TPS3839G33 circuit for read switch trigger - Page 1
Forum section: electronics / beginners

author: moci
author_rank: Newbie
timestamp: October 16, 2024, 05:47:37 am
content: Hi, newbie here, I would appreciate some help with the circuit attached.Circuit uses voltage supervision chip TPS3839G3, PMOS and passive components.Although, intuitively, I can say that circuit limits time P-MOSFET conducting and thus saves power, I have a problem understanding the circuit.Mostly capacitor is my problemHow it should charge/discharge, maybe more how it discharges...

author: moci
author_rank: Newbie
timestamp: October 18, 2024, 05:27:38 am
content: My understanding is that when RESET pin is pulled low, capacitor with start charging through rail Vcc - R220k - C - R100 - GND and PMOS will be open. While it is charging it will reach threshold voltage on the Gate and PMOS will stop charging, thus it will limit time period in which PMOS conducts. However when RESET pin is pulled high and we have charged capacitor, I don't know how circuit behaves.Anyone that could help with that?

author: Harry_22
author_rank: Frequent Contributor
timestamp: October 18, 2024, 03:51:05 pm
content: Hello!First of all we need to make the circuit more primitive and understandable.There are MOSFET keys at the output U3 (Reset). We replace them with a SPDT key.The gate of the transistor Q3 is an open circuit.Now we take the timing diagram from the data on U3 and draw a voltage line on the gate.1-2: the state of uncertainty U3. Let's skip it for now.2-3: The left plate of C2 is connected to ground (0V), the right one is charged to Vcc through R2 (Please see Pic. 1).3-4: The output of U3 is reconnected to the Vcc bus. The left plate is under voltage Vcc. Since the charge on the capacitor is inert, the right plate should theoretically increase its voltage at Vcc up to 2Vcc. But then there is a Schottky diode, which will open, and through it the voltage on both plates will be equalized by the amount of the drop on the diode.The capacitor discharges through resistor R1 which somewhat limits the current and therefore the pulse front on the gate is slightly delayed (Please see Pic.2)4-5: The diode closes and the capacitor discharges through R2. Since R2 is large, the time is stretched. Now both plates of C2 are charged to Vcc.Try to follow it further yourself using this approach.

author: moci
author_rank: Newbie
timestamp: October 18, 2024, 05:07:40 pm
content: Manny thanks.I am a bit lost with this part though:"The capacitor discharges through resistor R1 which somewhat limits the current and therefore the pulse front on the gate is slightly delayed (Please see Pic.2)"In step 3-4 I understand that RESET pin is = Vcc so not clear how C discharging that way?Btw, how do you simulate this, using Easyeda, do you create your own simplifications for PMOS and voltage supervisor or?

author: Harry_22
author_rank: Frequent Contributor
timestamp: October 18, 2024, 07:39:56 pm
content: I drew a curve for a circuit without diode to make it easier to understand how R1C2 and R2C2 circuits behave. Where R1<<R2.PSI was a little mistaken here, excluding the diode circuit automatically excludes resistance R1. As we see, it is much less than R2, which means it does not affect the behavior of the curve. This means that at the moment of switching - point 3, the voltage on the Gate will instantly change to 2Vpp.Next, C1 will be in a closed circuit with R2. On its left plate the voltage is Vpp, on the right it is twice as much. The charge from the right plate will flow through the resistor until the voltage decreases to Vpp. This will require more than 5*R2*C1, that is 11s.I will change the picture later on, unfortunately have no access to my comp.

