# Is this a D flip-flop circuit?

1. Dec 23, 2011

### genxium

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 $\bar{Q}$=1.

But I have no idea how to figure out Q when D=0&CLK=1, can anyone give me a hint? I wanna ask if the 2 left most BJTs with collector resistor are working as inverters?

#### Attached Files:

• ###### D flip-flop.JPG
File size:
77.2 KB
Views:
89
2. Dec 26, 2011

### es1

I can't see how this circuit would hold a value as the tail current is shared between the diff amps. One of them has to loose their bias depending on the state of clk. If I am right and it cannot hold a value it is not a flop.

Who labeled the inputs? My guess is that this is a comparator with an offset trim and hysteresis.

I only looked at this really fast so I could be totally off base though.

3. Dec 26, 2011

### es1

Actually, you know what. I think clk and nclk are supposed to be a select. You either select between a normal diff amp or a diff amp with both its inputs tied high.

I was thinking the other structure was for calibration or something but then I realized the bias is 1mA and the resistors and it dawned on me that this is not an IC.

I have no idea why one would do this though.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook