Summarised Solution:

The two configurations are close to equivalent.

Either would work equally well in almost all cases.

In a situation where one was better than the other the design would be excessively marginal for real world use (as anything so crucial to make the two differ substantially means the operation is "right on the edge"). .

\$R_{2}\$ or \$R_{4}\$ are needed only when \$V_{in}\$ can be open circuit, which in that case they are a good idea. Values up to about 100K are probably OK in most cases. 10k is a good safe value in most cases.

A secondary effect in bipolar transistors (which I have alluded to in my answer) means that R2 and R4 may be needed to sink Icb reverse bias leakage current. If this is not done then it will be carried by the be junction and can cause device turn on. This is a genuine real world effect which is well known and well documented but not always well taught in courses. See my answer addition.

Left hand case:

- Drive voltage is decreased by \$\frac{10}{11}\$, which means 9% less.
- Base sees 10K to ground, if input is open circuit.
- If input is LOW, then base sees about 1K to ground. Actually 1K//10K = essentially the same.

Right hand case:

- Drive = 100% of \$V_{in}\$ is applied via 1K.
- Base sees 10K to ground if \$V_{in}\$ is open circuit. (as opposed to 11K).
- If the input is LOW, base sees 1K, which is essentially the same.

R2 and R4 act to shunt the base leakage current to ground. For low power or small signal jellybean transistors, up to several Watts rating, this current is very small, and usually will not turn the transistor ON, but it just might in extreme cases - so say 100K would usually be enough to keep the base LOW.

This only applies if \$V_{in}\$ is open circuit. If \$V_{in}\$ is grounded, which means it is LOW, then R1 or R5 are from base to ground and R2 or R4 are not needed. Good design includes these resistors if \$V_{in}\$ may **ever** be open circuit (e.g. a processor pin during startup may be open circuit or undefined).

Here is as an example where a very short "blip" due to a pin floating was of major consequence: A very long time ago, I had a circuit controlling an 8 track open reel data tape drive. When the system was first turned on the tape would run backwards at high speed and despool. This was "very very very annoying". The code was checked and no fault was found. It turned out that the port drive went open circuit as the port initialized and this allowed the floating line to be pulled high by the tape deck which put a rewind code on the tape port. It rewound! The initialisation code did not explicitly command the tape to stop as it was assumed that it was already stopped and would not start by itself. Adding an explicit stop command meant that the tape would twitch but not despool.(Counts on fingers of the brain - hmmm 34 years ago. (That was at the very start of 1978 - now almost 38 years ago as I edit this answer). Yes, we had microprocessors back then. Just :-).

Specifics:

A 10K resistor is needed directly in the base to prevent the Q1 from switching ON unintentionally. If the configuration on the right, with Q1, is used, then the resistance will be too weak to pull the base down.

No!

10K = 11K for practical purposes 99.8% of the time, and even 100k would work in most cases.

R2 also protects VBE from over-voltage and give stability in case of temperature changes.

No practical difference in either case.

R1 protects from over-current to the Q1's base, and will be a bigger value resistor in case the voltage from "uC-out" is high (in example +24V). There is going to be a voltage divider formed, but that doesn't matter as the input voltage is high enough, already.

Some merit.

R1 is dimensioned to provide desired base drive current so yes.

\$R_{1} = \dfrac{V}{I} = \dfrac{(Vin - Vbe)}{I{desired\, base\, drive}}\$

As \$V_{BE}\$ low and you design for more than enough current, then:

\$R_{1} \cong \dfrac{Vin}{Ib_{desired}}\$

\$I_{base \ desired} >> \frac{Ic}{\beta}\$ - where \$\beta\$ = current gain.

If \$\beta_{nominal} = 400\$ (eg BC337-40 where \$\beta =\$ 250 to 600) then design for \$\beta \leq 100\$ unless there are special reasons not to.

For instance, if \$\beta_{nominal} = 400\$ then \$\beta_{design} = 100\$.

If \$Ic_{max} = 250mA \$ and \$V_{in} = 24V \$ then

$$I_b = \frac{I_c}{\beta} = \frac{250}{100} = 2.5mA $$
$$ R_b = \frac{V}{I} = \frac{24V}{2.5mA} = 9.6k \Omega$$

We could use 10k, as beta is conservative but 8.2k or even 4.7k is ok.

$$ Pr_{4.7k} = \frac{V^2}{R} = \frac{24^2}{4.7k} = 123mW $$

This would be ok with a \$\frac{1}{4}W\$ resistor **but** 123mW may not be totally trivial so one **may** wish to use the 10k resistor instead.

Note that switched collector power = V x I = 24 x 250 = 6 Watts.

On the right, with Q2, is my configuration. I think that:

Since an NPN transistor's base is not a high impedance point like a MOSFET or a JFET, and the HFE of the transistor is less than 500, and at least 0.6V is needed to turn the transistor ON, a pull-down resistor is not critical, and in most cases is not even needed.

