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I am not an engineer. I have researched basic information that can be found on optocouplers. I understand the theory of 4 pin optocouplers. I believe I even understand basics of 6 pin optocouplers. However, everything I have located about 6N137 or 8 pin optocouplers in general, assumes I understand more than I do. I have located two types of information on 8 pin optocouplers. Basic info: which does not seem to include the 8 pin configuration, or Advanced info: which assumes I have knowledge I do not possess.

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I have the Vishay 6N137 datasheet. Assume this 6N137 to be used instead of a 4 pin optocoupler or a relay in a simple circuit where closure times must be short. Small amperage DC input to trigger larger DC amperage output circuit. My questions are extremely basic.

Q1) Pins #1 & #4, may be used for what purpose?
Q2) Pin #5, I understand is Ground, but is it ground back to the negative side of the Vcc Collector DC supply?
Q3) Pin #6, Vo=Voltage Output. Is this output connected to the negative side of the circuit you are closing?
Q4) Pin #7 Ve=Voltage Emitter. Or is this the output connected to the negative side of the circuit you are closing? If not, what is it for?
Q5) Pin #8 Vcc Collector Supply. I assume the positive side of the circuit to be closed when the optocoupler is activated is connected here?

A direct answer to my questions would be wonderful. Direction to literature that describes specifically what I am asking would be helpful also.

I could understand 8 pin optocoupler layout better if someone had a diagram of the inside a 6N137 optocoupler showing internal routing of all 8 of the pins. The datasheet does not seem to account for every pin.

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  • \$\begingroup\$ This is actually a 6-pin opto-coupler in a 8-pin package. Pins 1 and 4 are not connected to anything. This is just like any other opto-coupler on the tranmitter side, which means it's just a LED. \$\endgroup\$ – Olin Lathrop Nov 2 '17 at 22:27
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As shown in the datasheet, pins 1 and 4 are not connected.

Pins 5 and 8 are the 5 V power supply (pin 8 is not directly connected to a collector).

Pin 6 is the open-collector output, i.e., it is either connected to ground or open. (Vishay's 6N137 actually uses a MOSFET, i.e., an open-drain output.) To get a valid output voltage when the output is inactive, you have to connect a pull-up resistor.

Pin 7 is not an emitter, but the enable input, i.e., forcing it low disables the output.

The original 6N137 was made by Hewlett-Packard; its innards are described in the Optoelectronics Application Manual:

6N137

[…] For high data rates, the transistors must be operated at lower closed-loop gain in order to achieve the required bandwidth. HP detector/amplifiers for digital applications have a high speed linear amplifier driving a Schottky-clamped output transistor, as shown in Figure 3.1.4-2. Bias for the photo diode is decoupled from VCC to reduce the possibility of "chatter" (oscillatory transition from one logic state to another due to regenerative coupling via the power supply line). The linear amplifier has a tendency to be unstable if the high-frequency impedance of the power supply is not low enough. For this reason, a low-inductance bypass capacitor (0.01 µF ceramic disc) should be installed adjacent to each isolator of this type. This, and other chatter-suppression techniques are discussed in Section 3.3.

The Schottky clamp is a metal-silicon diode in parallel with the base-collector junction of the output transistor. A metal-silicon (Schottky) diode has a lower turn-on voltage than a P-N junction,so when the transistor is driven into saturation, the Schottky diode bypasses the current which would otherwise enter the base-collector junction. With reduced current entering the base-collector junction, there is a proportionate reduction in the charge to be removed when the transistor is to be turned off, and thus the Schottky clamp reduces the turn-off delay. The attendant drawback is the 400 mV higher VCE(SAT) for a Schottky clamped transistor.

The "enable" input has threshold voltage and input current levels resembling a TTL input. However, it is not necessary to apply a pullup resistor to insure its remaining high. Unless the enable is connected to a strobe, it may simply be left open. The strobe applied to the enable input may be either open-collector or active-pullup.

With the enable high, analog operation is possible because there is no hysteresis. However, the dynamic range is limited. The lower limit is the threshold input current for operation in the active region – this threshold may be as high as 4 mA. The upper limit is the maximum dissipation rating on the output. Because of a touchy bias situation, analog separation is not recommended for designs to be mass produced.

