5 ~' Y* |+ A" v- o) a5 y0 a从图我们可以得知,R1取样A路电压,R2取样B路电压,R3取样C路电压.
' \8 u! Z) v6 X5 ~/ H计算步骤:
& N. S9 P2 e. `6 f6 W5 k$ O1 C1.计算总取样电流Iq值.
' b, @; M" ^( Y. d4 D0 U5 X3 Q- U, o: u总取样电流Iq=2.5/R4=2.5V/10K=0.25mA
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这里设总取样电流为0.25mA,所以R4取10K
$ s1 n: p8 E, T4 N. Q5 B& ]# c2.计算R1值.
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A路取样电流是Ia=Iq*[IoA/(IoA+IoB+IoC)]
8 h" o" I6 v/ j3 w" d Ia=0.25mA*[3.5A/(3.5A+2A+10A)]
" e: s# p3 s7 T* I% j Ia=0.056mA
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R1=(VoA-2.5V)/Ia=(12V-2.5V)/0.056mA
[" ?' v0 U4 z6 ?% Y( v7 ]# ? R1=170K(可用150K与20K电阻串联)
4 b; Q* I4 B0 f0 o2 o X4 v3.计算R2值.
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B路取样电流是Ib=Iq*[IoB/(IoA+IoB+IoC)]
# k! W* H1 P- l/ [' r& j Ib=0.25mA*[2A/(3.5A+2A+10A)]
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Ib=0.032mA
/ @6 r# @ }# Q$ T# q* A R2=(VoB-2.5V)/Ib=(5V-2.5V)/0.032mA
% l! V+ t1 A/ m& ^& e: x R2=78K(可用56K与22K电阻串联)
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4.计算R3值.
3 ` w% m1 \/ \ F- m7 B- o' R8 [ C路取样电流是Ic=Iq*[IoC/(IoA+IoB+IoC)]
6 f2 X. t9 m6 X% o6 T3 L8 c8 @ Ic=0.25mA*[10A/(3.5A+2A+10A)]
7 x+ W' i- p6 P" a- t" } Ic=0.16mA
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R3=(VoC-2.5)/Ic=(3.3V-2.5V)/0.16mA
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R3=5K
7 O0 ]0 e; d$ u5.验算.
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IR1电流=(12V-2.5V)/(150K+20K)=0.0558mA
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IR2电流=(5V-2.5V)/(56K+22K)=0.032mA
2 A4 G: S9 m( ]) ~9 Y5 e' B4 ^ IR3电流=(3.3v-2.5v)/5k=0.16mA
* \7 t/ L) |1 D+ m7 A( b) R+ h 实际取样总电流=0.0558mA+0.032mA+0.16mA
# R& K+ w: n$ W' W- U. ~7 \ =0.2478mA
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约等于理论计算值的2.5mA,说明理论计算OK!下一步就是调试环节了,再根据实际情况作相应的更改