What is a good Overdrive Factor for a BJT?

  • Thread starter saad87
  • Start date
  • Tags
    Bjt
In summary, the conversation discusses the search for a pre-biased dual NPN transistor to use as a buffer with specific criteria, including a small package size and a maximum collector current rating of 200mA. The PIMN31 transistor is considered, but there are concerns about overdriving it enough. The calculation of the BJT saturation is also discussed, along with the option of using a FDG6317NZ transistor. The conversation also touches on the use of resistors for gate and pull down resistance, and the need for 300 buffers on a board. The importance of simplifying calculations and using dominant components is emphasized, and the topic of using a darlington array is also brought up. The conversation ends with a question
  • #1
85
0
I've been looking some pre-biased dual NPNs to use as buffers. My criteria is to have a very small package (i would prefer their width be less than 3mm. 2mm is very good) and that they ought to have a max. collector current rating of 200mA.

I may have found such an NPN in the http://www.nxp.com/documents/data_sheet/PIMN31.pdf" [Broken] but I fear I may not be overdriving it enough. I'll first describe how I'm calculating if the BJT reaches saturation. The max. collector-emitter voltage at saturation for the PIMN31 is 0.3V. Hence, load current is:

My collector current is:
[itex]\frac{Vcc - Vce}{47}=\frac{3.3 - 0.3}{47}=63mA[/itex]

The min. Hfe for the PIMN31 is 70. Therefore, required Ib:

[itex]\frac{Ic}{β}=\frac{63}{70}=0.9mA[/itex]

The PIMN31 has two biasing resistors. R1 is the input resistor connected to the base and R2 is connected between the base and emitter. I convert this to a thevien equivalent circuit:

[itex]Vth = \frac{R2}{R1 + R2} * Vin =\frac{11K}{11K+1.3K}*3=2.68V[/itex]
[itex]Rth = R1||R2=\frac{11K*1.3K}{11K + 1.3K}=1.16K[/itex]

Therefore, actual base current:

[itex]\frac{Vth-Vbe}{Rin}=\frac{2.68-0.6}{1.16K}=1.78mA[/itex]

Now, if my calculations are correct (are they?), the transistor is saturating. But...

Overdrive Factor: 1.78/0.9 = 1.97 ~ 2.

Is this enough? I have read online online that a overdrive factor of 10 is recommended. The issue is, I can't find a transistor with a small enough input resistance and a high enough current rating. If I do, they aren't in a small enough package.

My other option is a http://www.fairchildsemi.com/ds/FD%2FFDG6317NZ.pdf" [Broken]. I have found one in a very small package, but the trouble is I will need external resistors for the gate and a pull down resistance between gate and source.

It's not that I'm hesitant to include these resistances, it's just I need 300 of these buffers on the board. If 2 per transistor are required, that's 600 on the board. Of course, this begs the question: are these resistances really needed?

Some of you will inevitably be wondering what the application is. Please see https://www.physicsforums.com/showthread.php?t=548271" where I provide quite a bit of information.
 
Last edited by a moderator:
Engineering news on Phys.org
  • #2
First you cannot count on [itex]\beta=70[/itex] when you drive into saturation. If you look at page 4, if you want 0.3 volt Vsat, you need 2.5mA to get collector current of 50mA. That means [itex]\beta=20[/itex]. For 63mA requirement, you better drive at least 3mA. I would drive 5mA.

First of all, it is not correct to use Thev. calculation. The base emitter diode turn on at 0.7V and essentially prevent the R2 from working. The more important thing, this is digital circuit, don't worry about Thev. stuff. R2 is 10 times R1, Just treat R2 not there. R1 is 1.3K max, so, drive the input of V=IR=5mAX1.3K=6.5V and you'll ok. The spec in page 2 give +12V is the limiting value to drive the input, so you are way within the safety range.

If you can only drive 5V, you still drive almost 4mA, so that should be good enough.

Learn to simplify calculation if possible, look for component that dominate the effect and simplify by ignore the part that is not that important. In your case, you are not doing analog, R2 is 10 times value of R1. Just look for the upper limit and overdrive the device. This is the practical way of design circuits even in analog. I always try to design with one dominant component and ignore the others. I can tell you in my almost 30 years design analog circuits, I can count on using network theorems with two hands! Just like opamp or closed loop control system, if you can get away with dominant pole compensation, don't add in other poles and zeros! You'll be surprise how much you can get away with simple assumption.
 
