Continuity Tester: A home-project query

[wirefree]

G'day!

I appreciate this opportunity to submit a question.

My concern is the branch from the Emitter (E) to and through the 100k Ohm resistor that connects to the upper probe.

I am unable to ascertain its utility.

The circuit performs its function assiduously even if I remove the link between E and 100k resistor, further instigating my suspicion.

Would somebody indulge me please.wirefree

 

[Tom.G]

That 100K resistor supplies a path from the Base to "Circuit Common", the Emitter of the transistor. Without it the Base would be floating, that is not connected to anything. With the Base floating, many transistors will partly turn on because of internal leakage current.

Also if there are any radio stations nearby, the upper probe wire, connected to the base, acts like an antenna and may supply enough voltage to turn the transistor on. Nearby power lines could also supply enough signal to turn on the transistor, but they would have to be very close to the probe wire, or very high voltage, to see any effect.

 

[wirefree]

That 100K resistor supplies a path from the Base to "Circuit Common", the Emitter of the transistor. Without it the Base would be floating, that is not connected to anything.

Appreciate the advice, Tom.

On second thought: of course. I have learned that PNP transistors need to have a potential difference of about 0.7V between Emitter and Base for it to amplify current.

Taking this further, I am keen to understand how the resistor values are arrived at.

My guess is that for the chosen transistor, collector current and gain is known beforehand; thereby, base current is easily calculated. Next, given base current and potential drop between base and ground (~5.3V), the 47k resistor value is calculated.

Carrying on with this train of thought, I am unable to perform a similar treatment to arrive at the value of the 100k resistor.

What factors determine the resistance of this resistor?

As an aside, which rows in your experience mention the gain (aka beta) value of this transistor?

Indebted for the advice,
wirefree

 

[Tom.G]

The resistance the Base sees must be low enough so the Collector Cut-off Current, I_CBO, does not raise the Base voltage to its threshold, -600mv for this transistor. So for this circuit, the maximum base resistor would be 600mV / 15nA = 40MegOhms; that's when operating at 25°C. Since the leakage current goes up at higher temperature, the maximum resistance will have to decrease.

The lower limit of the Base-Emitter resistor is determined by how much load it imposes on the incoming signal. A Continuity Tester is usually designed to detect a relatively low current flowing thru whatever is being tested. For instance if that 100K resistor were replaced with a 1Ω resistor, it would need 700mA flowing thru it to indicate continuity (turn on the transistor). Not healthy for testing a small-signal diode, LED, or a small-siganl transistor.

If testing a mechanical switch rated at an Amp or more, you might want that 700mA test current to see if the contacts are corroded or not making good contact.

 

[sophiecentaur]

I am unable to ascertain its utility.

The circuit is basically a slightly soggy switching circuit. (i.e. you may find the LED glowing dimly if there is a high resistance between the probes. In simple terms, the 100kR resistor keeps the base - emitter voltage at zero (you could call it a bias voltage to turn the transistor definitely off) when there is no continuity (i.e. no path to 'pull' the base volts to a 600mV or so, to make the transistor conduct (switch it on). You have found that the circuit works without the 100k R, which is not too surprising but, in electronics, we always try to give devices a definite voltage to work with, rather than relying on the natural tendency for the eb diode to hold the eb volts at zero.

Its a general principle. If you ever start to use discrete logic circuits, you will always need to connect unused pins to 1 or 0, depending and never leave them floating or you can get all sorts of unexpected results.

 

[wirefree]

A Continuity Tester is usually designed to detect a relatively low current flowing thru whatever is being tested. For instance if that 100K resistor were replaced with a 1Ω resistor, it would need 700mA flowing thru it to indicate continuity (turn on the transistor). Not healthy for testing a small-signal diode, LED, or a small-siganl transistor.

Much appreciated, Sir. Just what I was looking to hear!

The resistance the Base sees must be low enough so the Collector Cut-off Current, I_CBO, does not raise the Base voltage to its threshold, -600mv for this transistor. So for this circuit, the maximum base resistor would be 600mV / 15nA = 40MegOhms; that's when operating at 25°C. Since the leakage current goes up at higher temperature, the maximum resistance will have to decrease.

