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For a project I'm making I need to control a chip that works on 3.3V and my PIC who sends the control signal works on +5V so I need to downgrade this voltage so that the signal is not to high for the chip. I did this with a simple voltage divider, but because my chip needs to have a certain default state I also need a pull down resistor to provide this. Is this the right method how I did this? Or will this cause problems?

Voltage divider + Pull down

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2 Answers 2

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You seem to have the values wrong for what (I think) you want to achieve. There's no point in putting two resistors in parallel, might as well use one of an appropriate value.

If you want to drop 5v to 3.3v, then a 'top' resistor of 1.7k and a 'bottom' resistor of 3.3k (both conducting 1mA when on) is an easy to calculate ratio. Make the top one a standard 1.8k and you get slightly less than 3.3v. You could make them 10x that, 18k and 33k, if you want to take less current, as long as 33k is low enough for your 'pull-down' duty.

However, you cannot combine a pull-down function with a driven function, they are mutually exclusive. If, on the other hand, you disconnect the S terminal, or send it high impedance, then the resistor to ground will behave as a pull-down. So in that sense, the voltage divider is the pull-down, when the S point is not driven. When it is driven, the output is either 0 or 3.3v depending on whether S is set high or low. Do you see why now I say the two functions are mutually exclusive.

When drawing schematics, it's always worth using component numbering, the 'reference designators'. It's much easier to refer to R1, R2, than to 'the bottom one, no, the other bottom one'. Values may change with circuit refinement, so 'the 10k one' is not much better.

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  • \$\begingroup\$ Thank you, but I placed to second resistor because I need also a pull down resistor to provide a default state to my chip. How can I combine this with a voltage divider? \$\endgroup\$
    – Pierre
    Commented Feb 21, 2017 at 9:01
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    \$\begingroup\$ Which bit of 'you cannot combine the two, they are mutually exclusive' don't you understand? The voltage devider is the pulldown, when the 5v isn't driven. You may have to redefine what you want. Under the two conditions I state, it does act as a pull-down. Are they not sufficient? If not, then you are confused about how to control your IC, and it may be better to ask a question stating what you want to do, and what these two ICs are. \$\endgroup\$
    – Neil_UK
    Commented Feb 21, 2017 at 9:03
  • \$\begingroup\$ what do you think is stopping the current going through R from also going through R? Oh by-the-way, do you see the reason why we always tell people to name their components? \$\endgroup\$
    – Christian
    Commented Feb 21, 2017 at 9:04
  • \$\begingroup\$ Agreed, of course, about the "no point" point you bring up. But with different systems connecting like this, I might go with a pair of series resistors from 5 V output to 3.3 V input, with the center node of the two resistors tied to pin 3 of a BAV99 and the appropriate other two BAV99 (SOT23) pins tied to 0 V and 3.3 V. Limits output current of 5 V device, clamping limits risks to 3.3 V input, etc. If really nuts and application permits might include a pull-down and a capacitor at that center node, too. ;) \$\endgroup\$
    – jonk
    Commented Feb 21, 2017 at 9:05
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Your 10k resistor connected as a pull down resistor as well. Why you need additional 2k21 resistor as a pull down?

There will be another problem. Resistors in parallel 10k and 2k21 is influence your voltage divider circuit. R1 = 5k11 R2 = 10k R3 = 2k21 In parallel R(parallel)=(R2*R3)/(R2+R3)=1810 Ohm

So V(out) = V(in)*(R(parallel)/(R1+R(parallel))= 1.3 V So you can get 1.3 Volt in S input, if you use this circuit.

I recommend to use optocoupler. Connect the input (diode) side to 5V and the output (transistor) side to 3.3V It is easy and you do not need to calculate and find resistors with abnormal values.

There is one of it I made for you. Here your S input is normally pull up. enter image description here

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    \$\begingroup\$ This level of complication is not needed the simple voltage divider is enough remove the 2k2. \$\endgroup\$
    – RoyC
    Commented Sep 19, 2017 at 8:54
  • \$\begingroup\$ I know. I just advised an alternative method to get DI with optocoupler. \$\endgroup\$
    – IlgarSky
    Commented Sep 19, 2017 at 8:58

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