Last edited:
  • #3
Thanks yungman, your point about simplfying the calculation made sense and I'll follow that.

However, because now the input resistance is essentially 1.3K and my input voltage is 3.3V (I cannot increase this), my base current is just 2.5mA. Since beta = 20, then my required current is 3.1mA.

I think I will need a different transistor as this is not in saturation.
 
  • #4
Yes, it should be very easy to find a transistor. Look at a darlington also.
 
  • #5
yungman, I actually tried a darlington array before. The issue was that the collector-emitter voltage was 0.6V. This is not low-enough - the collector of the darlington is connected to a Max V CPLD. The max. voltage at which the CPLD is guaranteed to read a low is 0.8V. I felt that 0.6V was quite close to 0.8V and did not feel comfortable with that decision. Hence, my search for a suitable BJT or MOSFET.

With MOSFET, I'll need external resistors. With BJTs, the trick is to find a small enough package. Two BJTs in one package would be ideal really but a small enough package housing a single transistor is good enough as well.

But before I search again, I want to ask, for a collector current of 70mA is a Absolute Max. Rating of Ic 100mA ok or should it at least be 200mA? I feel 200mA is better or can the design be ok with 100mA? With 100mA, a lot of options open up.
 
  • #6
saad87 said:
yungman, I actually tried a darlington array before. The issue was that the collector-emitter voltage was 0.6V. This is not low-enough - the collector of the darlington is connected to a Max V CPLD. The max. voltage at which the CPLD is guaranteed to read a low is 0.8V. I felt that 0.6V was quite close to 0.8V and did not feel comfortable with that decision. Hence, my search for a suitable BJT or MOSFET.

With MOSFET, I'll need external resistors. With BJTs, the trick is to find a small enough package. Two BJTs in one package would be ideal really but a small enough package housing a single transistor is good enough as well.

But before I search again, I want to ask, for a collector current of 70mA is a Absolute Max. Rating of Ic 100mA ok or should it at least be 200mA? I feel 200mA is better or can the design be ok with 100mA? With 100mA, a lot of options open up.

I provide you with the link to DigiKey so you can look around to find one. You can get 150mA or 200mA easily.

http://search.digikey.com/scripts/DkSearch/dksus.dll

http://search.digikey.com/us/en/products/2PC4081S,115/568-5989-1-ND/2531276 [Broken]

http://www.nxp.com/documents/data_sheet/2PC4081.pdf
 
Last edited by a moderator:

1. What is an Overdrive Factor for a BJT?

An Overdrive Factor is a measure of how much above the minimum required voltage a transistor is being operated with. It is used to determine the level of amplification or gain of the transistor.

2. Why is it important to have a good Overdrive Factor for a BJT?

A good Overdrive Factor ensures that the transistor is operating in its most efficient and accurate range. This leads to better performance and less distortion in the output signal.

3. What is considered a good Overdrive Factor for a BJT?

The ideal Overdrive Factor for a BJT varies depending on the specific application and circuit design. Generally, a good Overdrive Factor falls within the range of 2-5, but it can also depend on the particular characteristics of the transistor being used.

4. How does the Overdrive Factor affect the performance of a BJT?

The Overdrive Factor directly affects the gain or amplification of the BJT. A higher Overdrive Factor results in a higher gain, but if the factor is too high, it can lead to distortion and instability in the circuit. On the other hand, a lower Overdrive Factor can result in lower gain and reduced performance.

5. Can the Overdrive Factor be adjusted in a BJT circuit?

Yes, the Overdrive Factor can be adjusted by changing the input voltage or the biasing resistors in the BJT circuit. It is important to carefully adjust the factor to achieve the desired performance without causing any damage to the transistor.

Suggested for: What is a good Overdrive Factor for a BJT?

Replies
19
Views
1K
Replies
5
Views
2K
Replies
43
Views
4K
Replies
3
Views
408
Replies
4
Views
1K
Replies
9
Views
568
Replies
9
Views
1K
Back
Top