Calculating resistors required

Hi all,

Sorry for more reeeally basic questions, but advice I've been given by different people conflicts, and some of the reading on the web is a bit over my head, whereas you guys always seem to explain things well :)

I've got a programmed PIC16F chip, and I want to wire it up to a few things... My power supply is 4AA batteries (about 6V), but my chip wants about 5V. The switches connected to my chip also want about 5V.

My motor is driven from an output pin of the chip (which I guess, is

5V?), but the motor is 3V. My LEDs are 3V too.

I guess I need a resistor to reduce my 6V to 5V, which can then be connected to the chip + input switches. The 5V output would be find for the 5V LEDs, but I guess I need another resistor for each motor.

It's also been mentioned that my switches should be connected to 0V via a resistor when "off", and 5V when "on". I don't understand the reasoning for this :-\

Also, if I was to need 6V to drive something, but the output from my chip is less, though my main supply is enough - how do I go about changing a 5V (I think!) output, to 6V? I've been reading about transistors thinking they do what I want, but I've just confused myself even more!

Many thanks!

--
Danny

The tradition of using pull-up resistors and a switch to ground comes from the characteristics of a TTL logic input. TTL inputs required a fairly low resistance to ground before they would consider it a logic low, and if the input was left open-circuit, would normally consider it a high (but it is still good practice to use pull-up resistors).

CMOS inputs, as on most microcontrollers, don't source or sink any current - if left unconnected, they will be in an unknown intermediate state. Since they don't supply any significant current in either state, you can use a fairly high value of resistor to pull the input either high or low - then use the switch to pull it in the other direction.

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
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You need a low-dropout regulator to convert 6V into 5V ("low-dropout" refers to the small difference between input and output voltages); you might get away with a diode drop (about 0.7V). Consult the IC datasheets: check the absolute max/min power supply voltages.

I find it hard to believe that a PIC output can provide enough current to drive your motor. Are you sure? You would normally need a transistor (e.g. MOSFET) with a back e.m.f. protection diode. You could also use a dropper resistor, in series with the motor, if the power supply is too high for it. Also: I would power the motor from the unregulated supply to reduce noise on the microprocessor side; and use power supply decoupling capacitors.

You can't use a resistor: the voltage drop across a resistor depends on the current flowing through it; the current is not constant. Consider using a diode or a (low dropout) regulator.

You mentioned "switches" that "also want about 5V" above. I wondered if you meant logic-level drive MOSFETs or something else? Switches, in the normal sense, do not "want" any voltage! Switches, in their simplest form, are two terminal open/closed mechanical devices which make or break a circuit. Logic inputs are driven from switches like this. View diagram using a fixed width font:

VCC + | | .-. | | R | | '-' | | +-----> Logic | | \ o \ \. o | | === GND

+6V (or higher) + | | .-. | | Load | | '-' | | | ||-+ ||-----||-+ MOSFET | Logic Level Drive | | === GND created by Andy´s ASCII-Circuit v1.24.140803 Beta

You take only a slight risk by powering the PIC directly from 4 AA cells. They put out 6 volts (or a little above) only when new. With any significant load, they sag, quickly. I would mess with a regulator, only is the chip was very valuable.

This loads have current requirements, as well as voltage needs. You have to take both into account when designing drivers. The PIC outputs have very limited current capability (read the spec sheet). The outputs may have enough capability ot drive an LED, especially in active low (on when the output is pulling negative), so a series resistor may be all you need for those. The resistor has to waste the extra voltage while passing the desired current. For instance if you desire that 10 ma pass through both LED and resistor, and you expect about 3 volts to be consumed by the LED, you have 2 or 3 volts that must be burned up by the resistor. 3 volts drop divided by 10 ma current through = 300 ohms resistance needed.

The motor will quite likely need a driver that provides current gain. A PNP transistor connected as emitter follower (base to PIC output, collector to ground, emitter to motor, other side of motor to battery

+) will provide about battery - .7 volts to motor when the PIC output goes low. If you need a high active driver, change to an NPN transistor and swap the collector and motor battery connections. Limiting the motor voltage to an accurate maximum voltage is a bit more complicated.

