Post by thread:[EE:] Interfacing an Active Low signal from a USB

I'm using a premade I/O board (Hagstrom USB board http://www.hagstromelectronics.com/keusb36.html) This board draws power from the USB port. The board has 3 pairs of pins that is meant to be hooked up to external LEDs. For each pair, one pin of the pair goes to the Anode and the other the Cathode of the LED. The Anode pin is held at 5V. The Anode pin is connected right after a current limited resistor (the resistor is ~350 Ohm). The Cathode pin is connected to their microcontroller. When we send a command to the board to light the LED, it brings the Cathode pin LOW. OK. So I want to use the active low signal to simply control a low voltage motor (1.5V to 3V probably) There are a couple of ways that I'm thinking about doing this. I'll be using an external power supply to do this. I guess I _can_ tap into the USB port's power, but I'm not sure how that will affect the I/O board. (1) Using an opto-isolator (because I can essentially hook up a pair of pins to the input of the opto-isolator) and have the opto-isolator's output transistor control another transistor that turn the motor on/off. (2) Using a relay. I guess I can use a solid state one...because I want to make sure that switching speed won't be an issue. These are quite expensive. (3) Using an NPN transitor...I'm not sure if this circuit will damage the microcontroller or not. Here's my guess, please tell me if this is correct or not. Connect the active low pin to the Emitter of the transistor. Connect the motor power supply to the Collector. (I'll probably use 5V here...if I use let's say 1.5V will this still work?) Connect 5V to the base...with a 4.7K resistor. The microcontroller will keep the Emitter at 5V. When it pulls it low, the transistor will turn on? The microcontroller will then be sinking the current that goes through the motor. Is this OK? Will the microcontroller be damaged? The motors we're looking at range from 1.3 - 3.0 VDC and draw between 65 to 95 mA. Another related question (this is probably a silly question) what is the best way to step down a voltage (let's say 5V here) to the operating voltage of a motor (the resistance of the motor is ~10 ohms)? If I use a 10 Ohm resistor in series, the motor will get 2.5 V right? But the current will be (2.5V/10 Ohm) which is 250mA...higher than the rating for the motor. And the power through the 10 Ohm resistor will be .250(2.5) = .625 W...that's a lot, no? When designing for the motor, should we just make sure that it gets the right voltage? The motor will try to draw as much current it needs to maintain its rpm's right? So a current limiting resistor won't do anything? Sorry for such novice questions. -James -- http://www.piclist.com hint: To leave the PICList spam_OUTpiclist-unsubscribe-requestTakeThisOuTmitvma.mit.edu

{Quote hidden}> I'm using a premade I/O board (Hagstrom USB > board http://www.hagstromelectronics.com/keusb36.html) > > This board draws power from the USB port. > > The board has 3 pairs of pins that is meant to be hooked up to external > LEDs. For each pair, one pin of the pair goes to the Anode and the > other the Cathode of the LED. > > The Anode pin is held at 5V. The Anode pin is connected right after a > current limited resistor (the resistor is ~350 Ohm). > The Cathode pin is connected to their microcontroller. > > When we send a command to the board to light the LED, it brings the > Cathode pin LOW. > > > OK. So I want to use the active low signal to simply control a low > voltage motor (1.5V to 3V probably) > > There are a couple of ways that I'm thinking about doing this. I'll be > using an external power supply to do this. I guess I _can_ tap into < the USB port's power, but I'm not sure how that will affect the I/O > board. > >>>>>>>>> I'm not sure how it would affect the I/O board, but it might tax the USB's output current capability and/or add noise to the USB port. In a word, I wouldn't do it. > (1) Using an opto-isolator (because I can essentially hook up a pair of > pins to the input of the opto-isolator) and have the opto-isolator's > output transistor control another transistor that turn the motor > on/off. > > (2) Using a relay. I guess I can use a solid state one...because I > want to make sure that switching speed won't be an issue. These are > quite expensive. >>>>>>>>>> Either # 1 or #2 above would work okay. It depends on how much complexity and/or cost you want to encounter. > > (3) Using an NPN transitor...I'm not sure if this circuit will damage > the microcontroller or not. Here's my guess, please tell me if this is > correct or not. > >>>>>>>>>> This method too would work, but not the way you describe. See below. > Connect the active low pin to the Emitter of the transistor. > Connect the motor power supply to the Collector. (I'll probably use 5V > here...if I use let's say 1.5V will this still work?) > Connect 5V to the base...with a 4.7K resistor. >>>>>>>>>> I would use a PNP transistor with the emitter connected to +v on the external power supply. Connect one end of the motor to the collector of the transistor, and the other end of the motor to ground. Connect the active low pin to the base of the transistor through a 500 ohms or so resistor. The way this works is when the unit is powered up, the active low output will come up high. The motor will be off. When the active low pin goes low (active), the transistor turns on and supplies power to the motor. > > The microcontroller will keep the Emitter at 5V. When it pulls it low, > the transistor will turn on? The microcontroller will then be sinking > the current that goes through the motor. Is this OK? Will the > microcontroller be damaged? The motors we're looking at range from 1.3 > - 3.0 VDC and draw between 65 to 95 mA. >>>>>>>>> I really doubt if the microcontroller can handle having 65 or more milliamps of current passing through it without letting the smoke out of the part. I would advise against it. Use a pass transistor as stated above. > Another related question (this is probably a silly question) what is > the best way to step down a voltage (let's say 5V here) to the > operating voltage of a motor (the resistance of the motor is ~10 > ohms)? >>>>>>>>>>>>> I would use a linear voltage regulator because they're cheap, easy to use, and in this case, not using a lot of current. Or you could put 3 or 4 diodes in series with the motor. The forward voltage drop of say a 1N400x diode is between .6 and .7 volts, depending on the current through it. So with this in mind, you would drop between 2.4 and 2.8 volts with 4 in series, or 1.8 and 2.1 with 3 in series. {Quote hidden}> > If I use a 10 Ohm resistor in series, the motor will get 2.5 V right? > But the current will be (2.5V/10 Ohm) which is 250mA...higher than the > rating for the motor. And the power through the 10 Ohm resistor will > be .250(2.5) = .625 W...that's a lot, no? When designing for the > motor, should we just make sure that it gets the right voltage? The > motor will try to draw as much current it needs to maintain its rpm's > right? So a current limiting resistor won't do anything? > > Sorry for such novice questions. > > >>>>>>>>>>>> I hope this answers your questions well enough for you to make a more informed decision. Regards and Good Luck, Jim > > -James > > -- > http://www.piclist.com hint: To leave the PICList > .....piclist-unsubscribe-requestKILLspam@spam@mitvma.mit.edu -- http://www.piclist.com hint: To leave the PICList piclist-unsubscribe-requestKILLspammitvma.mit.edu

