Small motor and the PNP transistor help please
grasshopper
Posts: 438
I need a bit of advice.
I am trying to calculate the bias resistor for a NPN transistor that will power a small motor using PWM. The motor specs are as follows. My problem is that I am not familiar with this terminology.
Nominal Supply Voltage 2.7
Armature resistance 11.5 Ohms +/- 15%
Maximum output power .15watts
Armature inductance .100mH
No load current 17mA
What is Nominal Supply Voltage mean? Is this the maximum voltage the motor needs? What is Armature resistance? I figured it was the resistance of the motor while not running? However I use the numbers and come up with a different Max power. Sorry if these questions seem stupid.
This is the schematic. How do I interpolate all the data as being a collector resistor (s1 in the picture) and thus calculate the bias resistor so that I can have god control over the motor. By the way the motor will have a very small weight attached so some opposition will be in effect.
In the picture the 5Volts will be 2.7 It is a typo.
I am trying to calculate the bias resistor for a NPN transistor that will power a small motor using PWM. The motor specs are as follows. My problem is that I am not familiar with this terminology.
Nominal Supply Voltage 2.7
Armature resistance 11.5 Ohms +/- 15%
Maximum output power .15watts
Armature inductance .100mH
No load current 17mA
What is Nominal Supply Voltage mean? Is this the maximum voltage the motor needs? What is Armature resistance? I figured it was the resistance of the motor while not running? However I use the numbers and come up with a different Max power. Sorry if these questions seem stupid.
This is the schematic. How do I interpolate all the data as being a collector resistor (s1 in the picture) and thus calculate the bias resistor so that I can have god control over the motor. By the way the motor will have a very small weight attached so some opposition will be in effect.
In the picture the 5Volts will be 2.7 It is a typo.
Comments
You need to take into account the resistance of the transistor when it is on. PWM is the usual way to control motor speed, using the transistor as a switch.
Leon
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Amateur radio callsign: G1HSM
Suzuki SV1000S motorcycle
Did I do this right? It seems as thought i need to include some other factors.
Comments are much welcome
·Depending in the PWM switching frequency, you may want to get rid of or reduce the size of the 1 uF cap on the base. It will drag out the transition through the linear region.
1.8 K should work fine. The concept of PWM is that we want the transistor to be completely switched ON or completely OFF. This minimizes heat dissipation while the transistor goes through its linear region. In switching circuits its usual to err on the side higher base current, so I'd go with 1K or even 680 Ohms
Also note , in your calculation, that the armature resistance of 11.5 Ohms will be in effect only while the rotor is stopped (at the instant of start-up). As the motor spins up, a back EMF is generated which has the effect of reducing the armature current. So, if you check the running current, it will be less than the locked-rotor current.(ie the no load current of 17 mA)
In perfectly pedantic terms, your equation can also take in to account the junction drops of your transistor and whether the Propellor can put out a full 3.3 volts when switched on.
Cheers,
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Tom Sisk
http://www.siskconsult.com
Post Edited (stamptrol) : 10/24/2008 5:55:50 PM GMT
·
As stamptrol·mentions... "...your equation can also take in to account the junction drops of your transistor and whether the Propeller can put out a full 3.3 volts when switched on."
·
The·C-E (Collector Emitter) junction of·a typical bi-polar transistor has a voltage drop of about 0.6V, so in order for your motor to "see" 2.7V·your supply needs to be at least 3.3V (2.7V + 0.6V = 3.3V).· Otherwise if the supply voltage is 2.7V then your motor will only receive 2.1V
·
Once you figure the acceptable voltage that will be across the motor and in turn the Current requirements of the motor, then it's a good design practice to double this current requirement.· This ensures good transistor saturation.· So your 234.8 mA becomes 469.6 mA
·
With an Hfe of 200 this means that 2.35mA should be applied across the·B-E (Base Emitter) junction.· Keep in mind that there is also a 0.6V drop here as well, so if your I/O can supply 3.3V, you should figure this drop in as if the I/O were supplying 2.7V.
·
2.7V / 2.35mA = 1.15K which I would round the closest standard value of 1K for your current limiting resistor.