This probably means that the program is set up to to watch the input switches (things you push to make the program do things) that are activated when they connect to +5, and when you are not pushing on them, a resistor pulls the input down to zero volts (active high). This is a completely arbitrary program choice.

The simplest way is to add an N channel mosfet to the output (gate to output, source to zero, load between +6 and drain). When an output goes high, the mosfet will switch on and connect the full battery voltage across the load.

--
John Popelish

bit

wants

You could use 4 NiCad or NiMH batteries, this will give you approx 4.8V. The PIC is fairly flexible, it will work from >3V - ~6V, though 6V is really pusing it. You could also use 3 alkaline batteries for about

4.5V. Or, as you should probably do, you could use a linear voltage regulator with a low dropout.

Unless your motor is really small ( I guess I need a resistor to reduce my 6V to 5V, which can then be

for

A dropping resistor is not the right way to lower your voltage to the PIC. Since the PIC will be drawing a varying amount of current (like when turning LEDs on and off), your voltage would fluctuate wildly. You will want to use a regulator of some kind as I mentioned above.

You *will* want to use resistors to limit the current thru your LEDs, otherwise the PIC will attempt to supply a very large amount of current to the LED. This is likely to result in damage to the LED and/or the PIC output pin circuitry. (you can kill a PIC one pin at a time) LEDs (like most diodes) are not voltage controled devices, they need current limiting or they will burn out.

via

The goal is to never have an input pin in a "floating" state. They are suggesting that you use a pull-up resistor on the pin so that when the switch isn't closed(pressed), the input pin sees a logic 1 (+5V) via the juice feeding thru the pull-up. Then you have the switch so that it shorts the pin directly to ground when pressed. The pull-up resistor will limit how much current is flowing to ground, and the input pin will "see" a logic 0. This way, the input pin is always connected to a voltage reference of 5V or 0V.

myself

You would want to use a transistor for this. For highish currents (like motors), I suggest looking into the wonderful and inexpensive logic-level n-channel MOSFETs. These are used allot like generic NPN transistors, but have extremely low "on resistance" (milliohms usually). The logic-level part means that you can connect the gate (base equiv) directly to the PIC output pin and you don't even need a current limiting resistor going into it since the MOSFETs gate input impedance is really high. You connect the source pin (emitter equiv) directly to ground. You stick your load (motor, lights, electric-chair or whatever ;-) between the drain pin (collector equiv) and your +6V supply. When the gate is high (+5V) the motor will run at full speed, when it's low (0V) it will be turned fully off.

Either can work, and the choice is up to the programmer. But most digital ICs pull down better than they pull up, so they can supply more load current in active low mode with less loss of voltage than when used in active high mode. I like inputs to be active low, because if the switch wiring shorts out to the case, the input line just gets grounded, instead of connecting the +5 volt supply to the case. But the choice is pretty arbitrary.

--
John Popelish

-

-> Flattery will get you a lot of places.

-

-LOL!

- -

-> -I've got a programmed PIC16F chip, and I want to wire it up to a few

-> -things... My power supply is 4AA batteries (about 6V), but my chip wants

-> -about 5V. The switches connected to my chip also want about 5V.

->

-> Switches generally don't care about voltage.

-

-I phrased that badly, but I meant I need to connect a supply to the

-input pins with the switches. Dunno where I got 5v from, but I assume it

-needs something!

If the switch is going into the PIC then 5V is right.

- -

-> First and foremost is that your 4AA batteries will last about 2 minutes in the type of

-> application with LEDs and motors that you are describing. I would advise working with

-> a DC wall wart of at least 8V and a 7805 voltage regulator until you get everything

-> working. The wall wart and regulator will solve most of the issues in terms of regulating

-> voltage.

