74hct9046 PC2 not correctly work - Page 1
Forum section: electronics / beginners

author: bobAk
author_rank: Contributor
timestamp: January 02, 2021, 06:27:42 pm
content: Hello, I'm trying to get pc2 74hct9046 to work. When working, only a capacitor or a low-frequency filter at the output of 0 volts. When working on a high-resistance load, for verification, the pulse has an amplitude of exactly 500 mV and, as expected, different widths.Why doesn't it work as a switching current source? Why a capacitor or low-frequency filter 0 volts? And where can there be 500 mV for a high-resistance load?A 1 Hz signal is applied to SIG_IN. At COMP_IN 1 MHz. 5 volt levels. Rbias 40K. VCO is not used. INH to the ground. Voltage measurement via opa365. Changing the input signal width, amplitude, frequency, edge speed or phase shift does not change the situation.

author: Benta
author_rank: Super Contributor
timestamp: January 02, 2021, 06:44:33 pm
content: The '9046 has a charge pump output, meaning current source and current sink.There's no reason to expect large voltage swing here, rather the opposite. When locked it should settle at a DC level somewhere between VDD and VSS, depending on your VCO.The charge pump output will "nudge" the filter cap as necessary with current, but you probably won't be able to see it in the output voltage.

author: bobAk
author_rank: Contributor
timestamp: January 02, 2021, 07:03:25 pm
content: The pc2 output is the current pump. I suppose that it should work like this - the transistors of the pump should switch the current direction from the low-frequency filter, cutting off a certain voltage level vco, which is necessary to lock the phase. If do not use a closed feedback loop and VCO, then the voltage across the LPF should change after the phase. But this simply does not happen, the output of the LPF is 0. There is no current flowing in or out. When connecting a high-resistance resistor, just see if there is any signal from the current sources (current pump) appears exactly 500 mV, which should not be. The current source in the pump must try to generate 1 mA.

author: Benta
author_rank: Super Contributor
timestamp: January 02, 2021, 08:41:21 pm
content: It's not clear to me how you are using this device, so I'm fumbling in the dark. A schematic, or at least a block diagram might help.

author: bobAk
author_rank: Contributor
timestamp: January 02, 2021, 09:23:21 pm
content: Forgive me, I'm already looking at it so much, I didn't even think that it might not be clear) By a high-resistance resistor I mean to replace C4 with 1Mohm, only for test. Also, replacing C4 with an RC LPF did not give anything.

author: Benta
author_rank: Super Contributor
timestamp: January 02, 2021, 10:16:48 pm
content: Please explain the purpose of D1.Also, please explain what you expect from this circuit. This is a complete mystery.The '9046 is a phase-frequency detector that will output an error signal when the input and reference frequencies are different.You have 1 Hz and 1 MHz. What on earth do you expect? Of course the output will be extremely close to zero.

author: bobAk
author_rank: Contributor
timestamp: January 02, 2021, 11:26:02 pm
content: D1 in this case does not allow c4 to be discharged when the pc2_out output is "off". I want to get a measurement of the phase offset by charging a capacitor with a constant current. How will the voltage across the capacitor be equal to or near 0? The current of the leakage of the capacitor + diode is ~ 100 pA, the time for measuring the voltage on the capacitor using the ADC is 10 μs, then there is a reset stage. For example, we support a phase shift of 400 ns, the current source outputs 1 mA, the capacitor is 1 nf, the rise time of the signal at the pc2_out output, according to the documents, to a supply voltage of ~ 60 ns, the diode opening time is 5 ns, which means in the worst case the capacitor will be 300 mV. Where am I wrong?

author: Benta
author_rank: Super Contributor
timestamp: January 02, 2021, 11:40:21 pm
content: Quote from: bobAk on January 02, 2021, 11:26:02 pmYou don't understand how a phase-frequency detector (PFD) works.That's where you're wrong.A PFD is designed to synchronize two periodic AC signals in frequency and phase.You supply the PFD with 1 Hz and 1 MHz. A recipe for non-functionality.I still don't understand what you're trying to achieve.