Why a pull down resistor for transistor base

It is not strictly required. It ensures the transistor is off under these conditions:

• no diver connected to the input
• driver at the input not active, e.g. a microcontroller's GPIO pin during reset before it is initialized as output by the boot software.

The value is kind of arbitrary. It needs to be high enough to present no noticeable burden to the driving stage. It needs to be low enough to ensure the transistor is off. 10 kΩ to 100 kΩ, whatever is available will usually work.

For one there isn't necessarily a pull-up, depends on the component. Verify by studying the datasheet.
Second a pull-up will activate the transistor which may not be what you want, see above.

 

Don't "think". Calcuate the resulting base current, multiply by the gain of the transistor, check if the resulting current is sufficient for the load.
That is very basic engineering and you should not need a forum to help you with that.

 

Which resistor do you mean? There is one labeled 4.7 k - 10 k as pull-up another labeld 10 k as pull-down.
When discussing a circuit, please label components with unique names (geek speak: reference designators) such as R1, R2, R3, C1, C2, ... and refer to these labels. This makes it absolutely clear what you are talking about.
It is also good practice to indicate any voltages or currents that turn up in the discussion with names and directional indicators (arrows) or polarity indicator (+, -). And most importantly between which points voltages are measured or within which connections currents are measured. This avoid ambiguities such as "voltage at the base of Qx" which we see rather often here but which makes no sense as this statement doesn't indicate the reference to which this voltage is being measured.

To give you an example I annotated (partly) your original schematic:

In this example the "voltage at the base of T1" is meaningless. Vbe is a different voltage than Vbase-gnd, although both voltages can be measured with one tip of the probe at the base of T1. You see the point?
That's just for this one voltage. I'll leave the other voltages and currents (as needed) to you.

 

Ic of T1 or T2?
Please re-draw the circuit with all useful information and be specific in your post. We could assume something and then calculate another thing but still be off from the real thing.
Also: the whole thing of T1 being on or off depends not only on the base-emitter voltage of T1 but also on the load current and we have no clue what this is. For T1 to be fully on (geek term: saturated), the base drive needs to be high enough to bring the collector emitter voltage down to the saturation voltage (a few 100 mV, typically).
Also: 2TY is not a transistor type. You could have us guessing that this is the marking on a MMBT8550 transistor, but better to provide us with the real transistor type used and possibly a link to the datasheet so we do not have to search for one.

 

Two problems:

1. The current drawn by the relay is still unknown
2. The green LED has no current limiting resistor. It will burn quickly as it is directly connected between 5 V and gnd by T2.

To get you started:

1. Add a series resistor to the green LED. I think you may have meant R6 to be this current limiter, but both sides of R6 are tied to gnd, thus rendering R6 useless.
2. Calculate the total worst case collector current through T2 (sum of relay current and current through green LED.
3. calculate the worst case base current for T2 by dividing max. collector current by the min. DC gain (as this is a slow on/off switching circuit DC gain is relevant, not AC gain).
4. From the base current and the voltage drops Vce(T1) and Vbe(T2) calculate the minimum required base resistor R4. If you want to add a safety margin, reduce that value by e.g. 10 % (increase current by 10%) and select the next standard value for R4.
5. Calculate the collector current of T1 (use the currents through R3 and R4) and perform the same steps as above to get a value for the T1's base resistor R1.

R3 can be 10 kΩ as it is. It is only required to draw the base of T2towards 5 V, thus making Vbe(T2) ~ 0 V. This is good enough to turn T2 off.

The same goes for R2. However, R2 will contribute to the current through R1 as I(R2) = Vbe(T1)/R2 needs to be taken into account. If you don't want this to happen, move R2 to the left of R1 where it will still work as pull-down. Whether 10 Ω is sufficient for R2 depends on the impedance of the output stage driving into the input (R1) of your circuit). R2 needs to fulfill these conditions:

• If the output of the driver is active High (I assume 5 V as the rest of your circuit), the driver needs to be able to supply current both into the base of T1 and through R2. For a typical µC output capable of driving a few mA this is no problem.
• If th eoutput of the driver is high impedance, the value of R2 needs to be low enough such that Vbe(T1) is much less than 0.6 V (typical value for an NPN) to turn T1 securely off. Design for 0.1 V max. and you're save. If, for example, the driver's output would deliver 1 µA (just a ballpark figure as I don't know what's driving into R1), this 1 µA would drop 10 mV across R2, definitely hood enough to turn T1 off.

 

Hello,

I have made a comment in your drawing:


Bertus