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I should start by saying I'm fairly new to electronics so bare with me as I struggle with some basic issues.

What I'm trying to do is trigger a 5V opto-isolator from the output of a series of logic gates, with the last one being an SN74LS08 AND gate. When I try to do this however, the opto-isolator doesn't seem to be triggering regardless of the AND gate being high or low. I've attached their datasheets and a schematic

AND gate: https://www.digikey.ca/product-detail/en/texas-instruments/SN74LS08N/296-1633-5-ND/277279

Opto-isolator: https://www.digikey.ca/product-detail/en/vishay-semiconductor-opto-division/VO4661/VO4661-ND/1738670

schematic

simulate this circuit – Schematic created using CircuitLab

When I hook the opto-isolator directly up to a 5 V supply, it works fine. This leads me to believe it's because the AND gate cannot supply enough current when logic high.

High level output current: -400 uA. Low level output current: 8 mA

Yet, the opto-isolator desires a high level current of 5-15 mA and low level current of 0-250 uA.

I don't know how on earth I'm supposed to satisfy both of these current ratings, or maybe it's something else entirely. I received this design from an undergraduate thesis that supposedly said this worked.

Any help is greatly appreciated! I will try to respond quickly if more info is requested.

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High level output current [of the AND gate]: -400 uA. Low level output current: 8 mA

Yet, the opto-isolator desires a high level current of 5-15 mA and low level current of 0-250 uA.

I don't know how on earth I'm supposed to satisfy both of these current ratings,

There are 100's of different AND gates to choose from in the world. You're not stuck using 74LS08.

If you choose a CMOS gate instead of TTL, you will most likely be able to get the required 5 mA. If you need TTL input levels for your gate, you can use the HCT family which has TTL input levels but CMOS outputs.

If you can't find an AND gate that can provide 5 mA output while meeting your other requirements you can use a buffer gate or a transistor buffer between the AND gate output and the optocoupler input.

Or you could use a NAND gate and connect its output on the cathode side of the optocoupler LED instead of the anode side.

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  • \$\begingroup\$ Thanks. I looked through a chip manufacturer site and found many gates that can output 5 mA. The problem is, the low level output current is always higher than the high level output current. Am I misinterpreting the datasheet? Or is this mismatch quite minor. \$\endgroup\$ – Physean3 Jan 8 at 18:40
  • \$\begingroup\$ @Physean3, for low output, the gate is sinking current. For high output it is sourcing current. (On TI datasheets, sourced current is given as a negative value, which is matching the passive current convention). In your configuration, for low output, the sinking current will be limited by the external circuit, not the gate capability. \$\endgroup\$ – The Photon Jan 8 at 19:50
  • \$\begingroup\$ We typically use something like 'ACT240's for driving optos. Oh, and we typically drive the cathode of the diode, pulling to GND (or close to it) to turn on the opto. The anode is connected to +5V through an appropriate value resistor. \$\endgroup\$ – SteveSh Jan 8 at 20:07
  • \$\begingroup\$ Slight correction - we've used both high side and low side for driving the opto. \$\endgroup\$ – SteveSh Jan 8 at 20:16
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Given :

(LS-TTL) High level output current: -400 uA.
Low level output current: 8 mA ...
(VO661) opto-isolator input high = 5-15 mA and low level = 0-250 uA. with R1 pullup to 5V and R2 pulldown to 0V.

Translation:

All the LED current is provided by R1,R2 and TTL gate turns it off on the Anode side. The (IR) LED Vf = 1.3V +/-0.3V specified @ 10mA. Thus for R1=1.3k and R2=0.5k

  • If= (5-1.3)V / (1.3k+0.5k)=2.06mA which is grossly under recommended value.

  • 74HC CMOS gate with 50 Ohm drivers the drive current would be If= (5-1.3)V / ( 50 + 0.5k) = 4.6 mA which is still under recommended 5~15mA value but might work.

Suggestion: It is not TTL compatible as designed, but is CMOS compatible.

  • Use 74HC CMOS logic not 74LS TTL

  • remove redundant R1 and

  • change R2 from 500 to 320 ~ 350 Ohms for nom. 10mA
    • If=10mA , R2=(5-1.3)V / 10mA = =370 Ohms incl. 50 Ohm +/-50% CMOS gate (Rol=Vol/Iol)
  • Remove R_out
  • Change R3= 1K to 350 on output if you want rated speed at 10MHz.

Triage

You show the student's paper with a voltage divider on the output. This was possibly to increase Iol to TTL requirements.

enter image description here

With CMOS you can invert logic by choosing to drive the high side or low side of IR LED and R in series to current limit.

Conclusion:

schematic

simulate this circuit – Schematic created using CircuitLab

In future use Datasheets and OEM App notes not student papers.

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