-

-A wall wart? What's that? I guess something I could connect batteries or

-to the wall would be good while building. I guess I need something that

-plugs into the wall, converts to DC, and drops to around 6V (same as

-batteries)?

a Wall wart is a DC power supply. Here's a typical picture:

And you don't want 6V because then you'll have a problem regulating it. For a 7805 you generally want at least 2.5V of headroom. So if you want regulated

5V then you'll want an input voltage of at least 7.5V. That's why I said 8V in the original post.

- [SNIP]

-> You'll need a transistor driver for your motor.

-

-Is an N-type MOSFET the same thing?

Yup.

- -

-> Also you can drive your motor directly for your input

- > voltage (8V) even if it's a 3V motor as long as the average voltage

- > presented to it is about 3V. You do this using pulse

- > width modulation.

-

-Didn't know that, but it does sound a little messy. I'd rather just give

-it 3V so I can leave it on/off as needed :)

Then you'll probably want to have a second regulator. Look up LM317 adjustable regulators as I said in my previous post. Both the 7805 and LM317 are available at Radio Shacks.

-[SNIP]

BAJ

Yikes.. Looks like things aren't quite as simple as I thought! I've responded to each point below...

The power is 3.0 - 5.5 Volts. My supply is *around* 6V (from 4AA batteries). I'm not sure how consistant this voltage is as the batteries wear. I thought diodes were "one way" components, but I guess you're suggesting a diode also drops the current. So I could just put a diode in between the battery box and the chip?

What value(?) diode should I use (and where would I find information on choosing this value in future?)?

Ignorance is not bliss! The output pins connect to LEDs fine, so I assumed I could stick a small motor in. I guess not (told you I'm a n00b! ;)). So, how would you recommend wiring the motor? Would a transistor allow the output pin on the chip drive a higher voltage/current to the motor? How do I calculate the values of components I need? (the motor is 3V)

I guess my understanding of resistors is a bit flawed! I know my batteries are 6v, and my chip wants 3.0 - 5.5, but nothing about the current the batteries can produce or the chip consumes :-\

I see. I thought the switch would be between the +ve and pin, not to ground! How does this work? Also, how do you calculate the value of the resistor being used there?

So the N-Channel MOSFET will connect the motor to ground when the PIC outputs a logic 1? I always thought switches where +ve side, not ground. Is there a difference?

Thanks for your help with my newbie questions!

--
Danny

Excellent. I'm just looking @ Rapid Electronics, and they seem to be available in lots of different packages! Some look like ICs, and some have 3 legs. How do I figure out what I need?

Righto. I've seen a formular for figuring out what resistors I need for LEDs as:

1000 * (supply.voltage - voltage.needed) / current

Does this look right? And is this for LEDs only, or can it be used for other things too?

My only question here, is that a switch would then give logic 0. I thought it would be reversed, and a 1 would mean the switch is pressed. Is that the usual way of doing things?

Right - getting somewhere now! :)

So, for a circuit with two switches, one driving an LED and one driving a motor, would this be correct? (I know the PIC is redundant, since the inputs directly drive the outputs, but this wouldn't always be the case!)

Excuse the bad ascii, I've never used that program before!

VCC + | o----------o-------o----------o----------------------o--------- | | | | | | | ,---. | | | | X | MOTOR | | | PIC '---' | | | + __ - | .-. .-. '-o| |o---------------)------. R | | R | | '-------o| |o--------. | | | | | | | .----o| |o-----. | | | '-' '-' | | .o|__|o--. | | | | | '----' | | | .-. | ||-+ | | | | | | | | | || |=|> | | | | ,---. | | | o | o '-' '-' | X |LED | | | | Unused Unused '---' | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | o----------o-------o-----------o--------o------o--------o---o--' | === GND (created by AACircuit v1.28.4 beta 13/12/04

Two switches on the left. Two pins unused, one LED, and one motor. That right?

Many thanks,

--
Danny

Whoops, I had 6V connected to the PIC there. I guess I'm missing a voltage regulator... Do I also need one for the motor? (motor is 3V)

--
Danny

LOL!

