[ OPA454 ] E/D and E/D Com Connection

Hello Kenichiro,

It appears your understanding of the OPA454 E/D for enabled state is correct. The first circuit you show is one I would not recommned as well based on the comments mentioned in datasheet sction 9.3.6. The second circuit, that provides ac bypassing of the pin, should provide a simple, usable solution.

The third soultion using the 1-Megohm resistor relies on the point that the E/D draws more current as the applied voltage on the pin is increased. This can be seen in datasheet graph, Figure 45. Section 9.3.6 indicates the Ip will be 50 uA, but the graph indicates that the current may be closer to 25 uA, which would result in only a 25 V drop across the 1-Megohm resistor. It is apparent that an Ip current higher than 25 uA flows and the voltage on the E/D pin assumes a level somewhat higher than 5 V. It musn't be high enough to cause voltage breakdown of the E/D circuit. However, even if something did go into breakdown the 1-Megohm resistor would limit the current to 50 uA which isn't likely to allow any damage. I have actually used the 1 Megohm Rp in test circuits operating from +/-50 V supplies and observe any problems. However, I understand your concern.

It would appear that increasing Rp would provide a safer solution. Using the Figure 45 curve we see that with a Venable of 4.5 V the Ienable is about 20 uA. If Rp is made a 2.2 Meg resistor about a 45.5 V voltage drop will accur across it. The E/D pin would then be biased to achieve a level about 4.5 V.

The other method of biasing the pin to a known level is to apply a zener diode as shown in the OPA454 applications circuit. In that application a 1N5225, 4.3 V zener is connected between E/D and E/D common. Rp is reduced to 100 kOhm to provide reverse current for the zener diode. Using the zener removes the risk of over-voltage on the E/D pin and keeps it biased at about 4.3 V.

Regards, Thomas

PA - Linear Applications Engineering