# BJT NPN pull-down Resistor

#### wuz

Joined Oct 4, 2010
8
Hi, I know that it's good practice to include a pull-down resistor R2 between the base - emitter junction to make sure that the transistor remains off when small leakage current is present, and understand that it has to be much larger than the base resistor R1, but I don't fully understand the effect of adding R2 to R1.
When calculating the base current, are R1 & R2 in parallel? E.g. if Vcc = Vin = 5V, R1 = 1K, R2 = 10K.
Is it correct that the input impedance = (1/R1+1/R2)^(-1) = 909, so Ib = 4.3V/909 = 4.73mA? So if we have a smaller R2, it's effectly reducing the base current?
Thanks!

#### Jony130

Joined Feb 17, 2009
5,244
You are very close to the right answer.
You can use the Thevenin's theorem to find Ib current.
For example Vin = 5V and R1 = 1K, R2 = 10K we have
Vth = R2/(R1+R2) * Vin = 4.5V
And the base current is equal
Ib = (Vth - Vbe) / R1||R2 = (4.5V - 0.6V) / 909Ω = 3.9V/909Ω ≈ 4.3mA

• anhnha

#### wuz

Joined Oct 4, 2010
8
Hi Jony130,

Thanks very much for your reply. I still don't quite understand how you got 4.3mA from the Thevenin's theorem, and what 2 points you are using for Vth (is it between b & e)? Could you please show me your Thevenin equivalent circuit?

#### Jony130

Joined Feb 17, 2009
5,244
Sure I can show you.
We replace this part of the circuit ( in this gray rectangle)
With his Thevenin's equivalent circuit

Vth = Vin * R2/(R1+R2)

Rth = R1||R2 = (R1*R2)/(R1+R2)

#### wuz

Joined Oct 4, 2010
8
Thank you for your explanation. It makes perfect sense to me now! I can now go to sleep without feeling confused about the circuit #### Jony130

Joined Feb 17, 2009
5,244
I'm very glad that you can sleep well.

#### kmohanra

Joined Feb 4, 2008
7
Hello together,

Can you apply kirchoffs law in the same circuit and explain the base current concept from the Vin.

Thanks

#### kkaczor

Joined Aug 4, 2013
6
Sure I can show you.
We replace this part of the circuit ( in this gray rectangle)
Hi,

Could you tell me why we cannot calculate the base current by simple kirchoff's law:
1. Assume base-emitter voltage drop of 0.7V,
2. The same voltage drop is on R2, so this means the current which flows through R2 is 0,07mA,
3. The voltage drop on R1 is 5-0,7 = 4,3V. This gives us the current which flows through R1 of (5-4,3)/1k = 4.3mA

Finally base current could be calculated as 4,3mA - 0,07mA = 4,23mA

What I'm doing wrong?

#### Jony130

Joined Feb 17, 2009
5,244
For sure you can use kirchoff's law to find IB current. I don't see anything wrong in your solution.

#### kkaczor

Joined Aug 4, 2013
6
Okay, but maybe it would be stupid question - why the result values differ?

#### #12

Joined Nov 30, 2010
18,223
Post#2 did not account for the current through the 10k resistor subtracting from the 4.3ma. Your version is right, kk.

• kkaczor

#### Jony130

Joined Feb 17, 2009
5,244
I assume Vbe = 0.6V and you assume 0.7V
So for Vbe = 0.6V we have

IR2 = 0.6V/10KΩ = 60μA and IR1 = (5V - 0.6V)/1KΩ = 4.4mA
So Ib = 4.4mA - 60μA = 4.34mA

• kkaczor

#### kkaczor

Joined Aug 4, 2013
6
I used to assume 0.6V, but I've measured several NPNs yesterday and it does not want to be other than 0.7V.

Thanks guys for your explanations and help!

#### Jony130

Joined Feb 17, 2009
5,244
Post#2 did not account for the current through the 10k resistor subtracting from the 4.3ma. Your version is right, kk.
This is not true.

I used to assume 0.6V, but I've measured several NPNs yesterday and it does not want to be other than 0.7V.

Thanks guys for your explanations and help!
Well its all depend on the Vin value and Rb value.

I measure Vbe Vs Ib for BC337-40 for Vcc = 10V

RB = 680kΩ....Vbe = 0.614V....Ib = 13.8µA

RB = 470kΩ....Vbe = 0.616V....Ib = 20µA

RB = 220kΩ....Vbe = 0.624V....Ib = 42.61µA

RB = 100kΩ....Vbe = 0.639V....Ib = 93.61µA

RB = 50kΩ......Vbe = 0.659V....Ib = 187µA

RB = 10kΩ......Vbe = 0.719V....Ib = 928µA

RB = 5kΩ........Vbe = 0.748V....Ib = 1.85mA

RB = 2kΩ........Vbe = 0.787V....Ib = 4.6mA

RB = 1kΩ........Vbe = 0.819V....Ib = 9.18mA

RB = 500Ω......Vbe = 0.856V....Ib = 18.29mA

RB = 200Ω......Vbe = 0.989V....Ib = 45mA

• KLillie and screen1988

#### kkaczor

Joined Aug 4, 2013
6
Vcc was equal to Vin, both 10V?

#### screen1988

Joined Mar 7, 2013
310
I know that it's good practice to include a pull-down resistor R2 between the base - emitter junction to make sure that the transistor remains off when small leakage current is present
Could anyone help me explain it?

#### kkaczor

Joined Aug 4, 2013
6
Could anyone help me explain it?
When you drive transistor from microcontroller for example, when it is starting the state of output pins in unknown and there is a possibility a small current leakage. If so, a small current may drive your transistor.

To prevent this, you can place a resistor between base and emitter to force current not drive the base, but go through pullup resistor instead.

• screen1988

#### screen1988

Joined Mar 7, 2013
310
Thanks, kkaczor.
To prevent this, you can place a resistor between base and emitter to force current not drive the base, but go through pullup resistor instead.
I have just read that the pull-down resistor is very big (in this case it is much larger than R1). If so, in order for current flows this resistor to ground instead of flowing into base of transistor, we also need the pull-down resistance << input resistance of transistor?
Is that right?

#### kkaczor

Joined Aug 4, 2013
6
• screen1988

#### ramancini8

Joined Jul 18, 2012
473
The calculated 4.5V is actually 4.545V. One writer assumed Vbe=0.6V and you assumed 0.7V. Not theory, just quick, sloppy calculations.