I have a question on the BC517 transistor, which is a darlington pair transistor. My question is this: in order to enter "active" mode, doesnt the voltage at the base have to be greater than 1.4V, or does the 1.4V represent the voltage at which the transistor becomes saturated, and if THAT's the case, at what voltage (or current) does the transistor enter "active" mode?

 

You have two base-emitter PN junctions to forward bias, which is the voltage required relative to the base and emitter to enable current flow. Saturation is another situation, where no further current in the base circuit will result in a change in current in the collector circuit.

As soon as current starts to flow in the base-emitter circuit, the transistor has entered the active region.

You can follow this link - http://www.allaboutcircuits.com/vol_3/chpt_2/8.html - into our Ebook for further reading.

 

Here's a link to Fairchild's BC517 datasheet:
http://www.fairchildsemi.com/ds/BC/BC517.pdf
It says that Vbe(on) is maximum 1.4v; that is if Vce = 5, Ic will be 10mA when Vbe is a maximum of 1.4v. Thus, it will enter the active region when Vbe is somewhat less than 1.4v.

 

So does that mean that when Vbe is between 0V and 1.4V that it is in active mode? So basically as long as there is ANY current flow into the base, that it is amplified (and in active mode)?

 

Well, not really. Vbe varies from transistor to transistor, but it's typically somewhere around 0.63v. Since your Darlington has two base-emitter junctions, you need to double that to around 1.26v. You must understand that this is a "ballpark" number.

You might feed microamps into the base, and the transistor won't really be in the active region. You'll have to do some experiments to find out where a few of them enter the active region; your mileage will vary.

 

So with this transistor, the active region is between 1.26V (approximately) and 1.4V (listed as the ON voltage where it turns into the saturation regime). Is that right?

Well, not really. Vbe varies from transistor to transistor, but it's typically somewhere around 0.63v. Since your Darlington has two base-emitter junctions, you need to double that to around 1.26v. You must understand that this is a "ballpark" number.

You might feed microamps into the base, and the transistor won't really be in the active region. You'll have to do some experiments to find out where a few of them enter the active region; your mileage will vary.

 

No. Look at the datasheet for Vce(sat).

 

I'm not talking about Vce. I guess my original question may not be clear, because it should be a pretty straightfoward question:

In what scenario does BC517 go into active mode (from cutoff mode)? If I slowly increase the voltage (Vbe), what causes it to leave cutoff mode and enter active mode? Is it the voltage hitting a certain level? Is it when the current reaches a limit? Is it when Vbe >0?

 

It involves the base-emitter PN junction becoming forward biased. No forward bias (by whatever amount) and it is in cutoff.

Here is another instructive link - http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html.

 

Is there a minimum voltage necessary to enter active mode from cutoff mode. i.e. if I am starting with the Vbe =0 and slowly it moves up from there, will the transistor remain in cutoff mode until some condition is met? If so, what is that condition? Let us assume for the sake of argument that Vce is a constant 5V.

It involves the base-emitter PN junction becoming forward biased. No forward bias (by whatever amount) and it is in cutoff.

Here is another instructive link - http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html.

 

As before, yes. Until the PN junction has become forward biased, it will be cutoff. In general, this will happen when the voltage is at some value. Each device may show a bit of individual variance, though, so exceedingly careful measurement is urged to discover just what this forward bias voltage is for every case.

 

ON Semi have a detailed datasheet for the BC517.
They have a graph of the typical current gain at 3 temperatures, a graph of the typical collector saturation voltage at 4 collector currents and many base currents, a graph of the typical saturated VBE and VCE and the typical linear VBE.
They have a graph of typical temperature coefficients for hot and cold.
They list the max expected VCE at one collector current and list the max expected base voltage at one current.

Each transistor is different even if they are from the same batch. You don't know if yours is a sensitive one or a not so good one. Negative feedback can be used to make them more alike.

The manufacturer does not show the spec's where the transistor emerges from cutoff to beginning to turn on because nobody uses them that way. If you want a switch then use hysteresis in a Schmitt-trigger circuit to switch, not the variable threshold voltage/current of a transistor.

 

If you tell us what you are trying to do with the BC517, we might be able to be more helpful.

 

The Motorola datasheet says that the base-emitter voltage is 1.1 volts when the collector current is 5 milliamps and the collector voltage is 5 volts. VBE for lower collector currents or lower collector voltages are not defined.

 

What I'm really trying to do is figure out what the maximum resistance that water can be and still deliver some Ic (5 mA or so...) to the buzzer (which will be replaced by something else in my circuit).

The circuit I"m referring to is here:

http://www.velleman.be/downloads/0/manual_mk108.pdf

 

The water detector circuit is too simple.
It uses DC which will unplate one probe and plate the other probe. It should use AC for no plating and reduced corrosion of the probes.

The circuit's sensitivity depends on the resistivity of the water. Distilled water is almost an insulator and conducts very little current. Salt water and water with disolved metals and salts has low resistance and conducts a lot of current. My tap water has a fairly low resistance.

The sensitivity of the circuit also depends on the surface area of the probes in the water and their spacing.

 

If you want to live with the limitations Audioguru pointed out, have a look at the analysis below.
I have assumed you want the transistor to saturate, so I used forced beta=1000. See attached curves from Onsemi datasheet.

EDIT: It will probably work with higher resistance, since the forced beta is worst case. The 2.5uA through the 470k resistor is the limiting factor.

 

Thank you very much for your analysis Ron & audioguru. I never considered the plating/unplating issue. Would this be a serious concern? I mean, where would the plating "go". I"m assuming this would only plate/unplate when there would be actual water between the leads, is that right?

 

Yes, that would be the case.