I phrased that badly, but I meant I need to connect a supply to the input pins with the switches. Dunno where I got 5v from, but I assume it needs something!

type of

working with

everything

regulating

A wall wart? What's that? I guess something I could connect batteries or to the wall would be good while building. I guess I need something that plugs into the wall, converts to DC, and drops to around 6V (same as batteries)?

the process

Yep, got this sorted :)

Is an N-type MOSFET the same thing?

Didn't know that, but it does sound a little messy. I'd rather just give it 3V so I can leave it on/off as needed :)

Currently I'm just fiddling. I want to simply set up a circuit with a Danny-programmed PIC in it. The ultimate goal is just to learn as much as possible about this kind of thing (the more I know, the more I can do!), but it'll probably start off heading down the route of little robots that move around avoiding obstacles at home!

Thanks!

--
Danny

[snip]

Didoes drop voltage not current. Silicon diodes (e.g. 1N4148 or 1N4001) drop about 0.7V so one would probably be adequate; but two, in series, would be safer.

You need to know the current to calculate the dropper resistor. Use Ohm's Law. Or, find it by trial and error. Alternatively, you could use diodes to drop the voltage:

6 - 3 = 3V 3 / 0.7 = 4 diodes in series. [snip] [snip: diagram of SPST switch with pull-up resistor]

Yes, you could swap the switch and the resistor around to invert the sense of the logic.

10k (or more) is fine for CMOS inputs. [snip: diagram of N-channel logic-driven MOSFET with load]

Correct.

difference?

You would need a P-channel MOSFET to do it that way around.

But does it make any difference, switching +ve or ground? Is there any reasons for doing it either way? Not just for the motor, but the switches too?

--
Danny

That's determined by how much current you need to draw, how much power the regulator is going to have to dissipate (as heat) dropping the voltage to the output level and how much margin the regulator has to work with (i.e. how much greater the input voltage is than the output voltage). The LM2940CT looks pretty good for you for all around experimentation if you want to supply allot of current, but the LP2950CZ should work ok in your circuit (as long as it is not trying to drive the motors too). It is limited to 160mA output.

It's just a variation of Ohms law with the 1000 multiplier allowing you to specify current in milliamps instead of amps. For example, if you are using a diode with a 1.7Vf (forward voltage drop), and you wished to run it from a 5V supply, you will need to drop 3.3V thru the resistor. If you want 20mA of current then ((5-1.7)/20)*1000 = 165 Ohms.

It can be done the other way around if you like.

It looks ok to me. I assume you have MCLR disabled and are using the internal oscillator.

Yep! (When I can figure out how to stop the programmer going nuts when I disable MCLR! - If not, I guess I just connect the MCLR pin to the PIC Vdd pin?)

Going back to the regulators...:

So, would I be right in thinking that 2x LM2940CT-5.0 would drop 6V to around 5V?

I want to drive 2x3V motors, but it seems the LM2940CT-5.0 has a minimum input of 5.45 (to drop to around 5V). Motor specs say:

No load current 0.34A max. Rated load current 1.07A max.

But none of the low dropout regulators seem to fit... Was just looking at fixed voltage regulators, but the only one that'll allow a high enough current is an output of 5V, still too high for my motors :-(

Any suggestions?

--
Danny

Any chance you could look at my thread "Circuit & Component Check" to see if all looks well?

Thanks! :-)

--
Danny

Any chance you could look at my thread "Circuit & Component Check" to see if all looks well?

Thanks! :-)

--
Danny

Any chance you could look at my thread "Circuit & Component Check" to see if all looks well?

Would something like this work? (if I cut the end off and attach wires I can plug into my breadboard!)

Or do I need something that says "DC" in it?

Thanks! :-)

--
Danny

The dropout voltage of a regulator is the point at which it can no longer regulate. The LM2940-5.0 will maintain its output at 5.0 volts, as long as the input voltage is above 5.5 volts. (will probably work with up to 30 volts or so on the input - but I haven't checked the datasheet)

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca