南航考研模拟电子线路（模电）第四章03 - 图文

电子线路基础.........+...33.......hongfeng@nuaa.edu.cn2015/12/16-.电子线路基础在双平衡差分电路中，输出电压为：uo???iC1?iC3???iC2?iC4???RC???iC1?iC2???iC4?iC3???RC差分对管中电流和偏置电流的关系为：?u1??iC1?iC2?iC5th??2U??T?同理：?u1??iC4?iC3?iC6th??2U??T?所以：uO??iC52015/12/16?u1?iC6??RC?th??2U?T34????hongfeng@nuaa.edu.cn电子线路基础由图可得：iC5?IOY?iRY因此有：可得：iC6?IOY?iRYiRYuY?RY?????u12uYuO??RC?th??RY?2UT当电压u1??UT时，上式可简化为：2RCRCu1uO??uY???u1uY?K1?u1uYRY2UTRYUTRC式中K1?。RY?UT2015/12/1635hongfeng@nuaa.edu.cn电子线路基础非线性补偿电路中，有：uC8?uC7?u1二极管的压降和电流的关系为：uA?uC7iD1所以u1?UTlniD2则：iD1?UTlnISuA?uC8iD2?UTlnISiD1?iC7?IOX?iRXiD2?iC8?IOX?iRX所以uX???1???uX?IOX RX???u1?UTln?2UTarcth??IR??uX?OXX??1???IOX RX36?hongfeng@nuaa.edu.cn2015/12/16?uX?IOX?RXuX?IOX?RX