Questions about electronics

[esaj]

As I've said in a couple of posts, I've started to "re-learn" electronics. I did do some electronics studies over a decade ago (early 2000's), but have long since forgotten the little I knew. While it has started come back to me after lots of reading, I'm still not that sure about semiconductors... specifically, I've been looking at some current limiting/constant current source -circuits lately, and now I'm perplexed.

I picked this circuit up from here: http://www.instructables.com/id/Constant-Current-Source-with-Operational-Amplifier/step3/How-it-works/

And here's my simulation in LTspice:

(Btw, not that sure if op-amp should be used with separate +-15V supply voltages, but the simulation seems to work OK).

It's a constant current source made with voltage divider, an op-amp (operational amplifier) and a PNP-transistor. Simulated in LTSpice, it does what it's advertised to do: it limits the current across R3 and R_load to 9.1mA (with 150 ohm R3 and 15V supply).

Here's what I do understand about this (I think, do mention if I made a mistake here ), which is pretty much also what the linked "how it works"-page explains:

The voltage-divider R1 + R2 supplies 15V * (R2 / (R1 + R2)) = around 13.636V to the non-inverting input of the op-amp. There's a negative feedback to the inverting input, so it tries to keep the inverted input at non-inverting input voltage by changing the output voltage, so the voltage drop over R3 is 15V-13.636V = 1,364V. The actual output from the op-amp is about 0.7V lower than 13.636V, because there's a voltage drop over the base/emitter -junction of the transistor (over which the negative feedback goes). For the current over R3, the transistor and R_load to be 9.1mA (0.0091A), like the instructions say (and the simulation confirms), the resistance should be 15V / 0.0091A = 1648.35 ohms.

I simulated this circuit in LTspice with different resistances of the R_load, and yes, it does limit the current to around 9.1mA (well, there's slight variance, but LTspice takes some more phenomenons into account). Lowering the R_load resistance, I got the following results (I've rounded the numbers a bit, but it shouldn't matter here):

R_load, Voltage over R_load, Power dissipation in transistor
2k,   13.6158V,  1.914mW  
1.5k, 13.5214V,  1.091mW
1k,    9.0465V, 41.55mW
470,   4.2528V, 84.94mW
300,   2.7147V, 98.86mW
100,   0.905V, 115.25mW
10,    0.095V, 122.6mW
1,     0.009V, 123.4mW

With 2k ohm R_load, the current is still below the limit. With 1.5k ohm R_load, series resistance of R3 and R_load is pretty near the 1648 ohm (that would produce 9.1mA current if there were only resistors), so no limit needed still, the power dissipation over the transistor is slightly lower than with 2k R_load (because the current over R3 is still 9.1mA with 2k load, and the transistor burns the excess off somehow?). After this, the voltage over R_load starts to drop, and the power dissipation (and voltage drop) of the transistor goes up. The voltage drop over R3 stays the same (because the op-amp keeps the base & emitter-voltages at specific level?).

What I really fail to understand is what happens at the transistor. I do know that the base current can be used to control the current over the collector and emitter, but here the current is got from an op-amp and doesn't seem to vary much between simulations (a few µA) and the current over R3 + R_load (and of course collector and emitter of the transistor) stays around 9.1mA.  I guess the transistor is being driven in the active-area here (not saturated), and that's controlling the current or is the maximum current limit simply caused by the controlled voltage drop over R3? Why and how does the voltage drop/power dissipation over the transistor occur (how is it calculated?), I do have a faint memory of an effect like "dynamic resistance" in transistor (using transistor as an adjustable resistor) being discussed at class, is that at work here? Need to look at my books again, if any of those explain it...

If someone could explain to me how this circuit actually works (preferably with a few calculation examples), I'd be very thankful

[esaj]

Doh, finally stumbled upon this and I think that now I actually get how the values are calculated (and feel a bit stupid)...

Sure hope my books would have examples like that...   

In case anyone's interested (probably not ), here's what I figured out once I saw the above example:  Since I know the resistor values, current I_e (limited to 9.1mA) and voltage V_e , and since I_c = about the same as I_e (actually I_e + I_b), I can calculate the V_ce (voltage drop over the transistor) and from there the power dissipation (V_ce * I_c). 

V_e = 13.636V

V_b = V_e - V_eb = 13.636 - 0.7V = 12.936V   (Vb = Ve -Veb because it's a PNP-transistor, in the above picture that's an NPN and that's Ve = Vb - Vbe)

I_e = 0.0091A 

I_c = roughly I_e = 0.0091A   (Actually, I_c = I_b + I_e , but I don't know the value of I_b here, because I don't remember how op-amp output current is defined, but it's negligibly small in most cases anyway...)

Voltage drop over the transistor is V_ce = V_supply - I_c * R3 - I_e * R_load

Power dissipation of the transistor (neglegting small current in the base) is P_t = I_c * V_ce .

So for R_load values 1...2k ohm:

R_load,   V_ce,    P_t
1,    13.626V,  124.0mW
10,   13.544V,  123.3mW
100,  12.725V,  115.8mW
300,  10.905V,   99.2mW
470,   9.358V,   85.1mW
1k,    4.535V,   41.3mW
1.5k, -0.015V,   -0.14mW ?
2k,   -4.565V,  -41.5mW ?

Close enough compared to the simulation values. Although the last two are obviously off once the total resistance from the resistors alone goes above 15V / 9.1mA = about 1648ohm (it's not like the transistor could create power from nothing, and I don't think it can drop the voltage lower than ground?), the LTspice-simulation would suggest that a more meaningful current starts to flow towards the base, I guess it goes through the op-amp to ground when no actual current limiting is needed...