Ah ha, mystery solved!

The mouser part description is incorrect (here)

It is in fact an NPN after all.

I am curious about the Vbe being just 0.52v (I had expected closer to 0.7v)? Does that seem right?

 

Ah ha, mystery solved!

The mouser part description is incorrect (here)

It is in fact an NPN after all.

I am curious about the Vbe being just 0.52v (I had expected closer to 0.7v)? Does that seem right?

If R8 is 10k, then Ie≈ Ic=(3V-2.4V)/10k=60uA. 0.52V Vbe for Ie=60uA is very believable.
Let us know if you can get 700mA. You might want to change R8 to around 1k.

 

Ah yes, looking at the datasheet again, I can see on page three, Ib plotted against Vbe.

If my calculations are correct, with a gain of 150 and Ic of 60uA, Ib should be 60/150 = 0.4uA. Looking on the graph (for ambient temp = 25C), this gives a suggested Vbe=0.55V - pretty close.

Reducing the resistance from 10k to 1k, I'm assuming will roughly increase Ib from 0.4uA to 4uA. This would move us up the curve to about 0.61v.

Looking at the graph above (Vbe vs Ic), it appears that we need about Ic=3mA to get a Vbe of 0.7v. Is there a way to calculate the correct resistance needed in order to achieve 3mA?

(Edited out incorrect gibberish that was here!)

 

Here's the potential problem: When you increase Ids to 700mA, Vgs will have to be higher. You may not be able to get 700mA Ids with 2.3V Vgs (3V-0.7V).
You could increase your chances by using a MOSFET with lower Vgs(th), such as FDC637BNZ.
In any case, I would change R8 to a 10k pot (connected as a rheostat) in series with a 100Ω fixed resistor, and adjust the pot until the LED current is 700mA.

 

Many thanks, I now understand this circuit much better.
I've tried a 470 ohm resistor (instead of 10k). This does increase the Vbe to about 0.57v.
I completely agree, I need a MOSFET with a lower Vgs(th) to ensure it can be sufficiently on to have insignificant resistance.

I find it interesting that Vbe can be lower than 0.7v. While it does mean there is more heat to be dissipated by the MOSFET for any given battery voltage, it also means that the circuit can run to a lower battery voltage (hence run longer).

Any idea if there is likely to be consistency in the NPN Vbe at such low current levels?

 

Many thanks, I now understand this circuit much better.
I've tried a 470 ohm resistor (instead of 10k). This does increase the Vbe to about 0.57v.
I completely agree, I need a MOSFET with a lower Vgs(th) to ensure it can be sufficiently on to have insignificant resistance.

I find it interesting that Vbe can be lower than 0.7v. While it does mean there is more heat to be dissipated by the MOSFET for any given battery voltage, it also means that the circuit can run to a lower battery voltage (hence run longer).

Any idea if there is likely to be consistency in the NPN Vbe at such low current levels?

Low Vgs(th) does not necessarily imply low ON resistance. What you need is a MOSFET which can sustain 700mA at Vds≈1V without requiring high Vgs. This calls for a device with low Vgs(th) AND reasonably low Rds(on) while Vgs is low.
The only problem (that I can think of) with running low Vbe is that the temperature coefficient of Vbe is approximately -2mV/°C, pretty much independent of the value of Vbe. This means that the percentage change in Ids will be higher when Vbe is lower.

 

Thanks Ron.
So do you mean that as the circuit heats up, Vbe would drop more quickly (than if it started higher)? This could actually be desirable to protect the circuit from overhearing if I can tailor the parameters correctly.

 

Thanks Ron.
So do you mean that as the circuit heats up, Vbe would drop more quickly (than if it started higher)? This could actually be desirable to protect the circuit from overhearing if I can tailor the parameters correctly.

The percentage change per degree is greater when Vbe is lower.
2mV/0.5V=0.4%
2mV/0.7V=0.29%

The BJT will not get warmer when the MOSFET heats up unless you thermally bond the two parts together. The BJT alone dissipates very low power, and so will not heat up significantly.
Otherwise, if your ambient temperature does not change, the current will not change. This is ignoring the fact that Vgs will change as the MOSFET heats up. This will cause the BJT current to change, which will change Vbe, which will change the current. Not sure how significant this is.