The reason for omitting hysteresis was not to permit analog operation, but rather to permit maximum data rate. With hysteresis, there would be a higher immunity to both differential- and common-mode noise but the shifting threshold would reduce the data rate capability. The effect of hysteresis on data rate is the opposite of peaking.

(Vishay's 6N137 actually uses a Schmitt trigger to get hysteresis.)

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  • \$\begingroup\$ Great summation. This allows me to understand it much better! Since pin 7 is an enable input, why is at marked as Ve in the Vishay datasheet? Also, does having the 5V power supply connected to pin 8 (Positive) and pin 4 (Negative) cause a continued power drain, as it measures as hot all the time? \$\endgroup\$ – B. Varner Nov 5 '17 at 16:08
  • \$\begingroup\$ "VE" stands for "enable input voltage". The supply current is specified in the datasheet. \$\endgroup\$ – CL. Nov 5 '17 at 22:07
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Pins 1 and 4 are marked "NC" on the datasheet - that means "No Connection" - they are not used.

Pin 7 (Ve) is an enable input. If it is held High or left unconnected, the device works. If Ve is held Low, the output will remain High regardless of the LED input. See the "Truth Table" on Page 2 of the Vishay datasheet.

The 6N137 output side is something more than a simple transistor. The Vcc and Ground pins (8 and 5) provide operating power for the output circuit, and pin 6 is the Drain of the output transistor. The part will pull pin 6 to ground when the LED is lit, and let the output float when it is not lit.

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Q1) Pins #1 & #4, may be used for what purpose?

For securely mounting the part to the PCB/DIL socket. Electrically they are not used.

Q2) Pin #5, I understand is Ground, but is it ground back to the negative side of the Vcc Collector DC supply?

Pin 5 is the GROUND associated with Vcc (Pin 8).

Q3) Pin #6, Vo=Voltage Output. Is this output connected to the negative side of the circuit you are closing?

It is an open-drain used to pull-down some signal. So yes the "negative side of the circuit you are closing"

Q4) Pin #7 Ve=Voltage Emitter. Or is this the output connected to the negative side of the circuit you are closing? If not, what is it for?

The is an Enable Input - No pull up resistor required as the device has an internal pull up resistor.

Q5) Pin #8 Vcc Collector Supply. I assume the positive side of the circuit to be closed when the optocoupler is activated is connected here?

That is correct.

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  • \$\begingroup\$ Thanks! This seems to make sense. However, testing this on a breadboard.... If I connect positive to Pin 8 (Vcc) and an LED to Pin 5, then to ground, the LED lights whether the input side of the 6N137 is powered or not, In other words, all the time. I must not be applying this correctly. \$\endgroup\$ – B. Varner Nov 2 '17 at 22:24
  • \$\begingroup\$ Pin5 is ground so you have just connected an LED straight across the supply. NOTE this will stress teh device. Try connecting: Vcc -> Resistor -> LED_Anode -> LED -> LED_Cathode -> pin6 \$\endgroup\$ – JonRB Nov 2 '17 at 22:26
  • \$\begingroup\$ I did already have the resistor in the circuit. Ok, tried: Vcc (9V)->Resistors (620R)->LED Anode->LED->Cathode->pin6 + pin5 back to Vcc negative. This did not function when the input side was triggered. Just to be clear. On the input side I have 6V->Resistor (560R)->pin2 through 6N137 ->pin3->6V negative \$\endgroup\$ – B. Varner Nov 2 '17 at 22:41
  • \$\begingroup\$ The Absolute Maximum voltage for the output side (between pins 8 and 5) is 7 volts - if you applied 9 volts, you may have killed the 6N137. You MUST apply +5 volts (+6V would work) to Vcc (pin 8), with the negative side of the 5 volt supply to Ground (pin 5) to power the output circuit. \$\endgroup\$ – Peter Bennett Nov 2 '17 at 23:09
  • \$\begingroup\$ Think of the output side of the 6N137 as a logic gate, not as a plain transistor - it must get power via its Vcc and Gnd pins. When the LED is lit, the output (pin 6) will be pulled to Gnd. Look at the test circuits on page 5 of the Vishay datasheet. \$\endgroup\$ – Peter Bennett Nov 2 '17 at 23:13

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