simple switch

[LiteBright]

I have a very low power video camera (3.7V 150mA) that has two buttons: power-on and record.
Normally, as in figure #1 you tap power-on switch to turn it on and tap record switch to start recording.

Instead of using the buttons on the camera, I want to use a wireless controller with 2 outputs to operate power-on and record. When I press a button on the wireless remote, the wireless controller provides 3.7V on each of its 2 outputs. (sell figure #2)

I would just connect two relays to the camera switches and wireless controller (as in figure 3), but I do not have enough space for the relays.

Is there a transistor type (or simple circuit) I can use that will serve the same purpose as the relays?

[pebe]

Have you got enough room for an 8pin IC? If so you could use a dual optocoupler like the ILD615. See these

http://www.technogumbo.com/projects/Lea ... oIsolator/
http://www.datasheetcatalog.com/datashe ... D615.shtml

[LiteBright]

The ILD615 is large but would fit. It is much smaller than two relays. There are two things I don't understand however:

[1] I only have 3.7V and could not see in the ILD615 datasheet if this would work.

[2] The ILD615 appears to output voltage to power something like an LED, but I need something that will switch.
I could not find an example circuit with the ILD615 used for switching.
Are the connections to pins 1-8 in this illustration for switching correct?

[pebe]

Looking at just the left hand half of the coupler, when current is passed through the diode it emits IR light. That is directed to the transistor connected between 8 and 7 and switches it on. There is, in effect, a switched connection between 8 and 7.

There are only two considerations.
1. The current through the IR diode must be limited to a safe value. Ideally, it should be 10mA but can be up to 60mA max. The data sheet gives a nominal voltage drop of 1.15V (@10mA). As you have a 3.7V supply then you need to drop the surplus 2.55V with a resistor between +3.7V and Pin1. For 10mA that would be 255ohms. The nearest preferred value is 220ohms and that would give an emitter current of 11.6mA.

2. Polarity across the transistor terminals should be observed. As it’s an NPN transistor the connections should be so that Pin8 is more positive than Pin7.

The same for the other half.

Hope that helps.

[LiteBright]

A thousand thank-you's. I need to order the coupler which will take some time, but will let you know how it works out.