Congratulations, Your explanation is very detailed and much more elaborated than mine. Best regards, Isaac Em 6/1/2012 17:03, Byron Jeff escreveu: > It's days like this I need to remember that I'm a computer scientist, not > an electrical engineer. > > On Fri, Jan 06, 2012 at 03:22:48PM -0200, Isaac Marino Bavaresco wrote: >> Em 6/1/2012 12:28, Byron Jeff escreveu: >>> On Fri, Jan 06, 2012 at 11:00:23AM -0200, Isaac Marino Bavaresco wrote: >>>> Em 6/1/2012 10:26, Byron Jeff escreveu: >>>>> On Fri, Jan 06, 2012 at 07:20:08AM -0200, Isaac Marino Bavaresco wrot= e: >>>>>> Em 06/01/2012 06:34, William "Chops" Westfield escreveu: >>>>>>>>> Is there anything I can add to protect the chip from unwanted neg= ative voltage? >>>>>>> Well, to start with, there should be a substantial resistor; about = 10k. >>>>>>> This will rely on the internal protection diodes to clamp the volta= ges to levels that won't cause problems. There is plenty of reason to assu= me that this is insufficient, but microchip does publish app notes where th= is is done (apparently successfully.) >>>>>>> ... >>>>>> I prefer to use an external Zener diode plus two resistors to clamp = the >>>>>> voltages, forming a "T" with the Zener as the "leg" of the "T". >>>>> What's the purpose of the resistor between the zener and the PIC inpu= t? >>>> If the RS-232 is connected when the PIC is not powered, the pin will >>>> have 4.7V and the board will be powered by the PIC's internal parasiti= c >>>> diode and large currents may flow through it, perhaps violating some >>>> parameter. >>> The series resistor will limit the incoming current. With a 4.7V zener = and >>> a 4.7K series resistor, the maximum current the unpowered PIC will see = is 1 >>> mA. All the rest of current from the higher incoming voltage will be >>> shunted to ground by the zener. >> >> If I understand it right, you use a resistor with one terminal connected >> to the PC's RS-232 TX and its other terminal connected the the zener >> cathode AND to the pin of the PIC, right? >> >> In this scenario your circuit may clamp the voltage below the zener >> voltage and all the current will flow through the PIC's internal diode. >> If the resistor is connected to a higher voltage then the current may be >> much higher than your figures. > I'm trying to work this through but I keep confusing myself. Let's share = an > example so that maybe I can lift the fog. Take a look at this simple zene= r > regulator: > > http://en.wikipedia.org/wiki/Linear_regulator#Simple_Zener_regulator > > Fundamentally it's a resistor voltage divider with the zener as a shunt > between the two resistors. In operation R1, the top resistor, limits the > total current coming from the source, and the zener shunts current to kee= p > the midpoint between the two resistors at the zener voltage. The way I've > always thought about this circuit is that VS/R1 is the amount of current > available to the zener and R2, and that since the zener is forcing the > voltage between the two to the zener voltage, that the amount of current > through R2 is fixed. > > But I'm realizing from your discussion that's only the case if R2 is fixe= d. > If R2 is variable, then if it can accept more current, then it will > regardless of the zener. Let me clarify with some numbers: > > VS =3D 14.1V (3 * 4.7V) > R1 =3D 4.7k > DZ =3D 4.7V > R2 =3D variable > > The current available from R1 is 14.1V/4.7k =3D 3 mA. Now my (mistaken) > thinking is that the zener would shunt 2 mA of that current and that a > 4.7V/4.7K =3D 1mA supply would be available to R2. > > But that's only the case if the original voltage divider of R1/R2 develop= s > a voltage above the knee of the zener, such as if R2 is not present. But = if > R2 is a dead short of 0 ohms, then it would draw all 3 mA across it and t= he > zener would shunt no current since the voltage divider is 0V at the R1/R2 > junction. That's the part I didn't see. > > So if R2 is the PIC, and it can draw 3 mA of power, it could certainly be > powered through its inputs. > > So your T circuit splits R2 with a second resistor, say R2A and R2B and > puts the zener between R1 and R2A which limits current to 1 mA. > > Makes sense now. > >> Remember, the current through a resistor is proportional to the >> difference of voltage between its two terminals. The only thing you are >> guaranteeing is that the PIC's side terminal will never be higher than >> 4.7V (probably), but you cannot control the voltage on the other side. > But by added the second resistor, it guarantees enough resistance that R2 > in total cannot sink all the current. > >> >>>> With the additional resistor (say 1K) the current will be limited to a >>>> few mA, well inside the allowed range. > In my example R2A would need to high enough to force the R1/R2A junction = to > br 4.7V or higher so that the zener kicks in. So for example a second 4.7= K > resistor creates a half voltage divider to 7V, the zener would lower the > voltage to 4.7V and the maximum current would be 4.7V/4.7K =3D 1 mA even = if > the PIC (R2B) is a dead short. > >>>>> For >>>>> quick and dirty testing, I use a 4.7K resistor in series with the RS-= 232 >>>>> output and a 4.7V zener. For more permanent circuits I use this as th= e base >>>>> input to a NPN transistor inverter with a pullup resistor to protect = the >>>>> PIC's input from the -0.7V the zener allows when the RS-232 output sw= ings >>>>> negative. >>>> When using an NPN transistor I use a small signal diode from ground to >>>> base to protect the base-emitter junction from reverse biasing. The PI= C >>>> will never see the -0.7V because the base-collector junction will be >>>> reverse biased, but the transistor's base-emitter junction may be >>>> damaged as they are usually rated for -5V. >>> The zener serves the purpose of that small signal diode with the input >>> voltage is negative as the zener is now forward biased. So the base-emi= tter >>> voltage is never more that -0.7V which is nowhere near the reverse bias >>> breakdown voltage for the transistor. >> >> My point is that in this situation the small signal diode will work the >> same way as the zener, both will be forward biased and the base voltage >> will never be less than -0.7V (or close to this). When the voltage on >> the base is positive, the base-emitter junction will be forward biased >> and the zener will never conduct, thus the use of a zener or a small >> signal diode is indifferent. The advantage of the small signal diode is >> that it may have a much lower capacitance, making switching faster. > Got it. So a zener in a T configuration, or no zener, but a signal diode > instead, with the transistor. > > BAJ > >> >> Isaac >> >> >>> I haven't had any problems so far, and it's a heck of a lot easier than >>> trying to track down a MAX232 or equivalent. I haven't seen a USB seria= l >>> adapter yet that swings any more than +/- 5V, and every one so far work= s >>> fine with 0-5V going into their inputs. With the near universal death o= f >>> dedicated serial and parallel ports on PC's, USB serial adapters are my >>> current bit wiggling interface. >>> >>> BAJ >>> >>>> Best regards, >>>> >>>> Isaac >>>> >>>> >>>> >>>>> The last time this discussion came up, Russel noted to me that allowi= ng the >>>>> negative voltage to the PIC input, even though it's clamped by the >>>>> protection diodes, causes the part to operate outside of it's range >>>>> specified in the datasheet (-0.3V minimum). Hence the transistor in t= he >>>>> permanent circuit. I'm unsure what if any damage that 150 uA of curre= nt at >>>>> -0.7V can do to the input. >>>>> >>>>> BAJ >>>>> >>>> -- >>>> http://www.piclist.com PIC/SX FAQ & list archive >>>> View/change your membership options at >>>> http://mailman.mit.edu/mailman/listinfo/piclist >> -- >> http://www.piclist.com PIC/SX FAQ & list archive >> View/change your membership options at >> http://mailman.mit.edu/mailman/listinfo/piclist -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .