Is this a D flip-flop?

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genxium

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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?
 

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... 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)
 

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 ~
 

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