I am seeking some help with choosing resistor values for a high side switch circuit.

Firstly, a bit of context: I am adding a bluetooth receiver circuit to my old Sony HiFi. The bluetooth circuit (KRC-86B) comes with little information, but requires 3.3V to 5V DC supply and I have measured about 30mA maximum current draw when the device is paired. I am taking the 5V DC supply from a suitable point on the stereo main board, but have discovered that this rail is permanently at 5V, even when the stereo is in standby. I am therefore adding a simple transistor switch circuit to power up the bluetooth circuit only when the HiFi is switched on. This will be triggered by tapping off the STBY signal which enables the amplifier chips. The STBY signal goes from 0V when off to 6.9V when on.

Attached is a sketch of my proposed high side switch circuit. The transistors (2N3904 and 2N3906) are the only ones I currently have, so wish to use these unless there is a good reason not to. I'm assuming a maximum load current of say 40mA to be on the safe side and believe that the transistors have a modest saturated gain of about 10. I can estimate R3 at about 1k to give a Q2 base current of about 4mA (40mA/10) but am struggling to know how best to calculate the other two and not sure if R2 is even required??

Any advice would be gratefully received, specifically on whether I'm on the right track, how to calculate the resistor values and the necessity of R2.

 

The 1kΩ value for R3 is fine.
R2 connected from the base of Q2 to 5V would absorb the small leakage current of the transistors, but it's probably not needed for room temperature applications.

Calculate the value for R1 the same as you did for R3, to make Q1's base current 1/10th of its collector current.
Thus R1 = (6.9V-0.7V) / 0.4mA = 15kΩ.

 

Thanks crutschow, that's really helpful. Just to be clear, if I were to include R2, do you know how I would calculate this value?

 

The 1kΩ value for R3 is fine.
R2 connected from the base of Q2 to 5V would absorb the small leakage current of the transistors, but it's probably not needed for room temperature applications.

Calculate the value for R1 the same as you did for R3, to make Q1's base current 1/10th of its collector current.
Thus R1 = (6.9V-0.7V) / 0.4mA = 15kΩ.

Why wouldn't you need R2? Seems to me, when Q1 is on, wouldn't there be a direct short to ground? (I'm a NOOB and still learning....)

 

Why wouldn't you need R2? Seems to me, when Q1 is on, wouldn't there be a direct short to ground? (I'm a NOOB and still learning....)

You eliminate R2 and replace it with no connection to +5V.
Only R3 goes to Q1's collector.

 

The 1kΩ value for R3 is fine.
R2 connected from the base of Q2 to 5V would absorb the small leakage current of the transistors, but it's probably not needed for room temperature applications.

Apologies if this is a daft question, but my electronics knowledge is a bit rusty and I'm just trying to further understand the function of R2, if included. I understand that, when switched off, Q1 may still have a small leakage current (max 50uA) from C to E. I'm not sure that this would be sufficient but, in the absence of R2, this would flow down via Q2 and R3 and presumably could start to turn on Q2? I'm guessing that the effect of adding R2 would "pull up" the base of Q2 to prevent this?

Is my understanding correct and, if so, how would I calculate the value of R2?

 

Is my understanding correct and, if so, how would I calculate the value of R2?

That's correct.
You want R2's value low enough so that the maximum leakage current would generate no more than a couple tenths of a volt across it.

 

Brilliant... thanks again for your help crutschow.

 

You eliminate R2 and replace it with no connection to +5V.
Only R3 goes to Q1's collector.

Thank you, Crutschow!