Can I do this

[DetlefSchmitz]

It is a wifi security camera that runs off a USB socket

FYI the 5V USB supply is specified as being between 4.75V and 5.5V. So this is the range you need to aim for to guarantee operation.

 

[JohnW2]

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.

Is that not only really relevant at very low currents? As I understand it, The Ib vs Vbe curve for a transistor becomnes almost vertical (i.e. voltage changes little for further large increases in current) beyond a certain (low) current? Am I wrong again? (I'll see if I can hunt out some curves ).

.

I'm not quite sure what that means ('change' from what?) but, anyway, should it perhaps read 0.03 V ??

 

[DetlefSchmitz]

I'm not quite sure what that means ('change' from what?) but, anyway, should it perhaps read 0.03 V ??

Change from slightly over 0mA to 10mA (i.e. twice the current change while reducing the voltage drop change by nearly seven times compared with the transistor b-e diode example). And yes, sorry about the typo.
I'm not sure why anyone would consider not connecting the base to collector when it costs exactly nothing.

 

[RandomGrinch]

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.

Why not simply use a USB power bank?
They are easy to charge, have a large capacity and provide a steady 5V output.

 

[JohnW2]

Change from slightly over 0mA to 10mA (i.e. twice the current change while reducing the voltage drop change by nearly seven times compared with the transistor b-e diode example). And yes, sorry about the typo.

Thanks, but I still don't really understand your point. Are you perhaps assuming that the voltage would be exactly 0.6V at a current 'slightly over 0 mA (and,if so, why?) ? The graph only goes down to 10 mA, but the way the curves is heading suggests that it would hit 0.6 V at a current of about 7 mA- not all that much closer to 0 mA than is 10 mA.

One conceptual difference between the curves for diodes and transistor BE junctions is that one is 'concave' and the other 'convex'. I haven';t yet found a BE one which goes up to reasonably high currents but ones like this one illustrates my point, since it shows that Vbe changes little for increases in Ib beyond a very small number of mA...

I'm not sure why anyone would consider not connecting the base to collector when it costs exactly nothing.

I cannot disagree with that.

 

[JohnW2]

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.

I'm not sure why anyone would consider not connecting the base to collector when it costs exactly nothing.

I cannot disagree with that.

As I said, I couldn't disagree with your statement, but I also don't think it really makes any significant difference whether or not one connects the collector,given only that the transistor can stand the current concerned through its BE junction.

To reassure myself about that view, I've just experimented with the BE junctions of a couple of BC107s in series, with a 6V supply, with and without the collector connected to the base - and, as you can see, there is virtually no difference between the two in relation to the 'voltage drop'.

 

[SUNRAY]

As I said, I couldn't disagree with your statement, but I also don't think it really makes any significant difference whether or not one connects the collector,given only that the transistor can stand the current concerned through its BE junction.

I've just experimented with the BE junctions of a couple of BC107s in series, with a 6V supply, with and without the collector connected to the base - and, as you can see, there is virtually no difference between the two in relation to the 'voltage drop'.

View attachment 336181

It's so different to

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

Where the current was essentially upto 10mA I didn't note actual readings to plot an accurate graph but I think my additions are approx representative of my very quick test except I've predicted what would have happened with 2 transistors from my test with one.

 

[jj4091]

Why not simply use a USB power bank?
They are easy to charge, have a large capacity and provide a steady 5V output.

I have never heard of them and they never came up in my search for a power source for the camera. If using the type that I have purchased proves not to be acceptable ie the batteries do not maintain a charge for long enough then I will purchase one of those devices. Thank you.

 

[SUNRAY]

first hit on google other than ebay:

Amazon.co.uk : USB Power Bank

 

[jj4091]

first hit on google other than ebay:

Amazon.co.uk : USB Power Bank

Yes, thanks. your original suggestion of 3 batteries is working well at the moment, but as I said if it doesn't continue for an acceptable length of time then I will source one of those.

 

[JohnW2]

It's so different to .... Where the current was essentially upto 10mA I didn't note actual readings to plot an accurate graph but I think my additions are approx representative of my very quick test except I've predicted what would have happened with 2 transistors from my test with one.

Indeed- very (and inexplicably) different.

If I understand correctly, you have done something slightly different from me, in as much as it sounds as if you varied the 'supply voltage',whereas I had a fixed (6V) supply and varied the resistor to emulate changes in thee load - but I can't see how that could explain what you observed. In particular,I can't really see how, with C and B connected, thee 'voltage across the transistor' (C-E and B-E) could bee as low as 0.3-0.5 V,and nor can I see how connecting C and B could result in the large change in that voltage which you report!

 

[JohnW2]

Indeed- very (and inexplicably) different. .....

@SUNRAY ... I should perhaps have added - things start changing markedly when current reduces below about 10 mA - so you only really looked at a pretty atypical part of 'the curve'

 

[SUNRAY]

Indeed- very (and inexplicably) different.

If I understand correctly, you have done something slightly different from me, in as much as it sounds as if you varied the 'supply voltage',whereas I had a fixed (6V) supply and varied the resistor to emulate changes in thee load - but I can't see how that could explain what you observed. In particular,I can't really see how, with C and B connected, thee 'voltage across the transistor' (C-E and B-E) could bee as low as 0.3-0.5 V,and nor can I see how connecting C and B could result in the large change in that voltage which you report!

Yes I did it differently but the effect is the same, ie pass a current through a device. to me the green curve emulates your findings perfectly. As mentioned I didn't not any stage reading, literally varied the voltage between 0-10V several times and mentally noted max and min voltage with collector isolated, then connected. Below 1V it seemed to follow the supply V so there was an early hump, very vaguely this idea but in no way to scale:

Hence the reason I only mentioned 1-10V earlier. If I get the chance I'll have another play later.

 

[SimonH2]

Have you looked at the power consumption ? I strongly suspect your battery life could be measured in hours rather than days.

 

[JohnW2]

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

@DetlefSchmitz ... I've just been looking back through this thread and realised that I never answered this question of yours, so,for completeness ....

...once I started 'thinking straight' (remembering the conditions required for saturation), I realised that one cannot achieve C-E saturation if C and B are joined - since one of thee requisites for that saturation is that Vbe > Vce (which, in practice, occurs when Vce falls to below Vbe).

Kind Regards, John.