Why is my 2N2222 transistor not biasing?

[hugoender]

I have a Parallax passive infrared motion sensor (**broken link removed**) and am trying to have it activate a wireless transmitter (the type that is found in one of those cheap wireless door bells).

Since the motion sensor only outputs 3.3V and around 2.2mA (from what I measured), I added a transistor (2N2222) to act as a switch.

I first experimented with the transistor by itself by connecting Vc and Vb to wall-wart supplies. The base I connected to a voltage divider (R1=12Ohms, R2=22Ohms) with an input of what is supposed to be a 5V 550mA wall-wart to give me 3.3V at Vb. I have 3.3kOhms between Vcc and Vc and Vcc is a 15V, 400mA wall-wart.

The wireless transmitter normally uses a 12V MN21 or A23 battery but I have it's + and - terminals connected to the 3.3kOhms and Vc respectively.

With this setup, everything works fine. However, my issue arises when I switch the 5V wall-wart and voltage divider with a battery pack (3 AAA) to give me 4.5-5V. I connect this pack straight to Vb with no resistor. I don't know why this does not work. I don't think it takes much to saturate the transistor.

Any help would be greatly appreciated. Thank you.

 

[Reloadron]

A drawing or two of what you have would help. However, something I don't get is the biasing of your 2N2222 since you are using the transistor as a switch.

Generally you would have a circuit that resembles the below image.

Ron

 

[hugoender]

Here is a picture:
**broken link removed**

The transistor in question is a 2N2222 or 2N3904 (tried both). The wireless transmitter normally holds a 12V battery (A23 or MN21 battery). I do not have a relay so that is out of the question. I thought if I removed the battery I could just control the voltage supply with a transistor acting as a switch (just soldered two wires to the + and - terminals of the battery holder in the wireless transmitter).

So my question is why does it work with the wall-wart and not the batteries? I thought maybe the three AAA batteries were not supplying enough current but they are only being used for the base and that should not draw too much current in order to get it into saturation. Or am I wrong? I am not quite sure how much current three AAA batteries are capable of sourcing.

 

[hugoender]

Alright I figured out what it was... it was the transistors themselves. Turns out I got VERY unlucky and all four transistors I tried (two 2N2222's and two 2N3904's) were messed up or burnt out. Fifth one was the charm

No wonder it wasn't working. I was starting to doubt everything I had learned about electronics all these years.

 

[QuietMan]

Without a current limiting resistor on the Base you are going to continue burning out transistors. The base emitter is basically a simple diode junction, there is no current limiting capability there at all. The 5th one is probably bad, but has failed in a way where you get some operation out of it. Did you forget to draw the resistor or what, this is pretty basic?

 

[hugoender]

Oops. Yes I forgot the resistor in the picture. I put a 1kOhm resistor for Rb.

 

[QuietMan]

I bought a bunch of surplus transistors many orbits ago that were pure junk. They were labeled PN2222A, and their typical beta was 3 or so. Counterfeit electronics seems to be alive and well.

You can buy PN2222A for around 7¢ each from places like BJ Micro, less for SMT.

 

[hugoender]

Yeah I am pretty sure that is what happened with me. I just got a bad bunch. I ordered them straight from the source... Taiwan lol

Mine are labeled 2N2222A.

 

[mvs sarma]

may not be so
the base current limits are crossed to much higher sides even with 1K resistor, so all the issues
2N2222 are made in volumes, and perhaps no firm would loose its face value of suicide

 

[QuietMan]

The transistors will be in saturation with 1KΩ and 5V, with a base emitter current of 4.3ma. Especially with a 3.3KΩ resistor and whatever the PIC offers on the collector (2.7ma). If the PIC can work with a 9VDC battery and a 3.3KΩ resistor it will work with the transistor. I'm wondering the wireless TX is limited so much though.

 

[hugoender]

I am not quite sure what you are saying Quietman. I think you are not understanding my circuit. I drew a picture to better illustrate what I have done:

**broken link removed**

 

[hugoender]

With the 1kOhm resistor, there would be (3.3-0.7)/1000 = 2.6mA going into the base of the transistor.

I measured the current draw of the wireless transmitter to be 2.2mA. I originally had a Vcc of 19V and so I used Rc = (Vcc - Vloaddrop)/Ic to find the value of Rc. This gave me Rc = (19V - 12V)/2.2mA = 3.3kOhms. I assumed a 12V drop since that is the battery that the wireless transmitter normally uses. However, now that I am using a Vcc of 9V, I can probably remove Rc since the load (wireless transmitter) can operate at the same voltage as Vcc. This will also allow more current to go through the wireless transmitter and thus have more power to transmit further.

Now assuming a minimum hfe of 100, the amount of base current required to drive the 2.2mA that the load requires would be Ib = Ic / hfe = 2.2mA / 100 = 0.000022A = 0.022mA. To determine Rb more exact, I could then find Rb = (Vin - Vbe) / Ib = (3.3V - 0.7V) / 0.022mA <= ~118kOhms.

Therefore, essentially as long as I have a base resistor of 118kOhms or less, I should be supplying enough current to power the wireless transmitter.

