The effect of combined resistors?

I need to convert 5v to 3.3v, easy. I need an 2k resistor though so i looked if two 1k resistors could achieve the same effect. I found pretty contradictory information on the subject though, could someone help me out? (sry im kind of a newbie in electronics) Source one: Are two (or N) resistors in series more precise than one big resistor?

EDIT: Yes, i am trying to power devices, an ESP8266-01 to be precise, the power will be coming from a Lithium battery.

• Out of curiosity, what will you be using the 3.3V for? Are you powering devices? Is it a reference voltage? If you're trying to power anything, using a resistor divider is a bad idea. Applying a load to a resistor divider changes the voltage. See this question. – Daniel Apr 16 '18 at 19:52
• What is your question actually? Is your question about resistance precision? Is it about building a voltage divider for a certain voltage? – Curd Apr 16 '18 at 20:06

Well yes, you can combine in series two 1k resistor in order to make a 2k one. However, if you want to draw some current from this new voltage point, you rather use a linear voltage regulator such as 7805 or a LM317 combined with resistor.

A little tips about resistors in series :

By using two resistors in series, the power is also split equally across both resistor.

The current passing through each resistor is still the same but resistor's value is twice lower, so the dissipated power for each resistor is also divided by two. This can lead you a technical choice in some case of high current or cost restriction.

• Good point about drawing current from resistive dividers, however, 7805 won't convert 5V to 3.3V and LM317 requires 3V of headroom so 5V to 3.3V isn't guaranteed to work. – Wesley Lee Apr 16 '18 at 19:39
• True, it was just a exemple of linear regulator. Those exemple didn't intent to match @Some Guy requirement. As 78xx series is very common, he should take a look of the basic of voltage regulation – M.Ferru Apr 16 '18 at 19:43

I need to convert 5v to 3.3v, easy.

You have not explained why you think this is easy. It can be rather difficult in certain applications.

• If you just need a reference voltage then a pair of resistors will suffice.
• If you want to decrease the voltage to a load of fixed resistance then one resistor will suffice.
• If the load current can change but a fixed voltage is required then it is unlikely that one or two resistors will do the business for you. A voltage regulator is required instead and this will adjust its 'resistance' to compensate for changes in the load.

I need an 2k resistor though so i (sic) looked if two 1k resistors could achieve the same effect. I found pretty contradictory information on the subject though, could someone help me out?

You need to specify what variation your application can tolerate and work from there. Most circuits do not require great precision to work satisfactorily and standard E12 (12 resistors per decade) 5% or 1% tolerance are satisfactory.

Yes, i am trying to power devices, an ESP8266-01 to be precise, the power will be coming from a Lithium battery.

A resistive divider is not suitable. The ESP8266 will have varying current requirements on power-up and during operation depending on what the CPU is doing and what outputs are on or off.

You need a low drop-out voltage regulator.

Actually, more resistors CAN be more precise than one resistor.

Assume you have multiple resistors, say N, of 1000 ohm. Each resistor has a tolerance (in %). If you add multiple resistors of the same type (with the same tolerance), the chance you are ON AVERAGE closer to 1000 ohm is higher.

However, since this is statistics, it doesn't mean it is always true.

If you want to be sure, you measure it with an ohm meter and select a resistor that is close to 1000 ohm.

Also, an advantage of multiple resistors is that the heat is spread over more resistors (most default type of resistors is 0.25 W, probably less for SMD types). So using 2 resistors gives you a bit more headroom.

• Yes this is a good point, having one $1k\Omega \ 1\%$ resistor distributes $N(1000,3\cdot (0.01\cdot 1000)^2)$ while having 1000 $1\Omega \ 1\%$ resistors distributes $N(1000,3\cdot 0.01\cdot (0.01 \sqrt(1000))^2)$ with slower standard deviation. Tolerance means $\pm3\sigma$ – Brethlosze Apr 16 '18 at 19:57
• I don't think this is quite correct. If there is a bias as a result of manufacturing, for example, you can end up with a mean that has significant deviation from the nominal value. Adding more resistors in parallel just gets you closer to that mean divided by the number of resistors. The dice analogy doesn't hold well because there are a discrete number of possible answers with a fixed max and minimum and normal distribution. – Transistor Apr 16 '18 at 20:04
• @Transistor ... I think that picking a resistor with some tolerance also has a normal distribution, unless it is evenly spread across the range within tolerance (1000 - tolerance ... 1000 + tolerance). I assume it is the first (normal distribution). If not I should remove that part. But if it if a discrete value or not, is not that important for my example. – Michel Keijzers Apr 16 '18 at 20:35
• Yes, your answer is correct if the tolerance is evenly spread from the nominal value. I work in manufacturing (but not of electronics) and for various reasons there will genarlly be a bias away from the nominal. Who wants nits? – Transistor Apr 16 '18 at 20:48
• If the resistors are from the same manufacturing lot, then I would expect a very tight distribution, but still somewhere within the tolerance region. Averaging wouldn't help much unless you have resistors from many manufacturing lots which may have different values within the tolerance region. – TimB Apr 16 '18 at 21:43

ESP8266 specs: V10 1.7~3.6 V
LiPo Specs: 3~4.2V

Poor man's solution Silicon diode drop 0.65~0.7V with 0.1uF cap.

5 to 3.3V logic inputs can use R divider. 2k series, 3k shunt to 0V or similar to get 60~66% of input voltage.