Flipflop's output voltages are 0V

Question

I'm building a 4-bit asynchronous up counter with D flip-flops. Unfortunately the 4 output voltages are always 0V. Can someone please tell me why?

I had already placed a digital source in front of the first ‘D’ input, and that didn't help either.

I just saw that there is a Power Pins visible checkbox in the settings of every flip-flop. I checked the checkbox and saw a VCC pin appear. I connected a 5VDC source there. That didn't help either.

Edit: At \$U_3\$, no signal is coming in. Thus, I guess \$U_2\$ doesn't work.

Edit 2: I separated the first flip-flop from the second. With the first one, I should also see the output voltage change. Well, nothing happens. Regardless of whether it is a digital or pulse source. No matter which Youtube video I watch. All interconnect the flip-flop as shown in the picture. Nothing. Probably because of the program. 🙄

I'm using PSpice for TI2020

Answer 1

Solution of the problem for future readers

That has now cost me 12 hours. I have to add that I'm a beginner and don't do it full-time. Apart from that – so much disinformation and at the same time problems in the program made the whole project difficult. Well, now I know how to do it.

Start a project and choose “Create based upon an existing project”; with “AnalogGNDSymbol.opj”. That was said in a forum post that I found while researching.

Now build the counter. To do this, select “DFFRH” for the flip-flops. I said in the question above that it should be enough to set the “DIGINITSTATE” to 0 or 1 in the settings—and you have to continue to do that—but according to another forum post I found, something is going wrong internally. Long story short: it just doesn't work. Therefore, please select the “DFFRH”. A pulse source is connected to the ‘reset’ inputs, which ‘fires’ once at the beginning to clear all undefined states. Imagine in practice a button that you press briefly once and which brings a voltage to the ‘reset’ inputs.

One more thing: By the way, it is like I show an up counter. I had first found a website according to which it is supposedly a down counter when you go from ‘Q’ to the ‘Clock’ of the next flip-flop. Unfortunately, that was wrong. What is pictured on the website is an up counter. That's a pity. I was so confused at times that I briefly thought this video was wrong – but it was the only right one. I mean, well, there could be other solutions – that probably depends on the flip-flop. Even in my old university documents—and fortunately I took a few screenshots with my smartphone at the time—it says that a forward counter \$\overline{Q}\$ has to be connected to the next ‘clock’ input.

Anyway, it looks good this way (it starts at 2.5 seconds).