Post by thread:[PIC]: Using a 2N3904 to switch a relay

I'd like to preface this question by saying that while I can program fine my electronics knowledge goes little beyond the absolute basics. I want to use a pic to switch an inductive load on/off and it was suggested (on usenet) that the following should work VCC (5VDC) | | .----o | | 5V Coil 1N4002 - _|_ Relay ^ |_/_|- | | | | |----o | | 6.8K | ___ |/ o------|___|---o-------| 2N3904 PIC Output | |> .-. | | | | 6.8K | | | '-' | | | GND GND I've assembled the circuit (with the input being a PIC pin pulled up to 5V via a 10K resistor), however the relay switches (on) as soon as power is supplied to the circuit and the output state of the PIC pin has no effect on the relay state. As I've said at the start, my electronics knowledge is basic in the extreme but I've been googling a bit to try to understand how the above circuit should work (& why it doesn't). I've found a similar (to my mind) relay switching circuit using a BC377 rather than 2N3904 and missing the 6K8 resistor between ground & the transistor base. Following that I removed the ground resistor & replaced the second resistor with a 2K6 one (using calculations provided with the BC377 circuit), but this doesn't work either! Could anyone explain how the above circuit works and how the values of the resistors are calculated? Should it in fact be functioning as in when connected to a pic as I've described? Any suggestions of suitable foundation electronics books would also be appreciated! -- Dave Cunningham PGP KEY ID: 0xA78636DC -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

{Quote hidden}> I'd like to preface this question by saying that while I can program > fine my electronics knowledge goes little beyond the absolute basics. > > I want to use a pic to switch an inductive load on/off and it was > suggested (on usenet) that the following should work > > VCC (5VDC) > > | > | > .----o > | | 5V Coil > 1N4002 - _|_ Relay > ^ |_/_|- > | | > | | > |----o > | > | > 6.8K | > ___ |/ > o------|___|---o-------| 2N3904 > PIC Output | |> > .-. | > | | | > 6.8K | | | > '-' | > | | > GND GND > > > I've assembled the circuit (with the input being a PIC pin pulled up to > 5V via a 10K resistor), however the relay switches (on) as soon as power > is supplied to the circuit and the output state of the PIC pin has no > effect on the relay state. > > As I've said at the start, my electronics knowledge is basic in the > extreme but I've been googling a bit to try to understand how the above > circuit should work (& why it doesn't). I've found a similar (to my > mind) relay switching circuit using a BC377 rather than 2N3904 and > missing the 6K8 resistor between ground & the transistor base. Following > that I removed the ground resistor & replaced the second resistor with a > 2K6 one (using calculations provided with the BC377 circuit), but this > doesn't work either! > > Could anyone explain how the above circuit works and how the values of > the resistors are calculated? Should it in fact be functioning as in > when connected to a pic as I've described? Well, the circuit is pretty basic. First off, ignore the 6.8k resistor to ground, it's not really needed. Now, to simplify things: think of the transistor as a current mode switch. When your input current into the base the collector and emitter are practically shorted together. When there is no current flowing into the base the switch is off, and the emitter and collected can be considered open. Your PIC's output pin outputs a voltage. The 6.8k resister in series with it acts as a "safety" to ensure that not too much current flows into the base (think of a base as a diode going to ground). The value of the resistor is not as straightforward to calculate. What you WANT to do is ensure the transistor is saturated. You have to get the transistor's datasheet to figure out it's saturation current for your case (it will depend on the beta of the transistor and your load current), then go a little high to ensure it's definitely in saturation. So say you calculate the saturation current is 1mA: The value of the resistor would then be: R=V/I=((5V - 0.7V(which is a diode drop))/1mA ~= 4300ohms = 4.3k Now, consider this the BIGGEST resistor you can use. There is no reason to go this high unless you need to (to save power or other reason). To be safe go lower. Your lower limit is defined by the max current a PIC pin can source. Say it's 20mA, then the SMALLEST resistor you can use is: R=V/I=4.3V/20mA = 215 ohms. So, the SMALLEST resistor you can use is 215ohms. I don't recommend going that low either since there is also a "per device" current output that you might violate on the PIC. So, what do I recommend? Don't bother with the calculations! :) Why? Because MOST of the time selecting something like a 1k or 2k resistor will "work". If you want go through the calculations to ensure you are OK, but most of the time I just go with a 2k. Obviously a 2k won't work in every case, and in fact if you are dealing with different power supplies 2k will likely be no go. But it usually is. Back to your circuit. Disconnect the base and see if the relay still turns on. If it does then you've likely got a bum transistor, either that or you've hooked it up wrong. If you've confused the base with the collector then your symptoms would match. Hope this helped, TTYL -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

