Help analyzing circuit - PNP transistor with Base Emitter biasing

[JustSquareEnough]

I'm experimenting using a darlington PNP  (tip105) in a current limiting circuit. I am hooking it to a 1 ohm power resistor just for ease of measurement and calculations.

Originally I built this circuit with out R1 / C1 and experienced instability which I eventually realized was thermal runaway (current through Rl would climb and climb and never settle).  after some googling I found a similar tip105 circuit that used the included R1 and C1 configuration this solved the problem but for the life of me I can not explain why.

here are my observations:

with Vcc = 12.3 volts and the pot at 100k:
V(pot1 across wiper): 12.3V
V(R1): 0V
V(R2): 0V, therefore curent of the base is 0A
V(Rload): 0V  therefore, Iload / collector is 0A

with Vcc = 12.3 volts and the pot at 0:
V(pot1 across wiper): -.007V
V(R1): .915V
V(R2): 10.72 therefore current of the base is calculated at .00107A
V(Rload): 1.87V therefore current if Iload / collector is 1.87 A


I have read a few articles about transistor biasing but they all give NPN resistors as examples and I am have a difficult time underastanding and explaing the circuit.

my main question is what roles does R1 / C1 play as the transistor changes temperature , etc, to help stabilize it?

How do I go about calculating the values for R1, C1, R2 given a desired maximum collector current?

thank you in advance for any help that can be provided.

-David

[T3sl4co1l]

You need an emitter resistor to make a current limiter of any useful accuracy; without emitter degeneration, a Darlington especially will just thermally run away (heat dissipation --> more hFE --> more current --> ... --> BANG).  Check the datasheet for hFE versus temperature.

Tim

[Zero999]

Yes, no emitter degeneration = instability and smoke.

I have read a few articles about transistor biasing but they all give NPN resistors as examples and I am have a difficult time underastanding and explaing the circuit.

Simple. Reverse the power supply and any other polarised components such as electrolytic capacitors and diodes and the circuit will work with PNP transistors.

[orolo]

Hm, it's a curious configuration. Looks like the capacitor provides a hard start for the transistor, and then the resistor R1 could somewhat limit the action of the transistor.

First, forget about R1 and C1, and compute the Thevenin equivalent circuit of the POT + 10K resistor at the base. You get a voltage between Vcc and ground, depending on the wiper, and an equivalent series resistance of 10K plus the parallel combination of the two sides of the pot. So, in your case with Vcc=12.3V, if the wiper is at the middle, your transistor is driven by a voltage Vcc/2 = 6.15V and a series resistance of 10K+ 25K = 35K. This is a bad biasing scheme, as explained in previous posts.

If you introduce the 1K resistor R1, but for now forget about C1, this resistor goes in parallel with the Thevenin equivalent computed before. Since the Thevenin series resistance is 10K+, this brings the voltage at the base at least  9/10 times closer to Vcc than it was before, possibly not enough to drive the transistor at all. For example, with the wiper at the middle as before, if you parallel 1K to the 35K resistor, the voltage seen by the base rises from 6.15V to 12.13V. With a Veb of 0.18V the transistor won't turn on.

To activate the transistor you will have to take the wiper almost to earth, so the biasing reduces to: Vcc - 1k - base - 10k - ground. Even in that case, the voltage seen at the base is 10·Vcc/11 = 11.18V, with a Veb = 1.12V. This is barely enough to keep the darlington in the active region.

Now add the capacitor. If it starts uncharged, when you power the circuit on, the transistor will turn on very hard, since the base is at 0V. The transistor will saturate instantly, and the capacitor will charge very fast. During this time, you will get a very strong pulse of current into the load. Then, when the capacitor charges fully, the effect of R1 kicks in, and the Veb will fall so low that the transistor will barely keep active, so the current into the load will be very low. The biasing can't keep the transistor saturated after the initial pulse, since the 10K resistor doesn't allow for that much base current (10V drop at 1mA current, and the 1K resistor provides most of that mA, little left to the base to keep the darlington saturated).

Grounding the base of the darlington, even for a fraction of a ms, is mistreating the transistor. For a TIP105 the absolute maximum Veb is 5V, and for a brief time you'll be driving it well over that limit. Sorry, this is reverse voltage. What matters here is instantaneous base current, which I don't know if will be over 1A, the absolute maximum.

[JustSquareEnough]

thank you to everyone for the responses.

