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I have landscape lights. The load is 3 VDC 100 mA for each filament. I have two of them so the load is 200 mA at 3 VDC. I have solar panel which varies voltage from 0 V to 1.5 VDC. I want the landscape lights on when solar panel voltage is between 0 V to 0.5 VDC and off above 0.5 VDC.

Attached image. I am not sure what is the purpose of 100 kΩ resistor? Can anyone take a look and advise.

Also, I am not sure how the values of 100k, 470 ohms, 1k ohms values are derived. Can someone help me understand this circuit in depth along with my 100 kΩ resistor question what is its purpose?

Below image came from this site user name Michal Podmanicky post #33. Used here only to explain how the circuit works as the forum did not address fully how the circuit works exactly when this post was posted:

https://forum.allaboutcircuits.com/threads/bjt-as-switch-doesnt-work-why.200942/page-2

enter image description here

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  • \$\begingroup\$ I can't tell if the question is for an explanation of the circuit or if it's about getting a solution that activates a 3V system at 200 mA if and only if a solar panel (being used as a light/dark detector) is at or less than 500 mV. Which question is it? \$\endgroup\$ Commented May 20, 2024 at 18:49
  • \$\begingroup\$ sorry to confuse you. I have questions not a question. It is actually both understand the circuit and to achieve solution light to turn on/off at 500mV while doing hysteresis. \$\endgroup\$ Commented May 20, 2024 at 18:56
  • \$\begingroup\$ Is there an assumption being made that the circuit shown can be made to do the work, in practice, without changing its topology? \$\endgroup\$ Commented May 20, 2024 at 18:59
  • \$\begingroup\$ I don't follow your question. The circuit is in simulation and design phase not constructed yet if thats what your asking. \$\endgroup\$ Commented May 20, 2024 at 19:03
  • \$\begingroup\$ I simply asked if you already know for certain from prior experimentation that this topology can be made to work by only altering resistor values once you know how to calculate them. Or if you are assuming that's so. \$\endgroup\$ Commented May 20, 2024 at 19:26

2 Answers 2

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The 100 kΩ resistor provides some hysteresis to the circuit. It forms a potential divider with the 2k2 resistor. As VIN decreases Q1 turns off allowing the 470 Ω resistor to turn on Q2. This has the effect of turning off Q1 even more ...

As VIN rises again it will have to reach a voltage higher than the turn-off voltage before Q1 starts to turn on again.

Your last part when Vin rises again it will have to reach voltage higher than the turn off voltage is bit confusing. How so?

Let's say that Q1 turns on at 0.6 V.

  • VIN is low. Q1 is off, Q2 is on so Q2's collector is low.
  • R1 (2k2) and R2 (100k) form a potential divider with a ratio of about 98/100. To get to 0.6 V on Q1's base you would need to raise VIN to 0.6 × 100 / 98 = 0.61 V. Then Q1 turns on and Q2 turns off.
  • Now R2 (100k) is pulling high so we'll have the potential divider working the opposite direction. VIN will have to fall to about 0.59 V.

(This is a simplified explanation. The transistor switches gradually, etc.)

This is "hysteresis" and is used for all sort of things in electronics (Schmitt trigger), mechanical engineering (toggle-action clamps and latches) and magnetics (iron hysteresis curve). If you have a relay and adjustable power supply to hand check the pick-up and release voltages of the relay. Why are they different?

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  • \$\begingroup\$ Ahh I see regarding driving the voltage down quickly. Thanks for the explanation :). Your last part when Vin rises again it will have to reach voltage higher than the turn off voltage is bit confusing. How so? Also my turn on for 2SD8832 transistor is when Vin voltage is between 0VDC to 0.5VDC and above 0.5VDC 2SD8832 transistor should be off. How would the circuit post #1 affect the turn on voltage? \$\endgroup\$ Commented May 20, 2024 at 18:28
  • \$\begingroup\$ See the update. \$\endgroup\$ Commented May 20, 2024 at 18:49
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@Transistor pointed already some of the things you want to know.

I just added one simulation with the results you are awaiting.
Added a resistor ... and changed some.
Here is a simulation made "interactively" with microcap v12

You can change the threshold by changing, a "little", the resistor R8.

enter image description here

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  • \$\begingroup\$ Can you please explain exactly how your circuit works in detail? I am not sure what all these resistors do, how they exactly achieve threshold and hysteresis and how you got their value? Also, I was looking for between 0V to 0.5VDC Q2 to be on and above 0.5VDC Q2 to be off. Not sure if it exactly does that. Thanks \$\endgroup\$ Commented May 20, 2024 at 18:25
  • \$\begingroup\$ I added R6 (47k) to lower the voltage threshold of the circuit. I had also to higher the R5 to 4k7. And R3 to 2k (for lowering the current supply). Now, at 27 °C, the circuit switches the LED from 0 to 0.5 V (vo voltage is "low" in that interval which means that the LED is ON). The temperature changes a little this voltage to 0.45 V at 50 °C and to 0.53 V at 0°C. \$\endgroup\$ Commented May 20, 2024 at 18:57

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