As above - sort of, yes, BUT. ie base leakage will bite you sometimes. Murphy says that without the pull-down it will accidentally fire the potato cannon into the crowd just before the main act, but that a 10k to 100k pull-down will save you.

If a pull-down resistor is going to be put in the board, then the value of exact 10K is a myth. It depends on your power budget. A 12K would do fine as well as a 1K.

Yes!

10k = 12k = 33k. 100k MAY be getting a bit high.

Note that all this applies only if Vin can go open circuit.

If Vin is either high or low or anywhere in between then the path through R1 or R5 will dominate.

If the configuration on the left, with Q1, is used, then a voltage divider is created and may create problems if the input signal, that is used to switch the transistor ON, is low.

Only in very very very very extreme cases as shown.

$$ I_{R1} = \frac{V}{R} = \frac{V_{in}-V{be}}{R1} $$

$$ I_{R2} = \frac{V_{be}}{R_2} $$

So the fraction that R2 will "steal" is

$$ \frac{I_{R2}}{I_{R1}} = \frac{\frac{V_{be}}{R_2}} { \frac{V_{in}-V_{be}}{R_1}} $$

$$ \frac{I_{R2}}{I_{R1}} = \frac{R_1}{R_2} \times \frac{V_{be}}{V_{in}-V_{be}} $$

If \$R_1 = 1k \$, \$R2 = 10K\$ then $$\frac{R_1}{R_2} = 0.1 $$

and if \$V_{be} = 0.6V \$, \$V_{in} = 3.6V \$ (to make sums clearer) then
$$ \frac{V_{be}}{V_{in}-V_{be}} = \frac{0.6}{3.0} = 0.2 $$
So overall fraction of drive lost is \$ 0.1 \times 0.2 = 0.02 = 2\% \$

i.e even with 1k/10k the loss of drive is minimal.

If you can judge Beta and more so closely that 2% drive loss matters then you should be in the space program.

- Orbital launchers work with safety margins in the 1% - 2% range in some key areas. When your payload to orbit is 3% to 10% of your launch mass (or less) then every % of safety margin is a bite out of our lunch. The latest North Korean orbital launch attempt used an actual safety margin of -1% to -2% somewhere critical, apparently, and "summat gang aglae". They are in good company - the US and USSR lost many many many launchers in the early 1960s. I knew a man who used to build atlas missiles early on. What fun they had. One Russian system NEVER produced a successful launch - too complex.) UK launched one satellite ever FWIW.

ADDED

It has been suggested in comments that

R2 and R4 are never needed, because an NPN is a CURRENT-controlled device. R2 and R4 would only make sense for VOLTAGE-controlled devices, like MOSFETs

and

How can a pull-down be needed when the MCU output is hi-Z, and the transistor is controlled by current?

This suggestion in various forms has been repeated by enough people that it is worth emphasising.

If a bipolar transistor base is left floating then reality AND the relevant data sheet information both demonstrate that a small amount of collector current can flow under specified conditions.

The conditions where this typically can occur are described below.

I have personally seen real-world situations where this effect caused spurious turn-on problems.

If your worst case situation, using worst case (not typical) datasheet parameters, does not fulfill these conditions and/or the results do not concern you worst case, then a base pull down is not strictly essential.

There is an important secondary effect in bipolar transistors which results in R2 and R4 having a useful and sometimes essential role. I'll discuss the R2 version as it is the same as the R4 version but slightly "purer" for this case (ie R1 becomes irrelevant).

If Vin is open circuit then R2 is connected from base to ground. R1 has no effect. base APPEARS to be grounded with no signal source.

However, the CB junction is effectively a reverse biased silicon diode. Reverse leakage current will flow through the CB diode into the base. If no external path to ground is provided this current will then flow via the forward biased base-emitter diode to ground. This current will notionally result in a collector current of Beta x Icb leakage but at such low currents you need to look at the underlying equations and/or published device data.
A BC337 - datasheet here has a Icb cutoff of about 0.1 uA with Vbe = 0.

Ice0 = collector base current is about 200 nA in this case.

Vc is 40V in that example but the current approximately doubles per 10 degrees C rise and that spec is at 25C and the effect is relatively voltage independent. The two are closely related. At around 55c you may get 1 uA - not much. If usual Ic is 1 mA then 1 uA is irrelevant. Probably.

**I have seen real world circuits where omission of R2 caused spurious turn on problems.**

With R2 = say 100k then 1 uA will produce 0.1V voltage rise and all is well.

## Best Answer

You can take out the resistor completely and leave a short. As you have already pointed out, the emitter voltage is ~23.3 (one diode drop down), so the emitter current is 23.3/500 or 46.6 ma. Base current will be 46.6 ma/Hfe. This would be an "emitter follower" rather than a low-side switch.