Looking at datasheets for the 2N2222A, the maximum Ic is 800mA and maximum hfe is 300. Knowing that, the maximum base current you should be able to supply before burning the transistor should be Ib = 800mA / 300 = 2.67mA.

So in short, Quietman you might be correct in saying that I was probably burning out all my transistors myself since I was only using a 1kOhm resistor at the base. Thank you for your replies. I will now change that resistor to make sure I don't come anywhere near the limit.

 

[crutschow]

The maximum base current is likely well over 10mA before burn-out since, when used as a switch, a typical forced beta of 10 is often used to insure low collector saturation voltage, which would require a base current of 80mA. I think your first thought was correct, they were a bad batch of transistors.

 

[hugoender]

@crutschow Can you explain to me what you mean? How does the beta change when using it as a switch? Isn't the beta fixed for a transistor?

 

[vlad777]

@crutschow Can you explain to me what you mean? How does the beta change when using it as a switch? Isn't the beta fixed for a transistor?

It's like a rule. When using as a switch assume hfe=10..20 .
But this is probbaly because of those high power transistors that have hfe 20 to 70 anyway.
My experimental data tells me that if transistor has hfe 200+ this rule is not nesesery.

 

[crutschow]

@crutschow Can you explain to me what you mean? How does the beta change when using it as a switch? Isn't the beta fixed for a transistor?

True, the actual beta is fixed for a given transistor. Forced beta means you use a beta lower than the minimum beta (and certainly lower than the maximum you quote) and thus a higher base current to insure that the transistor is well turned on for minimum voltage drop across the collector emitter terminals. Normal beta values are used when operating the transistor as a linear amp, not a switch.

I would always use a beta value no greater than half the minimum value data sheet value when using the transistor as a switch to have a low switch voltage drop when on.

 

[hugoender]

True, the actual beta is fixed for a given transistor. Forced beta means you use a beta lower than the minimum beta (and certainly lower than the maximum you quote) and thus a higher base current to insure that the transistor is well turned on for minimum voltage drop across the collector emitter terminals. Normal beta values are used when operating the transistor as a linear amp, not a switch.

I would always use a beta value no greater than half the minimum value data sheet value when using the transistor as a switch to have a low switch voltage drop when on.

Okay so I am trying to understand this...

Here is the datasheet I am looking at: http://www.datasheetcatalog.org/datasheet/SGSThomsonMicroelectronics/mXyzzyw.pdf

Now, my understanding is that as you increase the base current, you are going from cut-off to active to saturation. So now looking at the datasheet, you can see that as Ic increases the beta increases. So does this mean that beta depends on the amount of current required to drive the load? So for example, if my load requires a current of 2.2mA (Ic) then the beta is approximately 50 (according to the datasheet)? And so to reach the 800mA max Ic I would need 16mA at the base (16 x 50 = 800)? No, this does not make sense because once I start increasing Ic, the beta changes. See this is what doesn't make sense to me.

crutschow, you are saying that you use a higher base current to have a lower than minimum beta... but the datasheet says that as you increase current the beta increases as well....

Oh wait, I think I know what you guys mean now. By having high base current, you are putting the transistor in saturation and in saturation your gain is flat. So you can keep increasing the base current without increasing Ic and thus your beta, or ratio between base and collector current, is much smaller. Is that it? If so, then how do you know, mathematically, when the transistor is in saturation? How can I calculate this?

A lot of websites have you calculate Ib = Ic_required/beta to find the base current needed to put the transistor in saturation but this does not make sense since the amount of base current needed to put a transistor in saturation does not depend on the load. Or am I wrong?

Sorry for going off on a rant. Just thought I'd type all my thoughts so that it made more sense as to where my confusion lies. Thank you all for your replies thus far.

 

[crutschow]

The beta does vary with collector current, but again, that's only of real concern if you are operating the transistor in the linear mode. When used as a switch, you want enough base current to insure the transistor stays well in saturation at the maximum load collector current. Thus a (forced) beta of 10 is often used for large transistors to insure that they stay saturated. Smaller transistor with higher beta will probably saturate well with a forced beta of 20 or so. Excess base current is not a problem when using a transistor as a switch (as long as the base current doesn't exceed the maximum transistor rating).

The required base current to keep the transistor in saturation does indeed depend upon collector current. Why do you think otherwise? If you are not providing enough base current the transistor will come out of saturation as the collector current increases. Thus the reason for calculating the required base current at the maximum collector current.

 

[audioguru]

The datasheet for all transistors list the beta (hFE) when the transistor has plenty of collector to emitter voltage so it is not saturated.
The datasheet for your transistor lists the hFE when the collector to emitter voltage (VCE) is 1V and 10V. It shows that the hFE is half when the VCE is only 1V. When it is saturated then the VCE is typically only 0.05V at low collector currents then the base current must be much higher than the hFE would calculate.

The max Collector-Emitter Saturation Voltage is listed with the base current at 1/10th the collector current because the hFE does not apply to a saturated transistor.

 

[vlad777]

audioguru I think this is the second time you stated that there needs to be plenty Vce for hfe to be correct.
Can you explain how this works?