Transistor switch without polarity

Question

I've successfully used an Arduino Uno to trigger Halloween props by closing the normally open circuit on an SRD-05VDC-SL-C relay. I am basically replacing a contact switch and tapping into each prop's "try me" circuit as depicted in the following 2.5 minute video:

Link to instructional video

I'm now trying to replicate the functionality on a breadboard by connecting the base leg of a BC337 transistor to a digital output pin on an ATMEGA328P-PU. I can make it work with every prop but as is suggested in the video I have to respect polarity when using the transistor. The relay works with every prop irrespective of polarity. Is there a way for me to make the transistor work irrespective of polarity too?

Answer 1

1) Put an optocoupler inside each treat when you modify it.

The H11AA1 AC isolator costs 25c and is AC input i.e. the led input side is unpolarised. If you fit one of these inside each treat across the switch (you still need to get polarity right when you fit it) then the wiring polarity to the opto is irrelevant. Cheap, one part, low on voltage.

You could just use a simple opto in the treat with a polarised plug, so that led polarity is correct.

You could also connect a single opto across two pins of your mcu (A,B), and drive it A hi, B lo, then B hi A lo. Thus it gets triggered no matter which way you wired the led up.

1b) If you want the switch side to be unpolarised, then use two simple optos (Opto1,2 below) (these seem to be 3c from CN) with the leds in series, and the transistors swapped. In this case put the optos on the MCU board.

BTW if you are triggering one treat at a time, then an 8 bit port is able to control up to 56 optos (1b) directly.

2) use an optofet isolator. Not really cheaper than relays, but lower on current, full isolation and no polarity or on-voltage issues.

I personally would go for optos or relays especially if any of the treats are not battery powered.

Bu if you really want to have problems...

a) Since you have to modify each treat anyway, just put a standard 2 pin polarised connector on when you modify them e.g 3.5mm jack, and then they will be correctly polarised when you plug them in. Drive with a transistor.

4) Here's a non polarised switch. You can also use it with optos for Q1 to get a non polarised output. The draw back is the on voltage is 1.4V using ordinary diode bridge, or 0.8V with schottky bridge.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

A quick search at Digikey shows several FET-output opto-isolator devices that are relatively low-cost and in stock. These have dual FET output stage, which makes the load polarity not relevant. In other words, they will handle either polarity of DC and will work with AC as well.

I sorted the available parts by cost and came up with the following two part numbers: TLP222G, TLP222G-2

The "-2" variant is two identical parts in a single package.

Digikey part number TLP222GF-ND costs Can $1.45 in quantity 10, dropping down to Can $1.29 in quantity 25. The dual versions cost almost exactly double, so no real savings in cost.

This is a 4-pin through-hole device. Max 350V, Max 120 mA, Max on-resistance 50 Ohms. The LED has a forward voltage less than 1.6 Vdc and the datasheet says the device is fully enhanced with 3mA LED current.

This is a very easy way to accomplish full isolation at low cost. You can put a whole bunch of these along with the LED resistors on a strip of Veroboard (stripboard).

Answer 3

You will need two transistors (one PNP, one NPN) to duplicate the polarity-independent switching functionality of your relay.

Your relay operates irrespective of polarity because it is a switch operated by electromagnetism. The relay does not care about the polarity (direction of current flow). When current is flowing, electromagnetism closes the contacts.

Your BC337 (NPN) transistor operates when you 'bias' the junction, by placing a positive voltage on the base. Add a BC327 (PNP) transistor, this will operate when you place a negative bias on the base.

Bottom line, you need two transistors to take the place of a relay when you cannot control the polarity of your input.

Since you can buy a dozen transistors for the price of a relay, you're still way ahead of the game.