Bet you've got the transistor connections wrong! George ----- Original Message ----- From: "Herbert Graf" <spam_OUTmailinglistTakeThisOuTFARCITE.NET> To: <.....PICLISTKILLspam@spam@MITVMA.MIT.EDU> Sent: Monday, October 20, 2003 8:24 PM Subject: Re: [PIC]: Using a 2N3904 to switch a relay {Quote hidden}> > I'd like to preface this question by saying that while I can program > > fine my electronics knowledge goes little beyond the absolute basics. > > > > I want to use a pic to switch an inductive load on/off and it was > > suggested (on usenet) that the following should work > > > > VCC (5VDC) > > > > | > > | > > .----o > > | | 5V Coil > > 1N4002 - _|_ Relay > > ^ |_/_|- > > | | > > | | > > |----o > > | > > | > > 6.8K | > > ___ |/ > > o------|___|---o-------| 2N3904 > > PIC Output | |> > > .-. | > > | | | > > 6.8K | | | > > '-' | > > | | > > GND GND > > > > > > I've assembled the circuit (with the input being a PIC pin pulled up to > > 5V via a 10K resistor), however the relay switches (on) as soon as power > > is supplied to the circuit and the output state of the PIC pin has no > > effect on the relay state. > > > > As I've said at the start, my electronics knowledge is basic in the > > extreme but I've been googling a bit to try to understand how the above > > circuit should work (& why it doesn't). I've found a similar (to my > > mind) relay switching circuit using a BC377 rather than 2N3904 and > > missing the 6K8 resistor between ground & the transistor base. Following > > that I removed the ground resistor & replaced the second resistor with a > > 2K6 one (using calculations provided with the BC377 circuit), but this > > doesn't work either! > > > > Could anyone explain how the above circuit works and how the values of > > the resistors are calculated? Should it in fact be functioning as in > > when connected to a pic as I've described? > > Well, the circuit is pretty basic. First off, ignore the 6.8k resistor to > ground, it's not really needed. > > Now, to simplify things: think of the transistor as a current mode switch. > When your input current into the base the collector and emitter are > practically shorted together. When there is no current flowing into the base > the switch is off, and the emitter and collected can be considered open. > > Your PIC's output pin outputs a voltage. The 6.8k resister in series with > it acts as a "safety" to ensure that not too much current flows into the > base (think of a base as a diode going to ground). > > The value of the resistor is not as straightforward to calculate. What you > WANT to do is ensure the transistor is saturated. You have to get the > transistor's datasheet to figure out it's saturation current for your case > (it will depend on the beta of the transistor and your load current), then > go a little high to ensure it's definitely in saturation. So say you > calculate the saturation current is 1mA: > > The value of the resistor would then be: R=V/I=((5V - 0.7V(which is a diode > drop))/1mA ~= 4300ohms = 4.3k > > Now, consider this the BIGGEST resistor you can use. There is no reason to > go this high unless you need to (to save power or other reason). To be safe > go lower. Your lower limit is defined by the max current a PIC pin can > source. Say it's 20mA, then the SMALLEST resistor you can use is: > > R=V/I=4.3V/20mA = 215 ohms. > > So, the SMALLEST resistor you can use is 215ohms. I don't recommend going > that low either since there is also a "per device" current output that you > might violate on the PIC. > > So, what do I recommend? Don't bother with the calculations! :) Why? > Because MOST of the time selecting something like a 1k or 2k resistor will > "work". If you want go through the calculations to ensure you are OK, but > most of the time I just go with a 2k. > > Obviously a 2k won't work in every case, and in fact if you are dealing > with different power supplies 2k will likely be no go. But it usually is. > > Back to your circuit. Disconnect the base and see if the relay still turns {Quote hidden}> on. If it does then you've likely got a bum transistor, either that or > you've hooked it up wrong. If you've confused the base with the collector > then your symptoms would match. > > Hope this helped, TTYL > > -- > http://www.piclist.com hint: The list server can filter out subtopics > (like ads or off topics) for you. See http://www.piclist.com/#topics > -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

