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# Voltage Divider?

I will need to convert 5v to 1.2v... I chose 10k and 2k because 10k/(10k+2k)=1.2v. Would these resistors work for this?

700Views13Replies

I will need to convert 5v to 1.2v... I chose 10k and 2k because 10k/(10k+2k)=1.2v. Would these resistors work for this?

## Discussions

Erm sage advisors...:

Vin => (black box) => Vout

where

blackbox= 0.24

Sure looks like a converter to me. Let's not be pedantic prigs. A voltage divider acts as a fractional voltage converter. Period.

It does have limitations. Ideally, the load presented to a voltage divider is high-impedance, or said another way, draws very little current.

In general, we design simple voltage dividers to meet

a) the required output voltageVout = Vin * R2/(R1 + R2)where R2 (the lower leg) is the resistor across which we deliver the result of the weighted division, or if you prefer "conversion" (or in simpler terms, where the output voltage shows up)

b) the required output current drawIout(MAX) </= I(R1)/100where the denominator 100 is often used by engineers to minimize (though not eliminate) a voltage drop on the output due to current drain through the load.

(basically, the load acts as a resistance in parallel to R2, reducing its effective value.)

In order to "boost" output current, one normally reduces the values of the voltage divider resistances, but it does so at a cost in terms of system load.

A diode can replace R2 in certain situations, but unless the diode is the load (LED for instance) , output current is still limited to relatively low values or is strictly limited to static loads (ie, loads that don't change how much current they're drawing...in fact, with voltage dividers, it's somewhat best practice to use them only with static loads, as there are better circuits and components for use in more dynamic environments.)

If you're intyo learnign, look up how to use zener diodes next...

"converter" to me implies that the output impedance is negligible.

Just sayin.

It doesn't to me. A converter simply converts one value (or one set of values) to another. And in the case of simple resistance-based voltage dividers, that's (in most cases) their basic purpose, to provide a reference derived by converting a source voltage to the reference.

In any case, the dismissive snap judgement a certain member made with regard to the OP's intent was unnecessary. The OP did not say he wanted to use the vlaue as a power source, he simply said he wanted to convert from 5V to 1.2V.

If you guys look at my Altoids USB charger, I used a voltage divider for the data pins of the USB. The resistors I said I needed earlier (10k and 33k) are for it.

+1.

A voltage divider will be fine for a data input, as long as your divider resistors aren't too high valued. Since most analog inputs on microprocessors have input resistances on the order of 1Mohm or more, you should be fine with the values you chose. Keep in mind what I said (below) about opamps if you're going to use it as a power source, OR if you're going to use it as a reference voltage for the A/D converter.

If you guys look at my Altoids USB charger, I used a voltage divider for the data pins of the USB. The resistors I said I needed earlier (10k and 33k) are for it.

And when you're ready, note that op amps make very nice little stable voltage "dividers" with very low output impedance, meaning that they CAN deliver "substantial" load current.

Sorry! I forgot 5x... Resistors are 10k and 33k.

5x 10/(10+33)

5 x 10/ 43

50/43=

1.16279069767

Just round up to the tenths place, and you will get 1.2v... Thanks!

Now if you are feeding that voltage point to anything but an Op-Amp you can expect a lower voltage as current flow increases the drop across the top 10k....

5v / 12k = xv / 2k .... xv = 5v ( 2/12) = 5v x 0.1667 =

0.83 volts........................ OR another way ............

Current = V / R = 5 / 12k => 0.41667 ma

Voltage = I x R = 0.00041667 x 2k =

0.83 voltsSolve for 1.2v at what resistor

Rx / 1.2v = (10k + Rx) / 5v ....

Rx = (10k + Rx) 1.2/5 = ( 10k + Rx) o.24 = 2.4k +o.24Rx

Rx (1-o.24) = 2.4k ....=> o.76Rx = 2.4k .... Rx = 2.4k / o.76 =

3.16kThat should be 5 x 2/(10+2), or 10/12, which is 0.83 V. Verence is right, what do you think you are trying to do ?

What do you mean by 'convert'?

A voltage divider does convert nothing. It gives you a part of the input voltage. And it only works, if the input resistance is low and the output resistance is high compared to the dividers resistance.

Is this for measurement purposes (low current) or as a power supply (relatively high current)? In first case, it may work, in second case probably not.

+1