带隙基准电压电路图.ewb

Multisim2001软件的仿真电路实例261例

Electronics Workbench Circuit File
Version: 5
Charset: ANSI
Description: 
"带隙基准电压电路图" 
"        在集成电路内部，往往需要建立低内阻点，以作为内部电压源。对这种电压源的要求是：" 
"         1. 等效微变电阻要低。" 
"         2. 直流电压对温度不敏感。" 
"         3. 对器件参数变化不敏感。" 
"         4. 有些情况还要对电源供电电压不敏感。" 
"        实现恒压的方法有：在电路中利用二极管工作于击穿状态；采用温度补偿；引入电压负反馈。" 
"        作为基准电压源，要求有精确的温度补偿，一种专作电压基准的带隙基准电压集成块，是利用UT的正温度系数和UBE的负温度系数相对消而获得零温度系数。两个结面积相等的发射结，当流过的电流不等时，其结电压之差便正比于UT。" 
"        图中A是一个高增益的放大器，输出电压和标有“－”的输入电压反相，和标有“+”号的输入电压同相，由图可见，这个放大器同时加有正反馈和负反馈，适当选择电路元件，使负反馈超过正反馈，而且是深度负反馈。这样，放大器的两个输入端，便可视作虚短路和虚开路。假定VT1和VT2具有相同的特性，即具有相同的工艺结构。则根据结面积相等，电流不等的特点。设流过R1、R2的电流分别为I1、I2。" 
"        有I1*R1=I2*R2                    ……（11-1）        Ube=Ut*ln(Ic/Is)        ……(11-2)" 
"         Ube1=I2*R3+Ube2              ……（11-3）" 
"        由（11-2）得：Ube1-Ube2=Ut*ln(I1/I2)        ……（11-4）" 
"        由（11-1）和（11-4）得：Ube1-Ube2=Ut*ln(R2/R1)        ……（11-5）" 
"        由（11-3）和（11-5）得：I2=（Ut/R3）*ln(R2/R1)        ……（11-6）" 
"        由（11-1）和（11-6）得：I1*R1=（Ut*R2/R3）*ln(R2/R1)        ……（11-7）" 
"        由图知：Uo=Ube1+I1*R1=Ube1+（Ut*R2/R3）*ln(R2/R1)        ……（11-8）" 
"        对（11-8）两边分别对温度T进行求导得常数（-2*10^(-3)V/℃)，(k/q=0.86*10^(-4)V/℃)" 
"        在室温下 ，UBE=660mV，UT=26mV，UR=660+23.25*26=1260mV=1.26V，由于这一电压十分接近硅的带隙电压，故这种基准电压称为带隙基准电压。" 
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