dave cunningham wrote: {Quote hidden}> I want to use a pic to switch an inductive load on/off and it was > suggested (on usenet) that the following should work > > VCC (5VDC) > > | > | > .----o > | | 5V Coil > 1N4002 - _|_ Relay > ^ |_/_|- > | | > | | > |----o > | > | > 6.8K | > ___ |/ > o------|___|---o-------| 2N3904 > PIC Output | |> > .-. | > | | | > 6.8K | | | > '-' | > | | > GND GND Looks mostly reasonable. The diode only needs to be 1N4001, which may be a little easier to find. However, the 1N4002 will work fine, it's just specified to a higher voltage than necessary. > I've assembled the circuit (with the input being a PIC pin pulled up to > 5V via a 10K resistor), however the relay switches (on) as soon as power > is supplied to the circuit That's because of the pullup resistor. Why is it there? Are you using the RA4 open drain output? The rest of the circuit is designed for a totem pole output. > and the output state of the PIC pin has no > effect on the relay state. There is something wrong in your code or the wiring. If the PIC pin is actively driving low, the relay should be off. What is the PIC pin voltage when the relay is supposed to be off but isn't? > I've found a similar (to my > mind) relay switching circuit using a BC377 rather than 2N3904 and > missing the 6K8 resistor between ground & the transistor base. Following > that I removed the ground resistor & replaced the second resistor with a > 2K6 one (using calculations provided with the BC377 circuit), but this > doesn't work either! I would lose the 6.8Kohm resistor to ground. > Could anyone explain how the above circuit works and how the values of > the resistors are calculated? Should it in fact be functioning as in > when connected to a pic as I've described? Let's lose the resistor between the base and ground, assume you are using a totem pole output (not RA4 open drain output) PIC pin to drive the "PIC Output" node on your schematic, and assume a 5V supply to the PIC. The 2N3904 is only rated for 200mA collector current, so this must be a small relay. If not, get a bigger transistor. I would use 2N4401 here, which can do up to an amp. The only value to figure out is the single base resistor. You want enough current to flow thru the base to keep the transistor solidly on at the maximum relay current. This transistor has a small signal gain of 300, so let's only require 100 of it so that we have plenty of margin. The max collector current is 200mA. 200mA / 100 = 2mA base desired base current. The base will be at about 700mV when on, so that leave 4.3V accross the base resistor. From Ohm's law, 4.3V / 2mA = 2.15Kohms desired base resistance. There's enough slop in the assumptions so the 2.2Kohms or 2.0Kohms will work fine. ***************************************************************** Embed Inc, embedded system specialists in Littleton Massachusetts (978) 742-9014, http://www.embedinc.com -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