author: Harry_22
author_rank: Frequent Contributor
timestamp: October 21, 2024, 08:43:35 am
content: Sorry for the delay.So, above we simplified the circuit by replacing the output U3 with a relay and Q3 transistor Gate with an open circuit. We replace complex circuit parts with primitives for a better understanding of their operation.In the absence of U3 power (Sensor pads don’t connected), its output is in an undefined state. We will consider this later, but for now will assume that it is connected to Ground bus.In such a steady state (the current does not flow), the left plate of the capacitor is connected to the GND through R1 resistor and the right one is connected to Vcc power bus through R2 resistor. Since the capacitor plates are not connected to each other, then in the state of rest we can replace it with an open circuit.Please see the voltage diagrams Reset, Left plate and Right plate as well as simplified circuit for this time period 2-3 at the picture below.At the next moment, at point 3, Reset output is connected to Vcc bus. I suggest that we first consider the operation of the circuit without a diode. In this case the series circuit Vcc-U3-P1-C2-P2-Vcc is formed. Since R1 resistance is much less than R2, we can replace R1 with a short circuit. This is another simplification.The fact is that the capacitor is an inert element in terms of voltage change (the degree of inertia of which is characterized by capacity). This means that it resists a change in the voltage difference on its plates. It follows from this that the voltage surge on Vcc left plate will cause an instantaneous change in voltage on the right plate by the same amount.As we mentioned earlier, during 3-4 the current is formed (see the following picture). The capacitor temporarily became a "battery" with an electromotive force of Vcc. As the "battery" discharges through the resistor R2, the voltage on the right plates decreases until it equalizes fixed left. In this case it is Vcc. The initial slope of the discharge curve corresponds to the slope of the line from the initial point to the point with the time constant t1. After time 5*t1, the curve will almost merge with Vcc.The greater the resistance R1 and capacitance C2, the longer the transient process continues and vice versa.Now we return the diode D4. When the voltage on the right capacitor plate is sufficient to open the diode, it stops increasing at the level of Vcc + V_D4drop. Now the resistor R1 plays a role. It limits C2 recharging current. The voltage changes only on the left С2 plate. Curve 3-4.The discharge curve of a capacitor with a time constant t2 is constructed in exactly the same way.When the left plate is charged to the drop voltage D4 it will close and resistor R2 will appear in the circuit again.The right plate of the capacitor will discharge through resistor R2. Curve 4-5.At points 5 the reset key switches back to ground. Left plate of C2 decrease by Vcc to 0V. The same happens on the right plate immediately.After the right plate will charge up to Vcc via R2 resistor. Curve 5-6 below.PSIf you do not have any questions about the described transition processes, we will move on.

author: moci
author_rank: Newbie
timestamp: October 21, 2024, 07:47:58 pm
content: I understand everything. Step 5 is a bit less intuitive to me, that instant drop. But since prior state was Vcc on booth plates, than it makes sense instant change to 0 on booth plates. It is nicely explained with those graphs. 🙏

author: Harry_22
author_rank: Frequent Contributor
timestamp: October 22, 2024, 08:25:32 am
content: Okay, now let's look at the switch transistor Q3. This is an induced channel MOSFET. P-N conductivity arrow points away from the gate, where P is positive and N is negative. This means that a P channel must be induced by the gate to open transistor. To create a P channel you need to attract positively charged particles (and repel negative ones).To do this we apply a negative voltage to the gate relative to the source.The threshold voltage level for CEM4435 ranges from -1 to-2.5V according to datasheet.Let's plot on our picture the area where the voltage at Gate will be less than at Source by this amount.Here transistor Q3 will be open (please see the picture below).

author: Harry_22
author_rank: Frequent Contributor
timestamp: October 23, 2024, 02:00:29 am
content: Hi!I still couldn’t stand it and transferred this circuit to the emulator.Not everything turned out to be simple. The 1N5817 Schottky diode has very high reverse leakage. That is, we connect a 2.5 k resistor in parallel with R2.That's why I replaced it with 1N4148

author: Harry_22
author_rank: Frequent Contributor
timestamp: October 23, 2024, 02:28:29 am
content: I would recommend you to use TPS3831. It has a manual reset. In this case, there will be no state of uncertainty as there is now before power is applied.And using a ground bus for the sensor is more correct.Please note that the power supply to the microcircuit is up to 6.5 volts.In attachment you will also find the file for Microcap

author: moci
author_rank: Newbie
timestamp: October 23, 2024, 05:28:15 pm
content: Finally have full picture on this circuit. Need to spend some time exploring simulator.... Thanks a lot for all explanations and suggestions.

author: Harry_22
author_rank: Frequent Contributor
timestamp: October 24, 2024, 12:41:15 am
content: You are welcome!PSI would not recommend using Schottky diode in this circuit. The high leakage current creates a large uncertainty in the delay time.