Extreme electronics newbie question

bxgirten

Please don't flame me.· I'm a software developer that wants to get more involved with electronics/circuitry.· To that end, I enrolled in a Circuits Analysis class this past semester.· I learned the usual: Ohm's Law, series circuits, parallel circuits, bridge circuits, series-parallel circuits, KVL, KCL, Mesh Analysis, Network Theorems.

I'm still left with one outstanding question: when creating a circuit, how do I know how much resistance to introduce into the series or branch?· Is the amount resistance driven by the component (led, motor, etc.).· Also, if I want to limit the draw of current going into a component, do I·connect that resistor before the component from the perspective of electron flow or conventional flow?

I learned a lot from the course but, you know how it is - its textbook knowledge.

Thanks in advance!

allanlane5

In most circumstances, the voltage you place across a branch is the 'independent' value. The 'current' in a branch is the 'dependent' value, and it's 'dependent' on the resistance in that branch.

For a quick example -- you want to light an LED. Now, an LED is a non-linear device -- the current-voltage curve for this device is not a simple straight line. At about 1.2 volts, the current through the LED is independent of the voltage across it. So you need some device (a resistor) to limit this current.

So, if we put 5 volts across the 'branch', and we have an LED in the branch, we need to size the resistor to get 12 mA.

V = I * R. 5 Volts - 1.2 volts (LED drop) is 3.8 volts. So the resistor must 'drop' 3.8 volts.

V = I * R. 3.8 volts == 12 mA * X ohms. X ohms == 3.8 volts / 0.012 Amps == 316.6666 ohms.

Now, 316 ohms is not one of the "standard" resistor values. 220 ohms is, and 470 ohms is. Let's see what those give us.

3.8 volts == X Amps * 220 ohms. X == 3.8 / 220, == 17 mA. 17 mA will work (looking at the LED data sheets, which says max current (burn-out current) is around 24 mA)

3.8 volts == X Amps * 470 ohms. X == 3.8 / 470 == 8 mA. 8 mA is kind of low, but might work.

So, there's the sort of analysis you would do to insure you don't burn out your LED's.

bxgirten

Thanks for the quick reply.

Two·last questions:

1) resistor placement, in relation to the LED example?
2) Since you are talking about voltage across the branch, should I be looking at voltage division?

Thanks, again!

Mike Green

The same current flows through any components in series, so the order of the components doesn't matter in a series circuit.

Voltage division? Most digital circuitry (and a lot of analog circuitry) is powered from one or a small number of fixed voltage power supplies that you can usually treat as a black box. Any of the Stamp related stuff is designed to work from a 5V supply and all the tutorials assume that the supply voltage is 5V. If you're powering motors bigger than a little servo motor, you may have a separate power supply just for the motors and the voltage there depends on the design and requirements of the motor itself.

If, by voltage division, you are talking about how the voltage drops across the various components in series, your standard series / parallel network design techniques apply. The voltage across the resistor can be calculated from the branch current and the resistor value. The voltage across the LED can be determined by the branch current and the characteristic forward voltage of the LED at that current (based on the datasheet graphs). The voltage across the branch is determined by the power supply design and would normally be 5V.

bxgirten

Thanks, Allan and Mike.

My question was driven by designing small things, like providing voltage to a simple circuit - like an alarm "blinker".· Instead of having the Stamp cause the LED to blink, it seems more prudent to feed the voltage to a branch that contains a 555 timer with a resistor-capacitor combination.

I'm slowly realizing that there are a lot of alternatives to "taxing" the Stamp by using flops.· Since I'm new to wiring a circuit, I just was unsure on how to wisely use the "proper" load·resistance in that circuit.

Thanks, again!

allanlane5

Bottom line -- the in-series resistor can be 'before' or 'after' the LED, the same current flows in both instances.