On Mon, Oct 20, 2003 at 06:47:46PM +0100, dave cunningham wrote: > I'd like to preface this question by saying that while I can program > fine my electronics knowledge goes little beyond the absolute basics. That's cool. The only way to learn is to do. {Quote hidden}> > I want to use a pic to switch an inductive load on/off and it was > suggested (on usenet) that the following should work > > VCC (5VDC) > > | > | > .----o > | | 5V Coil > 1N4002 - _|_ Relay > ^ |_/_|- > | | > | | > |----o > | > | > 6.8K | > ___ |/ > o------|___|---o-------| 2N3904 > PIC Output | |> > .-. | > | | | > 6.8K | | | > '-' | > | | > GND GND Everything but the 6.8K pulldown looks fine. It sets up a voltage divider that also limits the amount of current to the base of the transistor. While some may argue that is keeps the transistor off until the PIC output can be configured, if you're going to do it increase the value of that pulldown so that mode current can be available for the transistor base. I'd initally test with that pulldown off. All of my relay circuits function fine without it. > > I've assembled the circuit (with the input being a PIC pin pulled up to > 5V via a 10K resistor), however the relay switches (on) as soon as power > is supplied to the circuit and the output state of the PIC pin has no > effect on the relay state. Now that's odd. If the PIC output is low, then the transistor should be off and the relay should release. To test, simply ground the base of the transistor. The relay should turn off. Then try tying the base of the transistor to 5V using a 6.8K pullup. Relay should then turn on fully. If those two conditions work, then tie the base of the transistor to the PIC output (BTW which port pin are you using for your output?) and retest. It should work OK now. > As I've said at the start, my electronics knowledge is basic in the > extreme but I've been googling a bit to try to understand how the above > circuit should work (& why it doesn't). I've found a similar (to my > mind) relay switching circuit using a BC377 rather than 2N3904 and > missing the 6K8 resistor between ground & the transistor base. Following > that I removed the ground resistor & replaced the second resistor with a > 2K6 one (using calculations provided with the BC377 circuit), but this > doesn't work either! OK so you did test it. > > Could anyone explain how the above circuit works and how the values of > the resistors are calculated? The transistor is acting as a switched amplifier. This means that if you provide current through the base/emitter junction, then an amplified amount of current is allowed to flow between the collector emitter junction. The PIC output is supposed to be the switch. The purpose of the resistor is to in fact limit the amount of current that can be pulled through the base. OK I'm just going to throw out some numbers. The 2N3904 has a maximum continuous collector current of 200ma. Relay coils offer a resistance between 50 to 500 ohms depending on the model, so let's be conservative and say that it's 50 ohms and therefore draws 100 ma of current across it. So there will be 100ma of current flowing between the collector and the emitter (Ice) of the transistor. So you go to the datasheet that indicates that the beta (the amplification) is 30 for Ice = 100ma. That means that the base current will get amplified 30 times. So the base current requires is 100ma/30 -> 3.3ma. Now two things comes into play here. One is the fact that the base emitter voltage is 0.6V. The second is that Roman Black taught me that you should drive the base upwards of 10 times harder than the minimum specified to make sure that the transistor turns on solid. Since that's 33ma and overwhelms the PIC's output driver, let's back off and say that we'll use 5 times the base current: 16.5 ma. So here's the string we're trying to drive: Pic output ---R-----BE------GND 5V 4.4V 0.6V 0V So the resistor has to drop 4.4V @ 0.0165A. So we can compute the value: V = IR --> R=V/I -> R = 4.4/0.0165 -> 266 ohms. So now we see why there may be an issue. That 266 ohms is a lot less than both the 2.6K and the 6.8K that you've been testing with. So grab a 270 ohm and retest. >Should it in fact be functioning as in > when connected to a pic as I've described? Looking at these numbers above, it probably should never turn on. > > Any suggestions of suitable foundation electronics books would also be > appreciated! I hope this helps. I always use resistor values 1K and less. It's worked quite well for me. BAJ -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

Dave Cunningham wrote... >I've assembled the circuit (with the input being a PIC pin pulled up to >5V via a 10K resistor), however the relay switches (on) as soon as power >is supplied to the circuit and the output state of the PIC pin has no >effect on the relay state. Hmmm. Why are you using the 10K pullup resistor? Are you using the open-drain PIC output (RA4) ? Some suggestions: 1) Double-check (then triple-check, if necessary) the pinout of the transistor to make sure you've got it connected right. If you take a 2N3904 in the plastic TO-92 case and hold it such that the flat side is toward you and the leads are pointing downward, the connections will be E - B - C from left to right. 2) Are you sure your PIC pin has been properly configured as an output? Remember, on many PICs some of the pins are analog inputs until configured otherwise; and clearing the associated TRIS bits isn't enough to make them digital outputs. Use a scope or a DVM to monitor the PIC pin while the code toggles it HIGH and LOW, and confirm that it's functioning right. 3) As far as I can see, the circuit should work as you've got it drawn, though I would concur with others' comments about the 6.8K resistor from PIC output to transistor base being possibly too large (though this wouldn't be causing the symptoms you're seeing) and about the 6.8K resistor from base to ground being not very necessary. 4) If you **ARE** using RA4 as the PIC output driving this circuit, be aware the 10K resistor is too big; it probably won't give you enough base current to properly drive the transistor (again, this wouldn't be the cause of the symptoms you're seeing). Hope this helps a bit... Dave D. -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

> From: dave cunningham[SMTP:daveKILLspamUPSILON.ORG.UK] > Sent: Monday, October 20, 2003 1:47 PM > To: .....PICLISTKILLspam.....MITVMA.MIT.EDU > Subject: [PIC]: Using a 2N3904 to switch a relay > I want to use a pic to switch an inductive load on/off and it was > suggested (on usenet) that the following should work > ... > I've assembled the circuit (with the input being a PIC pin pulled up to > 5V via a 10K resistor), however the relay switches (on) as soon as power > is supplied to the circuit and the output state of the PIC pin has no > effect on the relay state. Hmmm. Looks like a programming problem. Are you sure that you set the output pin to output via the TRIS register? As long as the pin is an input (which is the default state immediately after power up), the 10K pullup will turn the transistor on, and writing to the output latch will not affect this. John Power -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics