# how to design a switch which works when an input of 0.10 is applied

i have an alarm clock , it's output voltage to the alarm buzzer is around 0.15 volt , i need to attach an electrical appliance to make it operate on scheduled time , for that i had used this circuit (removed the LPT port part from the circuit):

and attached to the alarm output. But the 0.15 volt is not driving the opto coupler, how can i design a circuit switch which will drive the opto coupler or the realy or anything with the 0.15 volt ..

• Is the 0.1 volt AC or DC? Do you have an image? Commented Feb 17, 2013 at 18:33
• @jippie - good question, confirming it's a DC out would be useful. cc4re - did you test with a multimeter? Do you have an oscilloscope? Could you also measure the signal voltage when the alarm is off? Commented Feb 17, 2013 at 18:43
• From three alarm clocks I've liberated, the buzzer signal is either a very high voltage high frequency AC if the sound transducer is a piezo, or a moderate voltage (3 to 12 Volts) AC if it uses a paper cone Speaker. Never a DC. Commented Feb 17, 2013 at 18:48
• @OliGlaser I suspect the voltage is so low because it actually is block signal, slightly above 50% duty cycle. Can't imagine you wake up from a buzzer controlled by 0.1 volt. Commented Feb 17, 2013 at 18:59
• @jippie - yes, it has me wondering. I suspect it may be the transistor saturation voltage (e.g. resistor, buzzer, then transistor switch) Or it could be AC and the OP is measuring on DC range. Hopefully we'll find out soon enough Commented Feb 17, 2013 at 19:16

0.15V won't drive the opto's input, since it's input LED has a typical forward voltage of 1.25V.

If 0.15V is really all you have access to, then you will need either a comparator or opamp with input range that includes it's lower supply rail.

Basically, you set the inverting input (-) to e.g. 75mV, then attach the signal to the positive input. Something a bit like this, but with the divider set for 75mV (e.g. 49.25kOmega; and 750Ω - using a pot to trim would make it easier to set):

If your power rail is prone to fluctuate, a voltage reference IC for the divider supply would be better.

If the output signal you have is e.g. 5V when the alarm is off, then you could do similar to the above but swap the inputs and have your reference connected to the + input at e.g. 2.5V. This would give you a lot more tolerance to supply variation or noise.

If it turns out to be an AC signal, then you would need to add a simple peak detect circuit in front of your comparator. You will suffer a drop in your output voltage, so given the small voltage to start with, a Schottky diode for D1 would be best:

Simulation - blue trace is 150mV 1kHz square wave switched on for 500ms at 100ms, the green trace is the output voltage at the top of R1/C1:

EDIT

Okay, you say your 150mV output is DC, which makes things a bit simpler. In your diagram you are using a MCT2E Optocoupler, which has a maximum input photodiode current spec of 60mA, and ~20mA typical in order to turn the phototransistor fully on. Since the phototransistor is only driving the base of the relay drive transistor in your circuit, we probably don't need it fully on, but unless you are trying to conserve power, we'll set things up for a 20mA drive to make things simple.
Note that in your circuit you have two layers of isolation - one from the opto and one from the relay, but that's fine if this is how you wish to do things.

Okay, you need a comparator with an input range that includes ground. There are lots of options of you check on places like Farnell, etc. We are using LTSpice to simulate, so let's pick one from LT - the LT1018 is a random choice in the LTSpice library which has rail to rail out and it's input includes ground - any similar comparator with an input which ranges to ground, and can drive the 20mA would do. This one can only sink current, so we need a pullup resistor on the output.
For the input, we need to set the reference voltage at the non-inverting input (+) to somewhere between 0 and 150mV - let's pick ~75mV. As long as it's far enough away from either 0 or 150mV you should be fine.
For the output, we need to limit the current to 20mA. To do this, we take a look at the typical photodiode forward voltage, note out supply voltage and calculate using this formula:

Rout = (Vsupply - Diode_Vf) / I_diode = (5V - 1.25V) / 20mA = 187.5Ω 180Ω, 200Ω or 220Ω is near enough.

So we end up with a circuit like this:

Simulation:

We can see in the top plot the threshold voltage and the input voltage from the buzzer. We can also see the hysteresis provided by R4, which helps to prevent glitches on switchover by changing the positive going and negative going thresholds (see here and here for some explanation)

• Hi , the explanations , totally confuses me , can you make it a bit clear :) as i am a fresher , can you provide me an exact circuit diagram with each components values.. Commented Mar 2, 2013 at 15:36
• @cc4re - Sorry about that, give me a little while and I'll update with some more detail. Commented Mar 2, 2013 at 19:36
• @cc4re - I added more detail, let me know if you have any more questions. Commented Mar 2, 2013 at 23:11
• @OliGaser Thanks a lot for this help :) in the above circuit Rout 220 is (220k or 220Ohm resistor ?) the opamp input is 5v volt , i am using a transformer of 12v for the relay , so can you adjust that for a 12v , else i need to buy another transformer and attach or shall i use a 100 ohm resister in between and apply the 12v to the opamp Commented Mar 3, 2013 at 17:20
• @OliGaser let me know which software are u using to draw this diagrams Commented Mar 3, 2013 at 17:23