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I want to connect my little RS232 to TTL converter to a 5V power supply but the converter needs 3v3. So I thought, even with my very little electronics knowledge, I could knock up a potential divider to do this...

... on considering it I realised I have two resistor values to calculate and one equation because I don't know the current that the device would draw.

R1/(R1+R2) == 3.3/5

So my question is should I just choose resistor values that are as low as possible so the current isn't limited significantly but I still achieve the same voltage across the load? Would this work? Thanks in advance :)

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    \$\begingroup\$ Voltage dividers are a terrible idea for supply rails. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 12 '13 at 12:31
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For an RS232 level converter the current drawn will depend largely on what it's attached to and the line state at the time. If you use low value resistors you'll end up generating a lot of heat / waste power and higher values won't have good regulation as the current changes.

A linear regulator such as the L78L33ACZ is in a hobbyist friendly TO-92 package and is only 40 cents in one-off quantities from Digikey. The drop-out voltage of 1.7V would put it just within spec. I'd suggest buying some of those or something similar from a local supplier and purchase a few extras for your parts drawer for future projects.

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  • \$\begingroup\$ You can never have too many '1117s (and associated adapters). \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 12 '13 at 12:50
  • \$\begingroup\$ A 1.7V drop out voltage seems cutting it very close... And 100mA is very weak. a 1117 is definitely a better choice, 1.2V dropout, 800mA, but 1.40 in single quantities. The MCP1702-3302E/TO is a 250mA, 0.6v Dropout TO-92 regulator, for 48 cents in single quantities directly from the manufacturer. \$\endgroup\$ – Passerby Dec 12 '13 at 17:48
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The voltage dropped by one of the resistors in a voltage divider (or anywhere for that matter) will depend upon the amount of current flowing through that resistor. If all of the current that flows through one resistor in a two-resistor divider flows through the other, the ratio of the voltages will equal the ratio of the resistances. Any current which flows through one resistor without flowing through the other will change the voltage drop of the resistor through which it flows; the consequent change in voltage drop will affect the voltage across the other resistor, which will in turn affect the current flowing through it.

If one has a resistor divider with RT on top and RB on the bottom, and a load connected from the junction point to the bottom supply rail, the voltage seen by the load if it doesn't draw any current will be RB/(RT+RB), as you are aware. If the load draws any current, its voltage will drop as though a resistor with value 1/(1/RT+1/RB) were connected between a voltage source at the "ideal" voltage and the load. Note that if the lower resistor is small relative to the upper one, the "effective resistance" seen by the load will be controlled mainly by resistor that's in parallel with the load, rather than the one in series. To understand why that is, imagine that the supply is fixed at 110 volts, and the resistors are 1K and 100ohm.

In the absence of any load current, there would be 100mA flowing through both resistors, so the top resistor would drop (100mA * 1000ohms) == 100 volts. The lower one would drop (100mA * 100ohms = 10 volts). If the load draws any current, then the top resistor will have more than 100mA flowing through it (since it will have to supply current to both the load and the bottom resistor). Each extra milliamp through the top resistor will increase its voltage drop by a volt, but a one-volt increase in the upper resistor's drop will reduce the voltage across the bottom resistor by a volt, in turn reducing its current draw by 10mA. The net effect is that for every additional milliamp supplied by the top resistor, 11mA will become available to the load (since any current which flows through the top resistor and doesn't flow through the bottom resistor will flow through the load).

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