I apologize for not responding right away, I got carried away with the fast enlightenment provided by the other users. (Sorry)No one has taken an interest in my earlier question, so here is another one.
With the proposed configuration you cannot connect one end of your load to ground.
Is this desirable?
If you require a ground connection you should revers the transitor polarity. ie use a pnp one.
O.1 amp pulses at 300 volts with american transistors?
MPSA 42 (NPN) or MPSA92 (PNP) will do the job nicely.
if you do not want to use power transistors.
MJE340 (NPN) MJE350 (PNP) would be a suitable choice of power transistor.
Also consider 2N6213 (NPN)
Vs is shown applied across R1 in series with the base emitter junction of the transistor by the follower amp.Vs goes from 0 to 3.3V in order to generate a current from 0 to 100mA(maximum) in R1
Is not 3.3 volts.3.3V + (2 x Vbe),
Ok, well I wasn't really looking to change the configuration anyway. I was thinking of enclosing the circuit in a small case or something, with no forced air flow or anything like that, but only basic breathing holes/vents..Studiot, I answered your post #6 question in #19.
jp is correct, the output of the opamp automatically compensates for the turn-on voltage of whatever the pass transistor is, up to the point where the output stage clips. For a darlington, it would be 3.3V + (2 x Vbe), or approx. 4.6V at max Vs. Vcc = 12 V will be plenty of headroom; in fact, the opamp output current is so low that a 5V rail-to-rail part would work. For a MOSFET the max opamp output voltage would be 3.3V + Vgs (@Id = 0.1 A). Depending on the FET this could be less than 4.6 V, but maybe not.
Note to jp - if you change to a current source circuit with a PNP transistor, the interface between the transistor and the opamp gets much more complicated since that interface now has 200V across it.
I don't think I'd run a TO-92 transistor at 200 mW dissipation without knowing more about the available cooling environment.
Don't think so. It is a *follower* loop, so the output follows the input (or tries to) no matter what is in between. The opamp output does whatever it takes to make V+in equal to V-in, no matter what the output is connected to. Even though the overall circuit is a current sink, the part is a voltage feedback opamp and that is how it thinks. The opamp compensates for the Vbe drops exactly the same way it does in an ideal rectifier circuit, where diode Vf drops are in the feedback loop.Is not 3.3 volts.
You obviously realise that both Vbe and Vr1 are within the follower loop, but Vs is shown as being applied to both so Vs will have to be greater by this amount, throughout its range.
Forgot my own post about average power. oops.The pulsed output with an average of 200mW is well within the capabilities of the stated small signal transistors.
You are right, I didn't read the initial circuit properly, I thought the op amp was ground referenced.Don't think so. It is a *follower* loop, so the output follows the input (or tries to) no matter what is in between. The opamp output does whatever it takes to make V+in equal to V-in, no matter what the output is connected to.
I don't see either age or temperature in my equation.In your equation, alpha varies with age and temperature,
I hate to step into someone else's argument but if the opamp has a lowish output impedance the gain of the transistor is not the main source of non-linearity and instability, it is the Vbe drop. (Or ratio of Vbe : Vrs)...
However the circuit is inherently non-linear since it depends upon the gain of the pass transistor, which in turn depends upon the collector current.
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by Jeff Child
by Jeff Child
by Jeff Child