Can I do this

[jj4091]

What is it you are running, Have you tried it with just 3 x 1.5V batteries?

No I have not thought of or tried that although I think it would work if I had a suitable container with a usb socket. It's a good idea though so I will try shorting out one of the battery sections in the 4 battery holder. Thanks for the idea.

 

[jj4091]

NO YOU ARE WRONG AGAIN...............
Why do you insist on muddying the waters, I already told you to think again. Stop confusing the OP with your misunderstanding of transistors!
I suggest you go away and build yourself an experimental circuit or try LTspice.

So would my second circuit work ok?

 

[DetlefSchmitz]

So like this then ?

Yes

 

[DetlefSchmitz]

What am I misunderstanding or "need to think harder about"?

If c-e saturates what will happen to b-e? What will be the result of that?

 

[SUNRAY]

Try this simple circuit with 2 ammeters:
I predict Ic will be very low, Ib will be a whole magnitude bigger, Vo will be some random voltage under 0.7V dependant on Hfe

 

[SUNRAY]

So would my second circuit work ok?

No I've just tried it with a single 2n3704 and 1kΩ resistor. Using a variable power supply of 1-10V the voltage across the transistor varied between 0.3 and 0.5

 

[DetlefSchmitz]

This is what Spice says:

 

[JohnW2]

OK. I've done my "thinking harder" as a result of which I think I essentially agree with what you all have been saying. It seems that I had been had been overlooking the fact that to achieve saturation, Vbe must be greater than (not just equal to) Vce. I have also think that I have more-or-less satisfied myself by trying with a random NPN transistor and a couple of near-to-hand resistors....

... with B and C joined, Vce (and,obviously Vce) was 0.68 V,around the expected BE junction 'threshold'. However, if I biased the base, then Vce fell to 0.14 V [(i.e. "Vce(sat)"].

Having said that, I still think that my original suggestion (to ignore the collector and just use the BE junction) would probably be an acceptable approach, given that Ib(max) is usually much the same as Ic(max) for small transistors.

 

[JohnW2]

This is what Spice says:

Indeed (now that I'm thinking straight ). As Sunray said,with B and C connected, one is in effect just using the BE junction,hence Vce will be equal to the 'threshold' value of the Vbe junction..

 

[DetlefSchmitz]

As Sunray said

I think he was saying the opposite.

 

[JohnW2]

I think he was saying the opposite.

Are we looking at the same post?...

In which case only the B-E junction is being used.

 

[DetlefSchmitz]

Are we looking at the same post?...

I was looking at his most recent posts 20 and 21.

Having said that, I still think that my original suggestion (to ignore the collector and just use the BE junction) would probably be an acceptable approach, given that Ib(max) is usually much the same as Ic(max) for small transistors.

It would be a lot less consistent.

 

[JohnW2]

I was looking at his most recent posts 20 and 21.

Fair enough - but what I wrote was correct in relation to the post I was looking at

View attachment 336114
It would be a lot less consistent.

Yes, I suppose so. In both cases it is Vbe that is 'driving' the 'voltage drop', but if one has C and B connected, then Ib (and changes in Ib) will be very much smaller, hence the VD will be less current-dependent.

Having said that, if one did what one would usually do (and (and as you suggested right at the start,in post #2) and used diodes, then one would be using just a single junction (in each diode), hence I presume not really any different from using a single BE junction, presumably with whatever consequences that means in terms of current-dependency of the VD ?

 

[jj4091]

Thanks to all of you gents, but I am having difficulty understanding the opposing views. I have just tried Sunrays suggestion of trying to run it off 3 batteries and it works but I guess it will result in the batteries draining quicker. It is a wifi security camera that runs off a USB socket so as it has motion sensing I don't think that will be a major problem, I'll just have to monitor the power indicator for a few days and work out how often it needs fresh batteries. Thanks for all your efforts to help.

 

[DetlefSchmitz]

hence I presume not really any different from using a single BE junction

Not really. A 1N4001 diode has a junction specifically designed to take a large current, unlike a junction transistor. So it will not vary in voltage nearly so much.
.