1、仿真电路
2、真值表
a |
b |
f |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
0 |
0 |
3、解析
3.1 已知:a = 1, b = 1
c = a & b = 1
d = a | b = 1
e = !c = 0
f = d & e = 1 & 0 = 0
3.2 已知: a = 1, b = 0
c = a & b = 0
d = a | b = 1
e = !c = 1
f = d & e = 1 & 1 = 1
3.3 已知: a = 0, b = 1
c = a & b = 0
d = a | b = 1
e = !c = 1
f = d & e = 1 & 1 = 1
3.4 已知: a = 0, b = 0
c = a & b = 0
d = a | b = 0
e = !c = 1
f = d & e = 0 & 1 = 0
4、用我们前三篇文章的电路来仿真一下试试
关于上图的解释,咱们只有一个5V的电源,或门是需要负电压的,所以只能用两个1K的电阻分成2.5V,0V和-2.5V,但是输出需要近似于5V,所以采用了一个NPN和一个PNP把状态放大,NPN只关心基极和发射极的压差,只有基极是高电位时,才会导通,而PNP只关心发射级和基级的压差,利用这点,我们才得以去选择更低的负2.5V。
上一版貌似虽然 LED亮了,但并不能输出5V电压,这一版就可以输出近似5V了。
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