbe ) / R1 – Vbe / R2 = ( VI – 0.7 ) / R1 – 0.7 / R2
由于集电极电流(IC)等于Ib乘以hFE,因此必须满足以下等式:
IC (max) > IC = hFE * IB = hFE * {( VI – 0.7 ) / R1 – 0.7 / R2 }
VI < R1 * IC(max) / hFE +( R1 + R2 ) * 0.7 / R2
根据上述两个因素计算得出的VI值中的较小者即为最大允许基极电压。
1. 串联基极电阻值(R1)
VI < √( R1 / 8 ) + 0.7
2. 最大集电极电流(IC(最大值))
VI < R1 * IC (max) / hFE +( R1 + R2 ) * 0.7 / R2
图2和图3显示了在不同电阻值条件下输入电压与集电极电流之间的关系。
尽管最大输入电压及其限制因素取决于BRT的hFE(IC/IB),但以下为正确情况: