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Help amplifying a tiny current

Avondale

Sep 4, 2016
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What's the operating Voltage and Current rating of your relay.
http://www.aliexpress.com/item/Free...IN-pcb-miniature-Power-Relay/32255500146.html

Here's a schematic of what I described in my last post.
thanks! i will try this! though i currently don't have any resistors! pnp isn't something i really understand yet

Relays are current-hungry devices. Could you use a FET instad for switching a load?
interesting! can you get them to work with 240v cheaply or is it just for lower currents?
 

CDRIVE

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Expanding on Alec's last post it has occurred to me that you haven't told us what the load is you're trying to switch on. Care to elaborate?

Chris
 

CDRIVE

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The link to your relay indicates that it's a 3V coil @ 20 to 25Ω. As Alec stated "Current hungry" because that computes to 120 to 150mA!

Chris
 

Alec_t

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It's beginning to look as though a solid state relay (SSR), with its lower input demands, would be better than your mechanical relay for this job.
 

(*steve*)

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A lot of excellent replies while I was sleeping.


I was heading along the path of a PNP transistor (which is just the evil twin opposite of an NPN transistor). This will require a base resistor.

On hearing the load you are switching, a solid state relay (SSR) is also exactly what I would have suggested. It has a massive advantage in your application of requiring a much lower current than a relay.

Also, the ssr will not need a transistor, just a series resistor of about 68 ohms (but this will depend on the characteristics of the SSR - it must operate from a low voltage, most are specced to operate from a 3V minimum. If you can use two 3V cells in series you will have no problems.
 
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(*steve*)

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Yeah, and you would have only read half the post. Don't you love autocorrect.
 

(*steve*)

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This is what I would advise (note it's simply the first result for the search SSR, not a specific recommendation). Note that it requires 3-32V and 3 to 25mA which is waaaaaay less than your relay!

So yeah, it will save heaps of current. 5mA vs 120mA? No question.
 
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hevans1944

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http://www.aliexpress.com/item/10pc...C-PCB-SSR-5VD-Out-240V-AC-2A/32660253659.html have i got this right?
they work like the ones i have but without the click? how much current does it save?
That is a fair choice. It's rated to operate with 4 V to 6 V DC, so 3V is a little on the iffy side. Can you provide your temperature sensing board with more than 3 V DC? Say, three AA cells in series for 4.5 V DC? It should tolerate that.

The input impedance of the SSR is a little on the low side at 440 Ω, so at 4 V DC input it will draw about 0.9 mA. You will need a pull-up resistor to Vcc to supply this current without dropping too much of the available supply voltage across the resistor, but large enough to limit the current to the open-collector comparator output when that output is low. Something on the order of 550 Ω will work, dropping 0.5 V from the 4.5 V DC supply when the comparator output is high (open collector not conducting) and turning on the SSR. When the comparator output is low, the resistor drops almost the full 4.5 V, passing about 8 mA through the now-conducting open collector output. This current is well within the current rating of the comparator and adds a small addition to the load presented by the LED and its current-limiting resistor.

This 550 Ω resistor is installed between Vcc and DO, effectively in parallel with the LED and its current-limiting resistor. The SSR control input is connected between DO and common.

Edit (09/09/2016): I miscalculated value for pull-up resistor. There is no value that will work without a transistor to increase the current capability because the LM393 can only sink about 16 mA max.

 
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(*steve*)

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This 550 Ω resistor is installed between Vcc and DO, effectively in parallel with the LED and its current-limiting resistor. The SSR control input is connected between DO and common.

I believe he wants the relay to turn on when the LED does. Thus the SSR is between Vcc and the output (with a series resistor)
 

hevans1944

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I believe he wants the relay to turn on when the LED does. Thus the SSR is between Vcc and the output (with a series resistor)
The LED turns ON when the output, DO, is low... so that would work. No series resistor is necessary. However, it was my understanding that the the OP wanted the relay to turn on when the temperature was above the setting temperature and the LED was OFF.

Whatever, I goofed on calculating the pull-up resistor if the SSR control is between DO and common. It would have to be 55 Ω to drop 0.5 V when DO is high and LED is OFF. But when the output transistor conducts, 55 Ω would require way more current than the output transistor is rated to conduct (typically 16 mA for 0.7 V saturation drop)... almost 82 mA! So back to the drawing board if my interpretation is correct. Of course you could wire the SSR like you said and use it to operate a REAL power relay with NO and NC contacts. Then just choose whichever pair of contacts gives you the control you need.
 

Avondale

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Can you provide your temperature sensing board with more than 3 V DC?
yes! it was listed to take up to 5v, though i don't have a spec sheet
Try using a little solder to hold the wires together.:p
i think i need an AA battery holder lol maybe i will solder some jumper cables on the ends of a few..
I believe he wants the relay to turn on when the LED does. Thus the SSR is between Vcc and the output (with a series resistor)
it (the LED) seems to go off when the temperature goes above setting, which is when i would like the relay to activate

i wish i could say i understood even half of that paragraph @hevans1944 : o I'm a real novice with component theory
 

(*steve*)

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Ok LED off, relay on :)

That board should be fine up to about 30V ish. 5, 6, or 7.5 should not be a problem.
 

CDRIVE

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Well much has been posted since my last contribution when I asked "What is the load that you're switching"? Unless I missed it we still don't know. This would seem rather important. This way we can determine if resistors should be left bare or fully clothed! :p

Chris
 

hevans1944

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Ok LED off, relay on :) ...
Well, I am glad that is settled.

So, IF we could find a SSR that didn't require a ton of current to turn on, then it could be connected between the output, DO, and common. The internal 10 kΩ pull-up resistor would supply current to the SSR when the output goes "high" and the open-collector transistor turns off. Unfortunately, even if the SSR input was a "dead short" when conducting, the current delivered through the 10 kΩ pull-up resistor would only be 0.00045 A or 0.45 mA with a 4.5 V DC supply. Not enough to operate any SSR input that I know about since these are usually an optical isolator with an LED and a series current-limiting resistor, about 440 Ω for the device under consideration.

The problem is, if you lower the value of the pull-up resistor to provide enough current to turn on the SSR when DO is high, it allows way too much current to flow through the open-collector output transistor when the DO is low. So... you need to insert a transistor in there to provide more current. Since output "high" turns the transistor on, we are back to the original NPN configuration with possibly a smaller "pull-up" resistor to provide sufficient base current drive to the transistor. I would start with 2 kΩ inserted between DO and Vcc, connect output DO directly to the base of the transistor, connect the NPN emitter to common, and insert the SSR or the original 3V relay coil between Vcc and the NPN collector. Connect a small 1N400x-series rectifier diode across the relay coil, cathode to Vcc and anode to the NPN collector. Use a small-signal transistor capable of carrying the coil current or the SSR control current. No heat sink necessary: the transistor is biased off when DO goes low and conducts in saturation when DO goes high.

This is basically the same circuit as the original post #1, except S1 is replaced with the circuit shown in post #7 and a 2 kΩ resistor is added from Vcc to DO to increase the base drive voltage. Try using a 2N3904 NPN transistor to operate the relay. Use a 4.5 V DC supply for Vcc. The 4.5 V DC supply will provide about 2 mA base current through the 2 kΩ resistor and the 2N3904 should be fully saturated and conducting when DO goes high.

And let's not get into any esoteric discussion as to whether the increased base current is responsible for driving the transistor into saturation and increasing the current available to operate the relay coil. We all know that it is the increased base-emitter forward voltage that decreases the base-emitter depletion layer width and allows more emitter electrons to drift across the depletion layer into the collector region to increase the collector current. The ratio of collector current to base current, called hfe or beta, is just a convenient observation sometimes used to incorrectly "explain" current "amplification" in a transistor amplifier circuit.
 
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Avondale

Sep 4, 2016
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thanks to all... bit drowned in information here but i'll have to take a break while china posts me some solid state relays, and amazon sends me some resistors : p
to be continued!!
 

CDRIVE

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