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I am trying to build a circuit using the signal from a Fuji XT-20 camera hot shoe to drive an interrupt input pin on a Teknic ClearCore controller.

My understanding is that all the hot shoe does is when the camera fires, the middle (trigger) and side (ground) pins are shorted (as also explained in this thread). I have confirmed this by connecting the contacts from the hot shoe to an interrupt input of my controller, and when I manually trigger the camera, it seems to behave just like a switch input.

However, it looks like if the ground from the camera hot shoe and the one from the ClearCore are commoned, the camera stops triggering when it receives a trigger signal from the controller, which is problematic since I am not looking to trigger the camera manually but programmatically.

I have built the circuit below using an optocoupler (the ISQ74) to keep the 9V camera power supply and the Clear Core power completely isolated.

enter image description here

My expectation would have been that when the hot shoe is in its normally open state, 52 mA flow from the camera power supply through the optocoupler LED, activating the phototransistor causing the ClearCore input to register a logic high. As the camera triggers and the hot shoe closes the contact, current will flow straight through ground and the optocoupler will stop conducting, making the ClearCore register a logic low.

This seems to work when connecting the circuit to a bench power supply - if I simulate the hotshoe using a jumper wire, I can measure continuity at the opto output when the switch is open and an open circuit when the jumper connects to ground. However when I try integrate it with the rest of the hardware, the LED seems to be conducting no matter whether the hot shoe switch is open or closed, and the ClearCore input is always "high".

What's even more bizarre is that manually shorting the hotshoe contacts with a jumper seems to yield the desired results.

Edit: below is the circuit I am looking to build to address the points made in the comments:

enter image description here

The potential divider would drop the camera power supply down to 5V at the point where the hot shoe and optocoupler connect. R3 Limits the current going through the optocoupler LED to 50mA when the hot shoe is in the open state. When the hot shoe shorts, 1mA is sunk to the ground and no current flows through the LED. Would this work as I have described?

user11271728
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1 Answers1

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Have a look here DSLR Hotshoe wiring to Arduino for details of someone else looking at a similar problem, which goes into how a modern hot-shoe is implemented.

I suspect the issue is that the camera doesn't use a relay between pins 1 & 2, but instead has a TTL compatible logic signal.

The absolute maximum current for the optocoupler LED is 50mA, whilst just a couple of mA should be sufficient for effective operation. According to the datasheet, at 50mA the forward drop of the LED is typically 1.2V, and whilst not speced, experience suggest it will be ~0.8V for 1mA.

You are using a 9V supply, and have selected a resistor which will give 52mA at a 1.2V drop, exceeding the maximum rating.

To turn off the opto-transistor, requires pin 2 to sink all of the ~55mA leaving <<1mA going through the LED. For a typical V/I curve means less than 0.6V across the LED.

Most TTL outputs will not be capable of this (however a bit of wire shorted across the terminals definitely will...)

I suggest you decrease your supply to ~5V, and increase the resistor to make the "open" current as small as possible. If you want to keep the 9V supply, split the dropper to give a 5V point at normal LED load, then connect the hot-shoe across that.

Update

In the new schematic, R2 is redundant, as the 5V operation point is set by R1, R3 & and LED forward voltage. If the optocoupler doesn't turn on normally, you may need to halve the values of R1 & R3 to increase the LED current to 2mA. Other than that, looks good.

colintd
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  • Thanks for your answer! If the hots hoe is effectively some kind of bipolar transistor, which makes sense, isn't it pretty common for those to accept pretty high collector voltages? And if 9V were too much for the transistor to handle, wouldn't it just break resulting in a short to ground which would immediately drive the controller output low? The reason why I cannot use the sync connector is that I am trying to use the hot shoe signal to assert that the camera has effectively taken a shot when being controlled from the sync input – user11271728 Feb 22 '23 at 14:36
  • There are lots of possible internal setups, but my guess is that it isn't able to sink the 50+mA required to "short" the OptoCoupler. Try a sensitive logic level optocoupler such as the H11L1 https://uk.farnell.com/isocom/h11l1/optocoupler-schmitt-trig-o-p-5kv/dp/1683200 which only needs 1mA, then adjust resistor accordingly. – colintd Feb 22 '23 at 15:20
  • Why would the camera's internal circuit have any role in sinking the 50mA the LED requires to short the opto? The camera is completely bypassed when the hot shoe is in the open state, and 50mA is the maximum forward current (the opto will still respond to much smaller currents). – user11271728 Feb 22 '23 at 15:49
  • I've updated answer with more details. – colintd Feb 22 '23 at 16:35
  • Thanks that makes sense and I am going to try what you suggested. Just slightly confused about where the 0.6V rise above ground you mentioned is coming from? – user11271728 Feb 23 '23 at 10:56
  • Please let us know if the suggestion works, and if so, mark the question as answered. – colintd Feb 23 '23 at 11:08
  • Thanks, before I go ahead with sourcing the resistors and building the circuit, I have edited the question with an updated diagram – user11271728 Feb 23 '23 at 12:39
  • Yes that's just to indicate that R1 and R2 drop the 9V supply to 5V and that's the voltage seen by the hot shoe and optocoupler - The 5V you see on the right hand side is the internal pin within the ClearCore, my bad should have made that more clear - will update everything once resolved – user11271728 Feb 23 '23 at 13:04
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/143111/discussion-between-colintd-and-user11271728). – colintd Feb 23 '23 at 13:21