CMOS inverter output to Biopolar

In other words, to an NPN BJT's base, with the emitter tied to ground? If so, that explains the 0.7V.

Jon

Only if the inverter is always sourcing current / can't pull to ground. We do not know this level of detail though.

Graham

Jonathan Kirwan =E5=AF=AB=E9=81=93=EF=BC=9A

to

You hit it.

Boki.

John Larkin =E5=AF=AB=E9=81=93=EF=BC=9A

It seems that what ever put this resisotr in base or emitter, we can't get full range ( 0v~3.3v) in inverter's output, if we use npn/pnp as next stage ... , am I right?

Best regards, Boki.

bipolar, not biopolar, good, now we're getting somewhere.

OK, use two resistors. Yawn. Either this is really boring, or it's my bedtime. Or perhaps both? Yawn. zzzzzzzzzzzzzz.

--
 Thanks,
    - Win

When your CMOS output is high, it is sourcing current into the BJT's base. If this is the output of a micro, chances are that the effective source resistance is in the rough 100 ohm range (50-500 wouldn't surprise me at all.) So this limits the current into the base. However, the base will only rise up to about one diode drop, though it may be more than .7V due to the neighborhood of 10mA or so.

This is not surprising. Think of the CMOS output itself as being two switches like this:

: +3.3V : | : | : o : \\ : \\ : o : | : \\ : / 100 : \\ : / : | : +-------> CMOS OUTPUT : | : \\ : / 40 : \\ : / : | : o : / : / : o : | : | : gnd

where exactly one of the two switches is closed at any one time. I specifically chose to use two different resistor values in order to highlight the fact that the low-side FET usually has somewhat lower resistance than the high-side FET -- although this is not necessarily the case and some CMOS outputs are actually configured in the opposite way (so that the sourcing side is actually beefier.)

Now, if you imagine this connected to your BJT, it looks like:

: +3.3V : | : | : o : \\ : \\ : o : | : \\ : / 100 : \\ ,------>

: / | : | |/c : +-------| : | |>e : \\ | : / 40 | : \\ gnd : / : | : o : / : / : o : | : | : gnd

Now, when the upper switch is closed (the lower one open, of course), current flows from +3.3, via the 100 ohm resistor, through the diode which is the base-emitter diode of the BJT. The current in the BJT's base rises exponentially with slight changes in the base voltage (at about 60mV per 10-fold current change. So quite quickly the 100 ohm resistor becomes the limiting factor. The voltage across the base- emitter junction will rise to some 0.75V or so and the current will be about (3.3 - .75)/100 or about 25.5mA. Actually, for 3.3V CMOS, the actual current will probably be less than that. But in any case, the base emitter voltage won't be a lot different from 0.75V.

Anyway, a resistor as others have said makes sense to further limit the current. It's not likely that you need that much base current and it's probably outside the specs for the CMOS, anyway. With a resistor going from the CMOS output to the BJT base, you still won't see the voltage at the base vary much from 0.7V. But, at least, the CMOS output will then be permitted to be much higher than before -- much closer to the value you expect from it.

Jon

No. You can not get full range unless the resistor was put in by what ever the base or emitter wants. An inverter will use the npn/pnp for the previous stage if the next stage does not work.

Can you post an ascii schematic showing what you have?

John

I assumed Boki was setting the CMOS output so that it was at about

3.3V before being connected to a BJT NPN base and that he was wondering why the output suddenly went to 0.7V or so, instead of staying at 3.3V where he was thinking it should be. One gets into their head the idea that a 3.3V output is supposed to put out 3.3V, no matter what. Not realizing that there are times when a very slightly deeper understanding is required.

Jon

Why not ?

I doubt it.

Graham

< snip >

Which is hugely wasteful of current ( power ).

Boki would do well to read the current thread 'transistor switch' here.

Graham

I follow you. I was puzzled by his comment that the output 'sticks' at 0.7V. I though maybe he meant despite being on or off. Usual problem with Boki English. That would explain his post.

Boki's not aware that there is effectively a 'hidden' resistor inside the inverter.

It makes me cringe to see the lack of knowledge of basics so prevalent today.

Graham

Just got back from gaming and drinking with coworkers, so this is especially funny right now. (Boki's "I don't know what a neuron is" makes it even funnier.)

LOL!

-Le Chaud Lapin-

base : emitter = 2 : 1

Right?

You mean should the resistance of the 'base resistor' be twice the value of the 'emitter resistor' - Not very likely !

Did you by any chance work in a former life as a cryptographer Boki ?

Please read the thread in this group called " transistor switch ". It will explain all.

Graham

Pooh Bear =E5=AF=AB=E9=81=93=EF=BC=9A

Very sorry, almos asleep.

I mean:

and I get:

No full swing, unless very large resistor, am I right?

Best regards, Boki.

John - KD5YI =E5=AF=AB=E9=81=93=EF=BC=9A

ous

Hi

and I get:

Boki.

Le Chaud Lapin =E5=AF=AB=E9=81=93=EF=BC=9A

which university ... =3D =3D ...

This is to drive an led - right ?

You do not need R2.

R4 should be chosen according to the required led current, say 330 ohms for about 3mA. The led of course goes in series with R4.

R6 can be increased in value to save power drain ( currently it 'wastes' about 1 mA - I suggest maybe 22k ).

I'm surprised you get as much swing in fact. You don't want R2 for this application.

Graham

Pooh Bear =E5=AF=AB=E9=81=93=EF=BC=9A

or

remove R2, I got it, thanks.

but it seems that I still have to use more than 500Kohm to get > 3V swing.

am I right?

If I use a MOS, it is easy to reach this specification without any resistor.

am I right?

Best regards, Boki.