This one is causing me some concern.

To my mind you suggest: while the transistor is 'cutoff', and has 15 nA flowing through from E to C, we, for some reason, don't want the transistor to turn on.

I am tempted to question at this point: Isn't the whole point of a transistor to turn on and amplify?

But I won't say that because, of course, we want to employ the transistor as a switch i.e. to turn it off & on at will.

However, I can't see that intention being implemented in the given continuity tester, as from the start itself the circuit is assumed to be 'active' i.e. amplifying from the word go.

Your continued guidance is requested.wirefree

 

[sophiecentaur]

The function of the 100k bias resistor is to turn the transistor off when there is no continuity. It turns on with only a small current through the test leads.
As an amplifier, the way the transistor operates is very non linear. It is more of a switch. The switch is 'operating' whether turned on or off. Both states give information about what is connected (or no) to the leads.

 

[Merlin3189]

As an aside, which rows in your experience mention the gain (aka beta) value of this transistor?

h_FE the DC current gain is your best bet here. Not quite the same as h_fe or β , but that doesn't matter since they have a very wide range, which is not unusual. This is why the required base current is not easy to define. Even if you measure it for your particular transistor, it will vary with collector current and temperature.

 

[sophiecentaur]

h_FE the DC current gain is your best bet here. Not quite the same as h_fe or β , but that doesn't matter since they have a very wide range, which is not unusual. This is why the required base current is not easy to define. Even if you measure it for your particular transistor, it will vary with collector current and temperature.

And that is why most well designed circuits use feedback to minimize the effect of the spread of transistor characteristics.

 

[Tom.G]

To my mind you suggest: while the transistor is 'cutoff', and has 15 nA flowing through from E to C, we, for some reason, don't want the transistor to turn on.

I have two comments here.

1)

low enough so the Collector Cut-off Current, I_CBO, does not raise the Base voltage to its threshold,

Note the "I_CBO" here. The "I" is current, the "CBO" shows where and under what conditions the current is measured. In this case it shows the measurement is from "C" (Collector) to "B" (Base) with the remaining terminal (the Emitter), "O" Open (no connection). This is the leakage current from Collector to Base. If the Base is left open-circuit this current can flow Base to Emitter and act like an input signal, partially turning on the transistor.

Other measurement conditions you may run across are I_CES . . . Collector to Emitter current with the Base Shorted (to the Emitter),
I_CEX . . . Collector to Emitter current with the Base connected to the Emitter thru a circuit defined in the data sheet (usually a resistor).

2)

I am tempted to question at this point: Isn't the whole point of a transistor to turn on and amplify?

Except when you want it off. @sophiecentaur covered it well in post #7.

 

[Merlin3189]

I've not commented on the main question here, as I thought the others, esp. Tom had gone into it pretty thoroughly. But it's rolling on, so I think I'll add my hand-waving thoughts, in case it helps you.

To my mind you suggest: while the transistor is 'cutoff', and has 15 nA flowing through from E to C, we, for some reason, don't want the transistor to turn on.

I am tempted to question at this point: Isn't the whole point of a transistor to turn on and amplify?

But I won't say that because, of course, we want to employ the transistor as a switch i.e. to turn it off & on at will.

However, I can't see that intention being implemented in the given continuity tester, as from the start itself the circuit is assumed to be 'active' i.e. amplifying from the word go.

Some thoughtful comments there. Yes, IMO, there isn't a "whole point" of a transistor. It is what it is. Sometimes we'd like a perfect switch - it isn't, but we can make it do a good enough job. Sometimes we want a perfect amplifier - it isn't, but we can make it serve. (I remember transistor radios where they just used it as a rectifier: presumably it was convenient not to stock diodes as well, so they just used another transistor. Or maybe they just used up the ones that were not good enough to work as an amplifier?)

