Is the voltage divider in this circuit backwards for what I'm attempting to achieve?

Question

This circuit is from another post (from long ago) where a user helped me build a circuit which would use a BJT as a switch.

^{simulate this circuit – Schematic created using CircuitLab}

The problem was that I needed to lower the voltage on the base pin of the BJT so the user suggested I add the voltage divider (R2, R3).

Target

The original target was to apply 0.5 - 0.7 V on the base pin.

Should Those Resistors Be Flipped?

I've recently begun looking at the circuit again and I now believe that the two resistors should be flipped.

It looks to me like this might put 4.54 volts on the base pin (instead of the @ .5 volts) that the user was trying to get onto the base pin.

I used the AllAbout Circuits voltage divider calculator and it seems to back this up. But, maybe I'm not looking at that properly?

Logically, it seems like you'd have the larger drop connected directly to the base pin too?

Should those two resistors be switched or is this correct? If they are correct, can you please provide a bit of explanation of how to think about this the correct way?

Answer 1

The problem was that I needed to lower the voltage on the base pin of the BJT so the user suggested I add the voltage divider (R2, R3)

That's not really what's going on here, but I guess that's what you're asking about.

To use a voltage divider to lower the applied voltage you'd need a resistor in series with the switch, so that when the switch closed it would be between the supply node and the divider node.

The original target was to apply 0.5 - 0.7 V on the base pin.

Your circuit as drawn should already achieve this, because the transistor itself will limit the voltage on the base pin to ~0.5 - 0.7 V, if you limit the amount of current supplied (which you are doing due to R2)

In the circuit you have, R3 is really just providing a path to discharge the base node when the switch is open. In a MOSFET circuit this would be important if you needed a quick shut-off after opening the switch. In a BJT circuit it's not as important since current can be sunk through the base-emitter junction. But it will still somewhat speed up the shut off. It will also waste about 2.5 mW when the switch is closed.

Edit

Now that you've linked the original question, I can point out that the answer you got there never said this was a voltage divider.

What he said was,

R2 will limit the base current when thet switch is closed. R3 pulls the base low when the switch is open, to ensure the transistor is not conducting.

Which is exactly the same as what I'm telling you now.

You don't need a voltage divider to reduce the voltage at the base of the BJT. If you limit the current (using R2), the properties of the BJT will take care of the voltage for you.

Answer 2

There are many beginner design errors in this question. So this is a good learning exercise.

• 1) We were not told your design goal. ( My guess.. Vcc=OK @ 5V indicator.)
• 2) no input specs
• 3) no output specs
• 4) no tolerances , no ambient temp range
• 5) No choice of LED or brightness, color , viewing angle ( which affects brightness)
• 6) you exceeded a 5mm max current spec rating of 20mA with 100 Ohms which is overkill
• 7) You use poor online calculator that does not show assumptions.
• 8) you do not know the correct Vbe @ 1mA dim output. (it is always 0.60V @ 1mA)

OK this is a good question how to improve design motus operandi.

my method.

Scope : Design an LED indictor for DC OK on 5V rail using -10% as the threshold for OK. Use 50% of max brightness for a 10Cd LED @ 20mA is adequate, e.g 5Cd when 5V is OK and 0.1 Cd dim as the threshold of not OK

INPUT: Vin: 4 to 5.5V with threshold of -10% = 4.5V = dim LED
@ 5.5V do not exceed power ratings of any part. Ambient 20~35'C

OUTPUTs: LED= ON @ 10 mA max , consider off threshold <= 1mA dim

Fact: any silicon diode or transistor Vbe =0.60V @ 1mA (not 0.7 which depends on transistor power rating and Ri,Rpi etc.)

So design switch for 0.6V threshold OFF/ON threshold.

• k*4.5V=0.6V =Vbe threshold @ 1mA
• k=Rb/(Rb+R1)=0.6/4.5=0.133...
• using White 5mm LED @ 10mA Vf=2.9~3.0V
  • Vf(red)~1.8+Rs*If Rs=~10 for 5mm Red rated for 20mA
  • Vf(white)~ 2.8+Rs*If , Rs = 15 ohms White rated for 20mA
  • your choice , but I choose 135 Ohms. (minimum) for White , 220 min for Red.

Let's make LED intensity grow from almost off @ 4.5V to change to very bright @ 5.5V.

Proof of concept