0
\$\begingroup\$

In a commercial optical trigger, for triggering flash units remotely using the on camera flash, there are very high resistors two 30Mohm in series. Why such a circuit would need resistors so big? This trigger works well with old flash units that have an high sync voltage (up to, and sometimes in excess of, 300V when the flash fires) or newer flash units usable with digital cameras that stay in the range of 2-6V (but also slave triggers built from the simplest circuits that have only a photodiode and an SCR work with both if the SCR can take an high voltage) In series with a photodiode they would be too big, as a divider they would leave nothing to the circuit to trigger the strobe... What am I missing? I cant undertand their function or benefit... (and i dont have one at hand to break apart) I attached two pictures, the first shows two slaves, same brand, same circuit, one is builtin in an hotshoe adapter the other has a plug at the back for connection to other adapters, cables or directly to the flash unit. The circuit has a photodiode, SCR, 4.7K resistor, 400V 0.047 capacitor, and two 30Mohm resistors that can be seen soldered in series! I suppose the 4.7k resistor and the capacitor are the RC to filter ambient light. The resistors are very visible in the closeup picture of the peanut slave. Triggering happens when the SCR shorts the negative rail of the flash unit and the positive center pin of the flash unit. closeup of resistors

optical slaves

\$\endgroup\$
  • \$\begingroup\$ Could you de-blob your question? \$\endgroup\$ – Samuel Jan 23 '14 at 16:54
  • \$\begingroup\$ de-blob? I am asking what could be the function of resistors of such high value in a circuit like that, I never seen something like that used in any other optical trigger and dont understand what could be the benefit of using them \$\endgroup\$ – user36012 Jan 23 '14 at 18:37
  • \$\begingroup\$ Could they be there to bleed-off the HV from the capacitor when not in use? \$\endgroup\$ – Tut Jan 23 '14 at 20:33
  • \$\begingroup\$ the capacitor in the slave? i thought the 4.7k would do that \$\endgroup\$ – user36003 Jan 23 '14 at 22:22
0
\$\begingroup\$

It's measuring the voltage on the high voltage supply, so any loading will drain the flash battery.

300VDC through 30M\$\Omega\$ is actually quite a large current (\$10\mu\$A), and would represent several mA being continuously sucked from the flash battery when the HV converter is operating.

\$\endgroup\$
  • \$\begingroup\$ You mean "controlling" the voltage to NOT drain the battery? \$\endgroup\$ – user36003 Jan 23 '14 at 20:20
  • \$\begingroup\$ Yes, controlling the current drawn so as to NOT drain the battery. The voltage- it can do nothing about. So they use a big R to keep the I low (and therefore the power low). \$\endgroup\$ – Spehro Pefhany Jan 23 '14 at 20:58
  • \$\begingroup\$ so is it in parallel with the SCR? because in series it wouldnt be able to have enough to have a trigger voltage at all... what happens with the flash units that give only 3V then? \$\endgroup\$ – user36003 Jan 23 '14 at 21:59
  • \$\begingroup\$ like this? s14.postimg.org/nmyyew167/paral.jpg \$\endgroup\$ – user36003 Jan 23 '14 at 22:17
  • \$\begingroup\$ @user36003 Yes, something like that, but the low end should go to some low-voltage circuitry rather than ground (but the current through the 60M will be about the same as if it was grounded. \$\endgroup\$ – Spehro Pefhany Jan 23 '14 at 22:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.