Flip Flop outputs 99% duty cycle when connected to Op Amp

Thread Starter

Signboy

Joined Jun 10, 2016
11
I'm trying to simulate a circuit in MultiSim as a proof of concept before putting parts to breadboard, but I'm baffled at the behaviour of the Flip Flop's output. In the pictures, the flip flop's job is to make a square wave, an octave below the input. The integrator at the bottom is supposed to turn that square wave into a triangle. The problem is that the flip flop works beautifully, until I plug it into the op amp. Then the square wave turns into an unintegratable 99% duty cycle mess.
Is this a multisim glitch, or something that I don't know yet about how things work?

not plugged in- good flip flop
screen1_working.PNG
plugged in- bad flip flop
screen2_not_working.PNG
 

crutschow

Joined Mar 14, 2008
30,137
Try reducing R5 and R7 to a lower value, such as 1kΩ.
Your virtual ground impedance may be too high.
You should also place a 0.1μF decoupling cap across each resistor.
 

AnalogKid

Joined Aug 1, 2013
10,100
Both R5 and R7 must have bypass capacitors in parallel. Start with a 0.1 uF ceramic and a 10 uF electrolytic in parallel with each resistor.

9 V is too high a power rail for 74LS devices. 5 V max. If the FF is operating between V+ and GND, the high impedance GND is acting as an unintended feedback path.

What is the purpose of D1?

Rather than connect it to the opamp, connect U3A pin 5 to a 100K resistor to GND and post that scope shot.

ak
 

Thread Starter

Signboy

Joined Jun 10, 2016
11
At least in MultiSim, connecting the FF output to a load resistor gives the exact same readout as in the first picture of the original post.
This shows the FF out to a load resistor, as well as scope A on the U1, showing the diode's purpose of amplified half wave rectification.
screen3_FF-RL_OscA-U1.PNG

What do the bypass caps at the power supply do? It somewhat fixed things, but I'm not sure why.

As for the power to the FF, isn't it at 4.5/0 V? The attempted idea here is to power the circuit with a single 9 V, as
(aw, crap. I realize I haven't said much about the circuit goal)
The goal is to take a bass guitar signal, make an octave down triangle wave out of it, then use that triangle wave to modulate a low pass filter.
I intend for it to fit in a nice little stompbox housing, and run off either a single 9 V battery, or AC adapter.
The circuit in the pictures is just trying to make the triangle wave, I haven't even tackled the filter yet.

After the suggested modifications, the integrator's output looks like a negatively clipped sawtooth wave. This must be because of some unintentional path allowing the C4 to discharge very quickly, right?
screen4_FF-Int.PNG
 

AnalogKid

Joined Aug 1, 2013
10,100
A fundamental assumption of all opamp circuits (and just about everything else) is that there is a ZERO ohm impedance between all voltage sources and GND at (here's the important part) all frequencies of interest. For a TTL gate the harmonics extend out to the 10's of MHz no matter what the switching frequency is. Also, opamps in the real world do *not* like non-zero power source impedances. What they do in simulation is entirely up to the program.

You need to confirm the relationship between the digital part and analog part power supplies. The software might be making assumptions or applying default conditions that are not correct for your circuit.

The U2 non-inverting input is tied to GND and the inverting input cannot go below GND - *assuming* U3A is powered by +5 V and GND. Again, depending on the model this probably won't work.

ak
 

Thread Starter

Signboy

Joined Jun 10, 2016
11
Thanks for the advice, Analog Kid. I'll look into some other ways to divide the frequency without mixing signals like that.
 

Audioguru

Joined Dec 20, 2007
11,248
How do we or Multisim know how the TTL IC is powered when it is not powered?
You cannot "ground" a fairly high current TTL IC with a couple of 10k resistors. A low dropout 5V voltage regulator IC should have been used instead.

Why use a 50 years old TTL IC? You could have used a single supply opamp and a CD4013 Cmos flip flop.
Then the circuit has everything grounded properly and does not have a virtual ground, it has a real ground.
The CD4013 works from a supply voltage from 3V to 18V.
 

AnalogKid

Joined Aug 1, 2013
10,100
Thanks for the advice, Analog Kid. I'll look into some other ways to divide the frequency without mixing signals like that.
I think you misunderstood. There is nothing fundamentally wrong with your approach, but the circuit needs a little more attention to detail. As AG says, changing some of the components might help.

Separate from that, consider this. A bass can make a range of different frequencies. Assuming a perfect pickup-to-square-wave converter and divide-by-two, that gives you a varying frequency square wave into the integrator. But the integrator has a fixed time constant, so the triangle wave amplitude will be inversely proportional to the frequency. For each octave change at the bass, the triangle wave amplitude will drop by 50%. I don't know if this is a feature or a problem for your overall system; just pointing it out.

ak
 

Thread Starter

Signboy

Joined Jun 10, 2016
11
Yeah, I was planning on just accepting that the filter would sweep a little differently at 30 Hz than at 500 Hz. It's not optimal, but if I can get this project to work in a remotely useful way, I'll be happy.
As per AudioGuru's suggestions, I'm having a look into some "rail-to-rail" op amps & CMOS flip flops, although I have never worked with either. A lot of the stuff in our school lab is... actually stamped USSR.
Once again, thanks for the input guys, I really appreciate it.
 
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