- #1
genxium
- 141
- 2
This component is from the ADS 2008 example(in Frequency Divider by 2, Design Guide), I don't quite understand how it works clearly.
By far I think that if D=0&CLK=1 , then the collector current Ic of BJT1 is small, resulting in that Vc of BJT1 is High, hence Vb of BJT14 is High and BJT14 is active, so [itex]\bar{Q}[/itex]=1.
But I have no idea how to figure out Q when D=0&CLK=1, can anyone give me a hint? I want to ask if the 2 left most BJTs with collector resistor are working as inverters?
By far I think that if D=0&CLK=1 , then the collector current Ic of BJT1 is small, resulting in that Vc of BJT1 is High, hence Vb of BJT14 is High and BJT14 is active, so [itex]\bar{Q}[/itex]=1.
But I have no idea how to figure out Q when D=0&CLK=1, can anyone give me a hint? I want to ask if the 2 left most BJTs with collector resistor are working as inverters?