NPN transistor still working? Strange ..

[n_tron]

Hello everyone!

I am a newbie hobbyist currently exploring more about electronic components and circuits. I have been exploring more about transistors (both NPN and PNP) and how they work. I came across this following very basic single LED blinker circuit built using single NPN transistor, capacitor and resistors.

Now, from what I have read and seen about NPN transistor is that the current starts to flow through it ONLY when a positive voltage is applied on base. (acting like a switch). However, in this circuit there is NO voltage applied on base and the emitter is connected to +ve and the collector is connected to -ve.

So how come the transistor is behaving like a closed switch in this state??

Circuit image is attached!

Would appreciate comments/advice.

 

[signalman72]

I may be wrong, but it appears the collector and emitter are shown connected backwards from what they should be.

 

[duke37]

The transistor is upside down. The emitter is normally connected to the negative supply and the collector to the positive.
Transistors such as the 2N2222 has a limit of about 6V negative, emitter to base, so the transistor will pass current when this is exeeded and discharge the capacitor into the led. I am not clear what determines the time constant.

 

[Audioguru]

I was told that allowing the emitter-base junction of a transistor to perform avalanche breakdown like that damages the fragile emitter-base junction due to it overheating, causing the transistor to perform poorly with low beta if it is later used normally.

 

[n_tron]

In above transistor, my question is that as the positive voltage is applied to its emitter (n) and negative is applied to collector (n) does that means electron can travel from collector to emitter overcoming the depletion layer? How can that be possible without applying any positive voltage on the base??

 

[signalman72]

I would imagine that in reverse as your print shows it is simply behaving like a forward biased diode.

 

[dorke]

In above transistor,
my question is that as the positive voltage is applied to its emitter (n) and negative is applied to collector (n) does that means electron can travel from collector to emitter overcoming the depletion layer? How can that be possible without applying any positive voltage on the base??

The answer to your 1st question is YES.
A transistor can operate in the "REVERSE ACTIVE mode",see picture below.
In this mode Ve>Vb>Vc (the revers of the active mode in which Vc>Vb>Ve).
We still have one of the Tr diodes in forward bias and the other reversed biased.

The answer to your 2nd question is NO.
To be in any active mode the transistor base has to be biased to satisfy the above voltage relations.
In this circuit you have Ve=12V.
Lets assume for the LED to be on a minimum of Vc=1.6V.

So we need 1.6V<Vb<12V ,but we have an open un-biased base .

Can Vb be in the desired range?
Strictly speaking ,NO.

But...
If you connect a long wire to the base it might pick-up noise like an antenna.
It might pick-up the mains signal (the very one that causes hum in audio circuits).
In which case,if that will get to the desired base voltage the transistor will conduct and the LED may blink(AC signal pick-up).

 

[Ratch]

In above transistor, my question is that as the positive voltage is applied to its emitter (n) and negative is applied to collector (n) does that means electron can travel from collector to emitter overcoming the depletion layer? How can that be possible without applying any positive voltage on the base??

You are forward biasing the LED diode and the base-collector diode. That leaves 10 volts or so to reverse bias the emitter-base diode. Is the e-b diode capable of holding up to that much reverse voltage? Or, did the voltage blow through and short it out. I would check the transistor to see if it still works.

Ratch

 

[(*steve*)]

We've looked at this before.

This circuit will probably cause the LED to emit short flashes.

The 1k resistor charges up the capacitor. At some point the voltage across the capacitor exceeds the sum of the forward voltage of the LED, the base-emitter breakdown voltage of the transistor, and the forward voltage of the base-collector junction of the transistor. At this point, the base-emitter junction breaks down allowing a pulse of current to flow from the capacitor, through the transistor and the LED, discharging the capacitor a little. The process repeats...

Does this damage the transistor, yes. Is it a thermal effect, no. The accepted reason is that hot carrier electrons are generated with sufficient energy to create point damage at the junction.

For more technical information, Google "base emitter hot carrier damage". You'll find a few papers.

If you hunny I'm this site, you'll find a thread where I should this damage to be the case -- I really didn't believe it at first.

The damage to the transistor is such that you wouldn't want to use it in a circuit requiring high gain or low noise, but it will work in this application for quite some time.

There is a similar application where you rely on the base collector breakdown. This operated at much higher voltages and can generate pulses with very short rise times.

 

[Audioguru]

Many years ago I used the reverse-biased emitter-base junction of a transistor as a 6V zener diode. It did not fully conduct and saturate like an SCR, instead it regulated the voltage like a zener diode.
A low voltage zener diode also has avalanche breakdown but it is designed to dissipate the heat but the emitter-base junction of a transistor usually never creates heat so it is fragile.

 

[n_tron]

We've looked at this before.

This circuit will probably cause the LED to emit short flashes.

The 1k resistor charges up the capacitor. At some point the voltage across the capacitor exceeds the sum of the forward voltage of the LED, the base-emitter breakdown voltage of the transistor, and the forward voltage of the base-collector junction of the transistor. At this point, the base-emitter junction breaks down allowing a pulse of current to flow from the capacitor, through the transistor and the LED, discharging the capacitor a little. The process repeats...

Does this damage the transistor, yes. Is it a thermal effect, no. The accepted reason is that hot carrier electrons are generated with sufficient energy to create point damage at the junction.

