# Light dependent resistor (LDR) switched LED-circuit

NOTE! The LDR has a minimum val = 2kOhm and a max. val. = about 5 MOhm

Im trying to build this circuit for my radio station. I would appreciate som help.

1. Your general opinion about the functionality of the circuit. (It is my first completely own-made circuit)
2. Some thoughts, let me explain.

Well basically I have more than one question, but lets start with the main focus. So the whole idea of using the Zenerdiode is that I want that when the light is of (the LDR is attached to a tiny LED-lamp on the mixerbord, wich goes on if someone is calling the radio station) and the LDR has a value about 5 MegaOhm, there is not enough voltage over the Zenerdiode, making it act as a open (no current passing through). In that case, in my opinion, there will be no current and likewise no potential relative to ground in the point between the diode and R2. This would close the transistor. No light on the mixerboad, no LED1 2 or 3 is lit.

In the case where the mixerboard LED is on, well the potential after the LDR is much higher (hopefully around 7-8[V]) so the Zenerdiode will now push trough current, and the potential after the zener will appear. This will open the transistor and the LED's will be lit.

My main problem, and the reason why I designed it like this is because it is really important that the transistor is acutally "closed", and the LEDs are not lit when they are expected to not be lit. Basically, I tried to quantify a "light on" (on the mixerboard) with a specific voltage on the gatepin of the transistor, via the zenerdiod.

Is this viable?

• your main problem is that you have the transistor configured as a follower instead of as a switch. Move the LEDs and resistors to above the transistor (in series with the drain). – Jasen Jun 22 '16 at 1:14
• Current?? what?? Please fix up your grammar. And that transistor you put in is a MOSFET and MOSFETs are voltage based, not current based. Also what that zener diode is doing is it's decreasing the voltage by 3 volts. – Bradman175 Jun 22 '16 at 1:55
• "In the case where the mixerboard LED is on, well the potential after the LDR is much higher (hopefully around 7-8V" - what voltage did you actually get? – Bruce Abbott Jun 22 '16 at 5:32
• So... you want the new leds to light up when the exist led lights up? You're trying to recreate the wheel. Glue/epoxy/heatshrink a photodiode or phototransistor to the existing led (making your own optocoupler). No need for a unwieldily LDR, Zener, etc. Just make sure there is no light bleed. – Passerby Jun 22 '16 at 7:27
• Basically this sound.westhost.com/p145-f7.jpg – Passerby Jun 22 '16 at 7:28

HINTS

• Choose R1's value so that

• When no light shines on LDR1 the voltage across R1 is less than M1's gate-to-source threshold voltage $V_{TH}$.

• When light shines on LDR1, the voltage across R1 is greater than M1's Miller Plateau voltage, but does not exceed M1's maximum allowed gate-source voltage $V_{GS(MAX)}$. The Miller plateau voltage is usually provided via a "Gate-to-Source Voltage vs. Total Gate Charge" graph in the MOSFET's datasheet.

$$V_{Miller Plateau} \leqslant V_{GS} \leqslant V_{GS(Max)}$$

• The zener diode and resistor R2 are both unnecessary and should be removed. Connect M1's GATE directly to the node between R1 and LDR1.

MEMORY AIDS

• When designing a MOSFET switch, always connect the SOURCE pin directly to the power supply (S⁠OURCE to S⁠UPPLY) as follows:

• For an N-channel MOSFET (NMOS) switch, connect the SOURCE pin directly to the power supply's NEGATIVE (-) terminal—i.e., N⁠MOS SOURCE -> N⁠EGATIVE terminal (-) on the power source.

• For a P-channel MOSFET (PMOS) switch, connect the SOURCE pin directly to the power supply's POSITIVE (+) terminal—i.e., P⁠MOS SOURCE -> P⁠OSITIVE terminal (+) on the power source.

• For both NMOS and PMOS switches, the load circuit is connected to the DRAIN pin.

• For both NMOS and PMOS switches, do not insert components between the SOURCE pin and the power supply.