Is this a D flip-flop?

[genxium]

This component is from the ADS 2008 example, 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 \[\bar{Q}\]=1.

But I have no idea how to figure out Q when D=0&CLK=1, can anyone give me a hint?

 

[erikl]

... how to figure out Q when D=0&CLK=1, can anyone give me a hint?

Then D_Bar=1 , node Q1=VCEsat(BJT4)+VCEsat(BJT11)+V(I_DC) < VBE(BJT15) , BJT15=off , Q=0 (via R16)

 

[genxium]

Then D_Bar=1 , node Q1=VCEsat(BJT4)+VCEsat(BJT11)+V(I_DC) < VBE(BJT15) , BJT15=off , Q=0 (via R16)

Thanks a lot~！ It really helps ~