@orolo: "To activate the transistor you will have to take the wiper almost to earth, " this is exactly what i was experiencing.  thank you for the detailed analysis of the operations.

I looked up emitter degeneration with very little understanding it looks like I have some more learning to do on the fundamentals around bjt's.

is a mosfet better suited in a current source configuration like the one I was trying to establish here, and BJT's better suited as on / off "switches" I know I just painted with a very broad brush but am curious on the response.

[LvW]

I looked up emitter degeneration with very little understanding it looks like I have some more learning to do on the fundamentals around bjt's.

Perhaps, you are stumbling on the term "emitter degeneration". Such a resistor RE in the emitter path provides negative voltage feedback - thereby, stabilizing the circuit against unwanted external influences (temperature) and parameter tolerances. Example: An unwanted emitter current increase due to rising temperatures causes a voltage increase at the emitter node - equivalent to a reduction of the base-emitter voltage, which brings the current back (nearly) to its previoous value.   

[Zero999]

Emitter degradation in this case would be a resistor in series with the transistor's emitter terminal. It is an example of negative feedback. As the current through the emitter increases, the voltage across the emitter resistor will increase, thus reducing the base-emitter voltage and base current.

[T3sl4co1l]

Note that a transistor measures its "on-ness" based on the base-emitter voltage.  Not the absolute base voltage (common beginner mistake ).

Tim

[JustSquareEnough]

Note that a transistor measures its "on-ness" based on the base-emitter voltage.  Not the absolute base voltage (common beginner mistake ).

This is the mistake I was making, and assuming the the difference in voltage presented through the pot and the voltage seen at the base was causing the turn on!  this also helps me better understand the comments by @hero999 and @LvW in that a reistor in series with the emitter helps to control / regulate the voltage DROP across base emitter.

so, is there a general set of equations on how to calculate the emitter resistor and base resistor for pnp setup with a known Vin and a wanted collector current, along with additional specs from the data sheet?

I don't expect to be spoon fed so pointing me in the right direction to what to read would also be great and I will circle back with what i learn.  i don't have AoE but i do have a copy of practical electronics for inventors. when i get home ill look to see if the subject of negative feedback for transistors is covered.

[LvW]

so, is there a general set of equations on how to calculate the emitter resistor and base resistor for pnp setup with a known Vin and a wanted collector current, along with additional specs from the data sheet?

No - there is a "combination" of equations and "selections". Normally, the sequence is as follows:
* Select the wanted collector current IC with respect to (allowed) power consumption and voltage gain requirements (the transconductance gm=IC/Vt determines the voltage gain).
* Select values for a collector resistor RC and emitter resistor RE - with respect to a reasonable remaining collector-emitter voltage VCE. At the same time, select a reasonable ratio RE/RC=0.1---0.2. (With RE you select the degree of negative DC feedback for stabilization of the operational point.).
* Based on the DC voltage VE=IE*RE and a good assumption for the base-emitter voltage VBE=(0.65...0.7)V you find the base voltage VB which must be provided (for example) by a voltage divider. Because of the negative DC feedback you have provided, the uncertainty connected with the VBE choice is of minor importance only.
* Duiring calculation of the resistor values you must consider the finite value of the DC current IB going into the base node. The resistors of the voltage divider should allow a DC current of app. (at least) 10*IB. Thus, all uncertainties/tolerances of IB=IC/B play a minor role only. For a rough calculation you even can forget the base current (divider current>10*IB).   

[Zero999]

Note that a transistor measures its "on-ness" based on the base-emitter voltage.  Not the absolute base voltage (common beginner mistake ).

so, is there a general set of equations on how to calculate the emitter resistor and base resistor for pnp setup with a known Vin and a wanted collector current, along with additional specs from the data sheet?

The normal way to do this is to assume the voltage across the emitter resistor R_E is the base-emitter drop V_BE, less than the base voltage V_B.

To put it another way: V_B = V_RE + V_BE

The current can then be controlled by varying V_B.

In your case you're using a Darlington transistor, which has double the base-emitter voltage drop of an ordinary BJT. The data sheet says if varies between 1.2V and 3V, depending on the collector current.
http://www.onsemi.com/pub_link/Collateral/TIP100-D.PDF




One way to vary V_B is to use a voltage reference before the potentiometer, so the current is less dependant on the power supply voltage. A crude method of doing this is to use an LED.