I am trying to make a 24V to 5V voltage translator for an arduino input. I found the following diagram online, but I have a few question about the resistor selection. So since I am using a 2N2222A NPN, I will use hfe,min=30 and Ic,max=800mA from the datasheet. This gives:

Rb1 > 30*Rc1
Rc1 + Rb2 < 30*Rc2
Rc1 < 24 / 0.8
Rc2 < 5 / 0.8

I am just curious why the designer would have used smaller base resistors than collector resistors, seems strange to me. Also, how do we know that the Vce(on) of Q1 won't cause the Q2 to turn on (would not expect Q2 to saturate, but I feel like we can't guarantee it won't turn on a little?).



Datasheet of 2n2222a : http://www.onsemi.com/pub_link/Collateral/2N2222A-D.PDF

 

It is not clear why the base resistor from the 24v digital signal. That could be 100k but, without knowing the rest of the circuit on that side.

The 1k resistor to the right NPN base can be smaller or even zero (bare wire) since the 10k it is connected to will limit any current to that transistor's base.

 

It is not clear why the base resistor from the 24v digital signal. That could be 100k but, without knowing the rest of the circuit on that side.

The 1k resistor to the right NPN base can be smaller or even zero (bare wire) since the 10k it is connected to will limit any current to that transistor's base.

That was my thoughts exactly, but I am still confused as to how we can garentee Vce(on) for Q1 will not switch on Q2. Perhaps this is the purpose of the 1k (Rb2)? You can just assume the LHS is an ideal 24V digital signal.

Edit: Datasheets for 2n2222a says Vce(sat),max = 1V, so the 1k does limit the base current to 1mA max in the case where Q1 is on. Does this prevent Q2 from turning on?

 

That was my thoughts exactly, but I am still confused as to how we can garentee Vce(on) for Q1 will not switch on Q2. Perhaps this is the purpose of the 1k (Rb2)? You can just assume the LHS is an ideal 24V digital signal.

Edit: Datasheets for 2n2222a says Vce(sat),max = 1V, so the 1k does limit the base current to 1mA max in the case where Q1 is on. Does this prevent Q2 from turning on?

You said...

"but I am still confused as to how we can garentee Vce(on) for Q1 will not switch on Q2. Perhaps this is the purpose of the 1k (Rb2)?"

I am not sure what you mean by this...?
Assuming 24v is digital (on or off), then the transistor it is connected to will be off or on, respectively (and 5V output will match the 24v input logic).

 

By the way, level translation is best done with a mosfet than NPN bjt. Commonly 2n7000 or smd equivalent 2n7002.

 

Datasheets for 2n2222a says Vce(sat),max = 1V,

But for what Ic current ??
In your circuit Ic1_max ≈ 24V/10kΩ ≈ 2.4mA so Vce(sat) will be lower than 0.1V. So Q2 for sure will be cut-off.

 

You said...

"but I am still confused as to how we can garentee Vce(on) for Q1 will not switch on Q2. Perhaps this is the purpose of the 1k (Rb2)?"

I am not sure what you mean by this...?
Assuming 24v is digital (on or off), then the transistor it is connected to will be off or on, respectively (and 5V output will match the 24v input logic).

What I mean is that when the first NPN is "ON", it still has a small amount of voltage across it (The saturation voltage). This voltage may be high enough to switch on Q2.

But for what Ic current ??
In your circuit Ic1_max ≈ 24V/10kΩ ≈ 2.4mA so Vce(sat) will be lower than 0.1V. So Q2 for sure will be cut-off.

I was working with the maximum possible Vce(sat) to be conservative. The on-semi datasheet only says the base-emitter cuttoff current is between 10n and 10u amps though. Datasheet does not show the Vbe threshold voltage. So I guess we can assume Vce(sat)@2.4mA << Vbe(threshold) ??

 

By the way, level translation is best done with a mosfet than NPN bjt. Commonly 2n7000 or smd equivalent 2n7002.

I think I just realized the reason why XD. With mosfets you will know with certainty the Vds(on) << Vgs(threshold).

 

I was working with the maximum possible Vce(sat) to be conservative. The on-semi datasheet only says the base-emitter cuttoff current is between 10n and 10u amps though. Datasheet does not show the Vbe threshold voltage. So I guess we can assume Vce(sat)@2.4mA << Vbe(threshold) ??

Data sheet give us Emitter−Base Cutoff Current ( base- emitter junction is reverse biased ) but this has nothing to do with base-emitter voltage and Vbe(threshold). In fact small-signal BJT will by but off if Vbe < 0.4V.
And also to ensure saturation always try to choose the resistor in such way that Ic/Ib = 10.

 

What I mean is that when the first NPN is "ON", it still has a small amount of voltage across it (The saturation voltage). This voltage may be high enough to switch on Q2.

Not at low collector currents. You already have all the clues.

When on, each transistor has a collector current of approx. 2.4 mA. Look at figure 3 on page 3 of your datasheet. For a collector current around 2-3 mA, saturation voltage of Q1 is below 0.1 V. Now look at figure 4. For the same collector current, the base-emitter voltage of Q2 is 0.6 V This is guaranteed turn-off for Q2.

Increase Rb1 to 10 K.

ak

 

Data sheet give us Emitter−Base Cutoff Current ( base- emitter junction is reverse biased ) but this has nothing to do with base-emitter voltage and Vbe(threshold). In fact small-signal BJT will by but off if Vbe < 0.4V.
And also to ensure saturation always try to choose the resistor in such way that Ic/Ib = 10.

I normally use the minimum value of Hfe for any Ic on the datasheet. Question, what is a good rule for thumb for selected the Ic in a signal conversion circuit? If we go soley on Ic,max, we get a minimum resistance of 24V/0.8A = 30 Ohms. Why do people always seem to use 10k?

 

10K is very common value because at 5V, it only draws 0.5mA. so we do not have to check the power dissipation in transistor and in resistor.
For your example 24*0.8A =19.2W. So how big the resistor you have to use? Also all this power must come from power supply.

 

10K is very common value because at 5V, it only draws 0.5mA. so we do not have to check the power dissipation in transistor and in resistor.
For your example 24*0.8A =19.2W. So how big the resistor you have to use? Also all this power must come from power supply.

I understand that using a larger resistor leads to less power usage, but then the question arises why not use 20k, 100k or even 1M? If 30 Ohms is the lower limit, what is the maximum, and is their an "ideal" value somewhere inbetween. I would assume that one of the factors would be the input resistance of the pin, but are their other considerations?

 

Well, the signal frequency, parasitic load capacitance, load resistance, noise immunity e.t.c. But your application is not so demanding, so you can use wide range of a resistor values. I suspect up to 1MΩ or even larger (10MΩ). But going lower than 1kΩ is a unnecessary waste of energy.

 

That was my thoughts exactly, but I am still confused as to how we can garentee Vce(on) for Q1 will not switch on Q2. Perhaps this is the purpose of the 1k (Rb2)? You can just assume the LHS is an ideal 24V digital signal.

Edit: Datasheets for 2n2222a says Vce(sat),max = 1V, so the 1k does limit the base current to 1mA max in the case where Q1 is on. Does this prevent Q2 from turning on?

Vce (sat) changes with collector current. At that 2.5 mA it will be around 250 mV.

 

Thank you for replying everyone, you have helped me understand this circuit much better!