Pump controller schematic breakdown

Thread Starter

James Tomy Abraham

Joined Sep 23, 2021
14
Pump Controller PCB Schematic.PNG

This is schematic for a pump controller circuit I got from Drawlogix. I would like to know why they've used diodes IN4150 in the circuit. Are they for preventing current flow in a particular direction? Please assist.

Also I would like to know the function of the 5.6KOhms plus 220nF capacitor that are connected to the input signals. Any signal conditioning ? Please help me with this too.

Cheers,
James.
 

MrSalts

Joined Apr 2, 2020
621
The 1N4150 prevents negative voltage applied to the "max" input from killing the IC.
The 5.6k in series with the 220nF capacitor turns a dc voltage transition (low to high) into a pulse (low-high-low).
 

Thread Starter

James Tomy Abraham

Joined Sep 23, 2021
14
The 1N4150 prevents negative voltage applied to the "max" input from killing the IC.
The 5.6k in series with the 220nF capacitor turns a dc voltage transition (low to high) into a pulse (low-high-low).
Hello MrSalts, thanks for the quick response. I was thinking about what you said after I saw your reply. You mean the IN4150 will take the negative voltage to ground as the diode will be forward biased by the negative voltage at the cathode and the relatively positive voltage (ground) at the anode? Is this a common arrangement while designing circuits in your experience? I read about reverse polarity protection on the internet. Can you tell me if that is the proper description for this arrangement? Looking forward for your assistance.

I will look into the step to pulse signal conversion you mentioned. If you have any resources, please share if not inconvenient.

Thanks and Cheers,
James.
 

MrSalts

Joined Apr 2, 2020
621
Yes, it is reverse polarity protection and many chips even have the reversed biased diodes included in the chip.
For the other question, DC current cannot pass through a capacitor so only a pulse snd dissipation of the voltage occurs.
 

Thread Starter

James Tomy Abraham

Joined Sep 23, 2021
14
Super, thanks.
Yes, it is reverse polarity protection and many chips even have the reversed biased diodes included in the chip.
For the other question, DC current cannot pass through a capacitor so only a pulse snd dissipation of the voltage occurs.
Hi Mr. Salts, I've got a few questions.

Doesn't DC current pass through the capacitor during the charging phase and till the voltage across the capacitor is equal to the voltage source voltage?

When you talked about a conversion to a pulse signal, it means that capacitor will send the current to a peak value and then discharge to zero like a low- high-low transition ? And I want to know if the capacitor will charge and discharge like a pulse waveform ( I mean a 90 degree climb up , then a steady signal, then a 90 degree climb down) . Why is it important for the Schmitt trigger to have a pulse waveform as a signal and not a normal voltage signal like a High signal or a Low signal?

Please help me with the answers.

Regards,
James.
 

MrSalts

Joined Apr 2, 2020
621
The charge across the capacitor will come to a steady state equal to the DC it sees. The pulse will occur on any resistive (or inductive or even another capacitive) load downstream. See the circuit below. The pulses occurs when I pressed the switch.
The simulated capacitor has no leakage so the 100k capacitor drains the positive side of the cap when the switch is released.
In real life, some types of capacitors can hold their charge a very long time and others drain over minutes. That means, the charge on the capacitor will not dissipate shortly after the button is pressed and, therefore, a second button press before the cap discharges will not pass a pulse across the resistor. If the cap is partially discharged, the pulse across the 10k resistor will be proportionally reduced.

14935EB9-C985-4232-B472-5FBBA6F5C230.jpegBE637602-7192-4C13-A94E-89BA4FF272A5.jpeg
 

Thread Starter

James Tomy Abraham

Joined Sep 23, 2021
14
The charge across the capacitor will come to a steady state equal to the DC it sees. The pulse will occur on any resistive (or inductive or even another capacitive) load downstream. See the circuit below. The pulses occurs when I pressed the switch.
The simulated capacitor has no leakage so the 100k capacitor drains the positive side of the cap when the switch is released.
In real life, some types of capacitors can hold their charge a very long time and others drain over minutes. That means, the charge on the capacitor will not dissipate shortly after the button is pressed and, therefore, a second button press before the cap discharges will not pass a pulse across the resistor. If the cap is partially discharged, the pulse across the 10k resistor will be proportionally reduced.

View attachment 249675View attachment 249676
Hi Mr. Salts,thank you very much for the prompt replies and kindly pardon my frequent questions.

As I understand from what you've written above, with the simulation pic., when the switch is pressed, the capacitor charges from 0V to 5V(rising edge of the pulse). When the switch is released, the 100K resistor drains the positive side of the CAP (falling edge of the pulse). Can you tell me what happens to the negative side of the CAP and the 10 K resistor? It might be a basic question, but I failed to understand what you were saying about the pulse across the 10 K resistor.

Regards,
James.
 

MrSalts

Joined Apr 2, 2020
621
capacitor
The yellow line is the voltage charge of the cap as you described.

Switch
the missing line is the graph is the voltage across the switch but you could see that the switching event occurs exactly when the capacitor voltage starts to rise (connection) or when capacitor voltage starts to fall (disconnect).

10k resistor
The green line graph above so the pulse I told you about in my first post. Th green line was the voltage measured across the resistor. In other words, a capacitor is an open circuit gap that prevents DC voltage flow EXCEPT when the DC voltage changes (as in a switch event). Tt change is passed through the cap as a pulse. Note that the resistor is only physically connected to the negative side so it will eventually reach the negative voltage (0) after a switch event.
 

Thread Starter

James Tomy Abraham

Joined Sep 23, 2021
14
capacitor
The yellow line is the voltage charge of the cap as you described.

Switch
the missing line is the graph is the voltage across the switch but you could see that the switching event occurs exactly when the capacitor voltage starts to rise (connection) or when capacitor voltage starts to fall (disconnect).

10k resistor
The green line graph above so the pulse I told you about in my first post. Th green line was the voltage measured across the resistor. In other words, a capacitor is an open circuit gap that prevents DC voltage flow EXCEPT when the DC voltage changes (as in a switch event). Tt change is passed through the cap as a pulse. Note that the resistor is only physically connected to the negative side so it will eventually reach the negative voltage (0) after a switch event.
Hello Mr. Salts, I read the answer a few times and thought about. I understand now what you mean. I saw the colors you used on the CAP and the 10K resistor and the colors on the graph. So it is very clear what is happening in the circuit you sent me. Thanks a lot.

I would like to ask why the pulse signal is important for the Schmitt trigger. Why not use a low- high transition instead of low-high-low transition as you have mentioned in your first answer in the thread? Is it to avoid erratic output generation due to fluctuating inputs?

Cheers,
James.
 
Top