Are activated open-collector and open-drain circuits wasting energy?

Question

Consider the following reference circuit for a 6N137 optocoupler. My understanding is that this circuit uses a pull-up resistor on the output node so when the optocoupler led is activated, the transistor gate opens and the output is pulled to ground. This seems to be a standard configuration for open-collector or open-drain circuits. What puzzles me is that it seems inefficient. In order to switch the output off, I need to effectively short the circuit in order to pull it to ground. Is my interpretation correct? If not, why not?

Answer 1

You are correct (except that when the LED is activated, the phototransistor gate is injected with current. It's the opposite logic of what you said). It has it's uses though such as flexible output voltages and inherent ORing of signals.

ADDED:

Using a large resistor minimizes current consumption when the output is pulled LO, but also increases the time it takes to charge the parasitic capacitances on the line and and input capacitances so your rise times get slower which might be unacceptable for some applications (the most common one is a serial bus like I2C).

Answer 2

What puzzles me is that it seems inefficient. In order to switch the output off, I need to effectively short the circuit in order to pull it to ground. Is my interpretation correct? If not, why not?

Figure 1. The opto-isolator's output is shown with R_L, the load.

What you are missing is that R_L is the load. When the output is pulled low there is voltage across R_L and power is consumed. When the output is high there is no voltage across the load so no power is consumed.

This arrangement works fine for certain designs but, as you suspect, it would waste power if, for example, it was only driving a CMOS input. In that case a push-pull arrangement would be much more efficient.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 2. Simplified representation of open-collector/drain output and push-pull outputs.

• Figure 2a has no internal means of pulling the output high so the pull-up resistor, R1 is required. Current draw is determined mostly by the value of R1.
• Figure 2b can pull high or pull low. In the steady state the current draw is determined by the input impedance of the next stage. This is potentially much more efficient.

Answer 3

What is output efficiency for an open drain or open collector?

Let's look at the output power dissipation.

When the LED input current is ON, it depends on pullup load \$P_R =\dfrac{(Vcc-Vout)^2} {R}\$.

When the LED input current is OFF, there is only 1uA load current MAX and typically less with no other loads.

Pasted from datasheet...

Low level output voltage VE = 2 V, If = 5 mA, IOL (sinking) = 13 mA VOL - 0.2typ 0.6max V

The output transistor has an equivalent resistance when ON

ΔV/ΔI= Vol/If= 0.6maxV/13 ma (sinking) = 46 Ω (max) , 200mV/13mA= 15 Ω (typ)

So have the following equivalent circuit for input If=5mA

^{simulate this circuit – Schematic created using CircuitLab}

Thus the datasheet recommends

 RL 330 (min)  4 kΩ (max)

This affects your speed and fanout but power dissipation of my 1st formula applies.

So one might use 3.3V with 330 Ω or 10mA for max speed and 5V with 1kΩ (5mA) for less power or 4kΩ for least power but slowest speed.