Thanks for all the useful information! To turn "on" the PNP transistor, the base has to be at least 0.6V less than the emitter (and also a small current needs to flow) right? So if my active low signal goes between 0 and 5V...and my motor power, which is connected to the emitter is 5V, then I'm set. What if my motor power was let's say 9V? Does that mean, even though the microcontroller is holding the pin at 5V, the transistor would still turn on, because the emitter is at 9V and the base is at 5V? -James On Apr 30, 2004, at 7:59 AM, Paul James E. wrote: {Quote hidden}> >>>>>>>>>>> I would use a PNP transistor with the emitter connected to > +v on the external power supply. Connect one end of the > motor to the collector of the transistor, and the other > end > of the motor to ground. Connect the active low pin to the > base of the transistor through a 500 ohms or so resistor. > The way this works is when the unit is powered up, the > active > low output will come up high. The motor will be off. > When the active low pin goes low (active), the transistor > turns on and supplies power to the motor. -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

On Sat, May 01, 2004 at 02:52:35PM -0400, James wrote: > Thanks for all the useful information! > > To turn "on" the PNP transistor, the base has to be at least 0.6V less > than the emitter (and also a small current needs to flow) right? Right. > > So if my active low signal goes between 0 and 5V...and my motor power, > which is connected to the emitter is 5V, then I'm set. Presuming that you've taken care of the power and back EMF issues. > What if my motor power was let's say 9V? Does that mean, even though > the microcontroller is holding the pin at 5V, the transistor would > still turn on, because the emitter is at 9V and the base is at 5V? That's right. Most times folks use a Darlington (or the complimentary pair whose name starts with a SZ, szuliki?) config where the PIC drives the 1st transistor base, which then drives the second base. Allows for higher voltage and stiffer current for the main driver. BAJ -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

> From: James[SMTP:.....jamesKILLspam.....2-BIT-TOYS.COM] > Sent: Saturday, May 01, 2004 2:52 PM > To: EraseMEPICLISTspam_OUTTakeThisOuTMITVMA.MIT.EDU > Subject: Re: [EE:] Interfacing an Active Low signal from a USB device with a Transistor for controlling a motor > Thanks for all the useful information! > To turn "on" the PNP transistor, the base has to be at least 0.6V less > than the emitter (and also a small current needs to flow) right? > So if my active low signal goes between 0 and 5V...and my motor power, > which is connected to the emitter is 5V, then I'm set. > What if my motor power was let's say 9V? Does that mean, even though > the microcontroller is holding the pin at 5V, the transistor would > still turn on, because the emitter is at 9V and the base is at 5V? > -James The emitter cannot be at +9V with the base at +5V unless the transistor is defective. Your first statement should read " . . the base can be no more than 0.6V less than the emitter . . .". You must use a resistor from the base of the PNP to a circuit which sinks current to ground. Typically an NPN transistor is used for this. The emitter of the NPN is grounded, and the resistor referred to above goes to the collector. A resistor from the base of the NPN is connected to the signal which turns on the NPN, and therefore, the PNP. This source would be the microcontroller output pin. When the I/O pin goes high, it sources current into the base of the NPN. This current is limited by the NPN's base resistor. That current is multiplied by the beta (current gain) of the NPN to produce an amplified collector current, which becomes the base current of the PNP. The current is amplified again by the beta of the PNP, giving a collector current which drives the load. The current from the I/O pin is multiplied by the product of the two betas, giving a result similar to that of a Darlington configuration, but the arrangement of the transistors provides for limiting the currents to reasonable limits. The NPN bipolar transistor can be replaced by an N channel MOSFET. In this case, the resistor from the I./O pin can be left out, but the first resistor, from the PNP's base to the MOSFET is essential. 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