cd4066 output current?

[Nigel Goodwin]

then perhaps you should read post #5 again, in it's entirety... especially look at the LAST sentence copied here.



An NPN driving a PNP.... this is the point where I asked Crutchow "wouldn't the PNP also drop .7V?" Then he responded, "not if they were both followers the BE drops would cancel", hence the "DUAL FOLLOWER". And then my posted schematic was trying to figure out what he meant or how they would be connected... Note the post doesn't say "HERE IS THE BEST SOLUTION AVAILABLE". Or presented as a solution at all...

Certainly no solution at all, but I was just amazed at how anyone could post such a poor 'circuit'? - it fails on absolutely every level.

sorry but you are wrong. The high side driver you present in your post does drop the .7V. Any time you switch a transistor like that, it's going to drop .7V emitter to collector.

I suggest you reassess your electronics knowledge, the entire reason for doing it the way I proposed (and the reason it's always done that way) is because it DOESN'T lose 0.7V in that configuration, only about 0.1V or so (which is the best you can do). This is also the reason you should switch low-side, as you only need a single NPN then to give the lowest possible loss. The configuration also gives the best possible base current, so ensures best switching (your 'solution' doesn't 'switch' at all, but runs both transistors in linear modes).

But apparently you weren't aware of that basic information?, which could explain your peculiar design effort.

And that is EXACTLY how you took it out of context... it was never presented as a 'design solution'... I was just trying to figure out what Crutchow was talking about by using the dual follower (NPN feeding PNP)...

Presumably he was talking about my solution, the standard method of doing it.

 

[4pyros]

What is the part number on the opto?

 

[Nigel Goodwin]

sorry but you are wrong. The high side driver you present in your post does drop the .7V. Any time you switch a transistor like that, it's going to drop .7V emitter to collector.

I suggest you try it and see, if your electronics knowledge is so limited as to not understand the simple basics?.

Here's one of MANY examples that google finds:

https://www.engineering.com/Ask@/qactid/1/qaqid/1513.aspx

 

[crutschow]

..........................................
An NPN driving a PNP.... this is the point where I asked Crutchow "wouldn't the PNP also drop .7V?" Then he responded, "not if they were both followers the BE drops would cancel", hence the "DUAL FOLLOWER". ....................................

............................The high side driver you present in your post does drop the .7V. Any time you switch a transistor like that, it's going to drop .7V emitter to collector.
.................................

Where exactly did I give you that response?

Here is a simulation of the circuit NPN-PNP configuration I was referring too (and Nigel also I believe). As you can see the emitter-collector ON voltage drop of the PNP is <50mv. The minimum drop is 0.7V if it's an emitter follower configuration, but this is common-emitter.

 

[Mike odom]

Where exactly did I give you that response?

Here is a simulation of the circuit NPN-PNP configuration I was referring too (and Nigel also I believe). As you can see the emitter-collector ON voltage drop of the PNP is <50mv. The minimum drop is 0.7V if it's an emitter follower configuration, but this is common-emitter.

View attachment 85034

you didn't, I was asking you if that was the circuit you meant, and/or to show me the circuit you were talking about

 

[crutschow]

you didn't, I was asking you if that was the circuit you meant, and/or to show me the circuit you were talking about

I'm confused. You said I responded with "not if they were both followers the BE drops would cancel", hence the "DUAL FOLLOWER", and now you are saying I didn't?

 

[crosslakeguy]

hmmmm..........

 

[crosslakeguy]

hmmmm.........

 

[crutschow]

hmmmm..........

Don't be so verbose.