# Adding Hysteresis to Supply Voltage Supervisor

I am going through this TI-APP note - Voltage Supervisor,with hysteresis.

I am not able to understand WHY they have derived the below equation 1 & equation 2 in the App note (page 2 & page 3)

I understand that the equation is obtained by using the KCL at the nodes.

My questions :

1. Why should I apply KCL there?

2. How is performing KCL there linked to finding the voltage at which RESET pin will change its state?  • Q1 - Personal preference but most EE prefer nodal analysis over the mesh. Q2 - the "switching" will occur when Vs = V_threshold. – G36 Apr 5 at 9:47
• Yes. But the threshold voltage is not included in the calculation. And Why does the RESET voltage state change when the current is in this condition (based on Equation 1)? This is what I am not understanding. Suppose, the current through R1 is like 3mA. And based on some value of resistance, the current through R2 and Rh, may split as 2mA & 1mA or vice versa or any other combination. What relation does this have with the voltage ? Please help me with an answer – Newbie Apr 5 at 10:20
• I don't know why they have gone to this; Vs is set by a simple voltage divider that only differs from when Vout is low or high. – Peter Smith Apr 5 at 10:54

In this example, the Ti engineer is using Nodal analysis at $$\V_S\$$ node.

Nodal analysis is used to find the voltage at a given node, and this method is base on KCL and some "mathematic magic".

But we can use any network analysis technique we know or like/prefer.

We have two cases.

1 - Input voltage increases

$$V_{S+} = V_{1+} \times \frac{R_2||R_{TH}}{R_2||R_{TH} + R_1} = V_{1+} \times\frac{\frac{R_2 R_{TH}}{R_2 + R_{TH}}}{\frac{R_2 R_{TH}}{R_2 + R_{TH}} + R_1}$$

2 - Input voltage decreases

$$V_{S-} = V_{1-} \times \frac{R_2||(R_{TH}+R_P)}{R_2||(R_{TH}+R_P) + R_1} +V_2 \times \frac{R_1||R_2}{R_1||R_2 + (R_{TH} + R_P)}$$

As you can see I used a voltage divider equation and superposition theorem.

Why is the voltage divider formula used in this circuit to find the terminal voltage?

Now in the example, we have

$$\V_{1+} = 2V\$$

$$\V_{1-} = 1.8V\$$

$$\V_{2} = 1.8V\$$

$$\V_{S-} = 0.4V\$$

$$\V_{S+} = 0.406V\$$

$$\R_{TH} = 1M\Omega\$$

$$\R_P = 100k\Omega\$$

So, we have two equations and two unknowns thus we can solve for $$\R_1\$$ and $$\R_2\$$. 