# Why are there so many resistors in a typical schematic?

I know that some resistors cause pull-up or pull-down or work as a voltage divider and/or current divider.

But sometimes I cannot understand their function in some simple circuits:

1. Take this motor driver (L9110) for an example. Why are there 2x4.7k resistors?
The L9110 is rated to be able to recieve 2.5v-12v according to the datasheet

2. Or take this zener diode voltage regulator, The 40Ω resistor, ohm's law says that V = I x R, increasing the resistance lowers the current. But why is that 40Ω resistor needed at all? Why not let it flow straight through (~0Ω)?

3. Or lastly this RS232 level shifter, I understand the 10kΩ resistors. But why is there a 4.7kΩ resistor between RX(Device) and TX(Device)? (Should they even be connected?)

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Too broad. If you're asking about these specific schematics, please say so. Nobody can answer on behalf of all schematics, or even 'typical' schematics. –  EJP Jul 23 at 23:28
"Too broad" @EJP, I'll agree, OP needs to get AoE and read it. I sometimes have the feeling that I should first put resistors between all the nodes and then ask, "How small should they be?" :^) –  George Herold Jul 23 at 23:50
In the first one they are pull-up resistors. You did say you understood pull-up resistors. –  immibis Jul 24 at 5:08
@GeorgeHerold: What is AoE? –  copper.hat Jul 24 at 6:39
@copper.hat: The Art of Electronics –  Dave Tweed Jul 24 at 13:52

All three circuits display examples of pull up, pull down, voltage dividers, and current dividers:

1. R1 and R2 are pull up resistors. You have to have both because you have two switches which can be in a different state (one high, one low).
2. The 40Ω resistor is the top half of the voltage divider. The zener is the bottom half of the voltage divider. The zener can be thought of as automatically adjusting its resistance so that the voltage will always be 6 volts. Without the 40Ω resistor, the top wire would be at 10V, and the zener would blow up trying to bring that wire's voltage down to 6V. If the 10V supply was current limited to a current less than the zener's capacity, then the zener would pull the wire down to 6V, putting the power supply into current limiting mode (rather than voltage regulating), and the circuit would work fine. Since the supply is a voltage regulator, though, then you need the 40Ω resistor so the zener can do its job without blowing up, and without getting a current limited power supply.
3. R3 is a pull down resistor. Ignore RX(MCU) and R2 for now, they just tap the TX(DEVICE) line. D1 and C1 form a negative power supply. RS-232 technically requires -12V for signaling. The TX(DEVICE) line will go to -12 occasionally, and the diode and capacitor store that charge so that the TX(MCU) line can use it without building a special -12V power supply into the circuit. It has some limitations, but for those RS-232 devices that demand obedience to older RS-232 specifications, it can work well. R3 therefore is a pull down - when TX(MCU) isn't high, then RX(DEVICE) will see a low current negative voltage. If the device uses -12V on its TX line, then the RX line will reflect the device's adherence to the RS-232 specification.
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In the case of the Zener regulator, the supply is 10 volts, and the Zener will do its best to limit the voltage across itself to 6 volts. If the 40 ohm resistor wasn't there, a very large current would flow as the Zener tried to do its thing, and the magic smoke would be released from many parts.

If I've done the math right, the 40 ohm resistor will carry 100 mA to drop the 4 volts from the 10 volt supply to the 6 volt Zener diode. Since the 100 ohm load resistor has 6 volts (controlled by the Zener diode) across it, it will pass 60 mA, and the zener will pass the remaining 40 mA from the 40 ohm resistor.

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1 and 3 have the same reason. It's a current limiting Pull Up or Pull Down.

In 1, the pull up resistors are there for 2 reasons. 1, to set the default state of the pins to Logic High, and 2, because if they were not, pressing the button would cause a short of VCC/5V directly to Ground. Bad stuff happens when you do that.

In 3 it is a pull down. Notice that RX(device) is connected above the resistor but below the transistor. When the transistor is off, the Pull Down (through the capacitor) brings the line to Logic Low. When the transistor is on, the RX(Device) line is pulled up to Logic High through the transistor, which is a lower impedance path to a Voltage Level (Essentially a diode) than the resistor is. Without the resistor, enabling the transistor would create a mostly direct path from 5V to Ground, again, a bad thing.

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the short answer is that an open input needs some finite current, to assure the voltage. If the current is zero, then the voltage is not determined. So the pullup and pulldown resistors, bias the pin to some fixed voltage.

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All of those are for current limiting. Ohm's law states that V=IR, so if you make R equal zero, then for a fixed V you get infinite I and your part blows up (because P=IV).

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