For more technical information, Google "base emitter hot carrier damage". You'll find a few papers.

If you hunny I'm this site, you'll find a thread where I should this damage to be the case -- I really didn't believe it at first.

The damage to the transistor is such that you wouldn't want to use it in a circuit requiring high gain or low noise, but it will work in this application for quite some time.

There is a similar application where you rely on the base collector breakdown. This operated at much higher voltages and can generate pulses with very short rise times.

Thanks so much for the detailed replies.... Just clarifying with some of the above information :

Here, firstly I am trying to understand the flow of the conventional current (which is from positive to negative). So in this circuit, I take it as a positive 12v charge starts from the beginning of the power source, it then passes on through the 1K resistor, now what I am confused is that the following 2 things can happen with the flow of the current at this point :

1. ALL Current will first go DOWN towards the 220uf capacitor, and when the capacitor is fully charged, it will then proceed TOWARDS the emitter of the transistor

2. Some of the current will go DOWN towards the 220uf capacitor and some will go towards the emitter of the transistor (now in this case, I don't know what exactly determines the division of the current). To clarify this, I have drawn two arrows (in red) in the below image :



Thanks.[/QUOTE]

 

[Audioguru]

The capacitor does not fully charge because the reverse-biased emitter-base has avalanche breakdown at about 6V and is in series with the forward-biased base-collector diode. Then the transistor conducts from emitter to collector and dumps some of the capacitor charge into the LED.

 

[BobK]

Current cannot flow both into and out of the capacitor at the same time. Once the breakdown voltage is hit, the current is flowing out of the capacitor. This contnues until the current becomes too low to sustain the breakdown state, the transistor stops conductung and current starts flowing into the capacitor.

Bob

 

[n_tron]

Ok .. thanks for great replies. I am able to understand all these a little bit! I first want to clarify only with the flow of the current (conventional current that flows from positive to negative)! So when the positive charge starts from the source and cross the first 1K resistor, then after that does ALL the current flow downwards towards the capacitor or some of the current flow down towards the capacitor and some towards the emitter (at the same time)? Please see the above image and the arrow I have marked in red to understand it better.

 

[Ratch]

Let's start with the last question first. Current will exist where there is a voltage and a conduction path. If more than one path exists, more current will be present in the path with the least opposition. The opposition of the capacitor is the back-voltaqe that occurs when the capacitor becomes energized to the voltage applied to it. The speed that this happens depends on the amount of current available to change the difference of the charge on the cap's plates. This in turn is dependent on the voltage applied, the resistance of the energizing circuit, and the value of the capacitance. At first, a lot of current will exist. Later as the cap energizes up to a higher back-voltage value, the current value will drop off. When the E-B junction breaks down, another conduction will be available and the currents in each branch will depend on the opposition of each branch path.

Conventional current is a mathematical convention. It assumes that the current direction from a source is from the positive to negative terminal. This is what will be physically happening If the charge carriers are positive. If the charge carriers are negative as electrons are, and it is necessary to know the true physical direction of the charge carriers, simply reverse the computed direction for assumed positive carriers. Never ask before you compute: "Does this charge carrier have a positive or negative charge"? That will engender confusion. Always compute using the mathematical convention, and change the direction of the result only if necessary to know the actual direction of negative charge carriers. You will notice that semiconductors and ammeters are marked by their manufacturers using the mathematical convention. The arrow on the diode shows the direction of conduction when a more positive voltage is applied to the side of the diode arrow. Same with an ammeter.

As I said many times before, the phrase "current flow" literally means "charge flow flow", which is redundant and ridiculous. Current is already charge flow, so you don't have to say it again. Current is present or exists, but charge doesn't flow twice.

Capacitors are not charged. They are energized. A cap energized at 1000 volts has the same net charge as when it was at zero volts. For every coulomb of charge added to one plate, the same coulomb of charge is removed from the opposite plate for a net charge change of zero. Applying a voltage across the capacitor rearranges the charge carriers so one plate has an excess of charge and the other plate has a deficiency of charge. It takes energy to do this, therefore you are charging the cap with energy. In which case you might as well say "energizing".

Ratch

 

[n_tron]

Oh! That was detail. Still trying to grasp. So in a nutshell, once the capacitor is energized, the current (positive charge) then flows towards the emitter and breaks down the e-b junction?

 

[Audioguru]

It is the reverse emitter-base voltage that causes the transistor to conduct avalanche breakdown current.
The voltage causes the breakdown and the result is current flowing in the transistor.

 

[n_tron]

So then how and why does the LED blink (on and off) ? Whats the reason behind that? What controls the timing?

 

[Ratch]

Oh! That was detail. Still trying to grasp. So in a nutshell, once the capacitor is energized, the current (positive charge) then flows towards the emitter and breaks down the e-b junction?

The cap builds up voltage as it is energized. When it reaches the breakdown voltage of the E-B junction, the E-B junction shorts out, and another path for the current from the battery and the current from the de-energizing cap is available. When the current in the E-B path becomes too low to sustain the breakdown, the E-B opens and the cap can energize again.

Ratch

 

[Audioguru]

It takes time for the capacitor to charge to the voltage that causes the transistor to conduct, then the capacitor is discharged by the transistor and its LED load to a voltage (or current) that is low enough for the transistor to turn off, then the capacitor slowly charges again and it all happens over and over.