Anyhow, the other bits & pieces, the resistors, are there to make it do the job we want. Here that job is a switch. We want the LED on when there is continuity, low resistance, between your test leads and we want the LED off when there isn't. (What counts as low enough resistance is another question.) And we don't want it to be "active" except for that brief instant when it switches on or off. The rest of the time we want it to be inert - either a perfect conductor we can ignore, or a perfect insulator we can ignore.

So in what way is our transistor not a perfect switch? Well, it doesn't switch off, not completely. V_CBO means that even when off, a tiny bit of current can flow. That (upto) 15nA getting from C to B isn't enough to worry about. It won't light the LED. But if it carries on to the E, it will get amplified (up to 800x) and become (upto) 12uA. My guess is that wouldn't matter either, because I don't think it's enough to light up your LED. (I think that's what you found, when you removed the 100k.)
But the designer thought, just to be sure we can provide it with another route - out of the base and through a resistor back to the battery - so that it doesn't get amplified. We want a big resistor, so that it doesn't confuse the issue of whether the circuit you're testing is conducting. But not too big, because we don't want the voltage across it to be more than a few tenths of a volt. If the voltage gets more than that, the current will start going through the E as well. Tom got 40M (max). The designer chose 100k, which is well on the safe side, but no one would confuse that with good circuit on the probes.

The other resistor deals with another limitation. If we connect a short-circuit to the probes, as we normally intend to do, current can flow from the battery straight through the EB and back to the battery. That's not an easy calculation, but it takes only 0.8V to drive 10mA through EB and the current rises exponentially with the voltage. So by the time it gets to 6V, it will be about.. a lot .. of current! And 6V x a lot of current is going to be way above the power capability of this little transistor. (It looks like it could manage about 500mW, so max current at 6V is about 100mA. The "a lot" of current would be well over that.)
The designer seems to have taken the view that, since the transistor current gain is at least 100, we only need to let 1/100th of the current through the base and has picked a resistor 100x as big as the collector resistor. (Let's hope he got the collector resistor right.)

 

[wirefree]

I have committed an error in not including in the picture in my very first post the list of materials used. It appears to the right of the circuit (and its top left corner is all that's visible).

Even so, the transistor is PNP type, as may be inferred from the E/C/B markings on the circuit.

As most of them pertain to NPN type, I fear I've stood to lose out on all your detailed and kind explanations on account of the above.

Having submitted that, I will attempt to translate your helpful explanations to PNP and would appreciate if you can stop by and check if I've got them correct.wirefree

 

[wirefree]

QUOTE="Tom.G, post: 5797768, member: 581973"]Note the "I_CBO" here. The "I" is current, the "CBO" shows where and under what conditions the current is measured. In this case it shows the measurement is from "C" (Collector) to "B" (Base) with the remaining terminal (the Emitter), "O" Open (no connection). This is the leakage current from Collector to Base. If the Base is left open-circuit this current can flow Base to Emitter and act like an input signal, partially turning on the transistor.[/QUOTE]

How could I have done without any understanding of the notation! Many thanks, Sir.

 

[Merlin3189]

I don't think you need worry too much here about whether it is PNP or NPN. If you changed it to an NPN, you'd just reverse the battery and the LED and everything would work just the same.
The crucial bit is whether the Base is 0.6 / 0.7 V different from the emitter. In one case it is +0.7 in the other it is -0.7, but I didn't even think about the sign, just the magnitude. The sign is always towards the collector supply.

For I_CBO I talked as if the current starts at the collector, goes to the base, then looks for where to go next. It isn't like that of course. Current flows all through a circuit at the same time. What we are really saying is that there are two routes between E and B, one through the EB junction and one through the resistors. The EB junction has an energy barrier such that no current can flow until there's about 500m V difference across it (in whatever polarity is appropriate for that transistor.) So the current will actually flow only via the resistors, as long as the pd stays less than the 500mV. If we did not provide the resistor path, the voltage would just rise until EB did conduct and the current would flow through that route (provided that the battery was above 500mV .).

If you join more transistors or other circuits together, then you do need to think about the polarities.