BTW) If you are going to supply a PWM signal to drive the motor, I would remove the 1uF capacitor.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
Post Edited (Beau Schwabe (Parallax)) : 10/24/2008 7:34:29 PM GMT
Thanks for the input. As far as the capacitor I am going to remove it and put in place a DAC. This will allow me to keep clock cycles and I wont have to run PWM. My initial drawing I was going to use the Resistor and capacitor as a filter to convert PWM into a voltage.
I am thankful for the suggestions and help.
The comment both Beau and I made about the capacitor was to reinforce the idea that the transistor must be either full ON or full OFF. That's what PWM is all about.
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Tom Sisk
http://www.siskconsult.com
·
Stamptrol is correct, the whole idea of making sure that the current limiting resistor had enough drive to saturate the transistor was to keep the transistor outside of it's linear region. The linear region of the transistor does have its place for small signal amplification, but not with relatively large current drives. For that you should use PWM.
The reason you want to minimize or remove the capacitor at the B-E transistor junction is because more capacitance can cause the transistor to stay in the linear region longer during the B-E transitions.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
I cant switch it on and off (PWM) it already creates to much noise. My 20bit ADC gets all out of wack. I must drive the motor with an constant voltage. The transistor is only pushing a small amount of current so it wont get to hot unless I mis calculated something.
I am confused now. Hum.
The data sheet is here www.st.com/stonline/products/literature/ds/9288/2n2222a%20.pdf
Post Edited (grasshopper) : 10/25/2008 4:47:26 AM GMT
If that is the case, you might consider driving an adjustable voltage regulator from a DAC, digital pot, or binary weighted I/O's directly from a BS1,BS2,SX, or Propeller.
cache.national.com/ds/LM/LM117.pdf
Page #22 hints on using a binary weighed Digital method.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
The motor you have chosen is a 3V motor that you are running on 5V so any of the figures you quoted need to be adjusted accordingly. A stalled motor at 2.7V could draw up to ~250ma but running without any load as little as 17ma. Assuming you are running from 3V, if the transistor is fully saturated then it will probably drop around 300 to 400mv across the collector to emitter which means less than 100mw power disipation under full load which even a tiny package can handle. I don't think that a large TO39 pack transistor is the best selection as all you need is a low Vce(sat), less voltage drop, less heat.
If you are using through-hole parts then there are plenty of transistors available, even the 2N2222 can easily handle this. Zetex make low Vce(sat) transistors for much higher currents in small packs also.
As regards the capacitor, definitely you should remove it. Besides the way you have it hooked up will not work as the RC circuit is fine for voltage output but the base of the BJT is effectively a diode to ground and will always clamp the base-emitter voltage to around 0.6V. It is current that drives the base, not voltage, which is why there is almost invariably a base resistor to “convert” the voltage to current. So the capacitor can never charge to any voltage above 0.6V and the transistor does not turn on unless that threshold is reached.
Don't want PWM? If you want an RC voltage to drive the transistor you need to drive the transistor as an emitter-follower circuit with the motor connected from the emitter to ground. This way the maximum voltage will be around 2.7V (3.3V(RC) – Vbe) across the motor and the RC circuit is now driving the transistor in a voltage-follower type mode (less 0.6V) but boosts the current drive. Power dissipation increases accordingly as the transistor is now operating in it's linear region and is dropping voltage across the collector-emitter which translates to simple maths: Iload * Vce = Power. If you run the motor at 1V under load than the transistor will have to “bear” the other 4V at probably close to stall current of around 100ma @1V so that the transistor now starts to get a bit warm dissipating 400mw.
Your call
*Peter*
Yes I am leaning towards a DAC so that I can adjust the speed of the motor. Size is of much importance. This device holds a small foot print and I need to keep it going in that direction. So I would still think that a DAC into a transistor would do the trick. The LM117 idea is a clever one, but I would be using lots of space and kind of over killing the section that drives the motor. It looks as though I would need a lm117 a digital pot and at least on more resistor. Perhaps...Then again the LM117 is good for a much larger voltage span I need a mere 3V-0V and less than 200mA.
Peter Jakacki:
You mentioned some interesting points Ill have to look up the specifics regarding the emitter-follower.
Thanks everyone. I cant stand Analog, makes one have to decide on too many variables.
Post Edited (grasshopper) : 10/25/2008 1:53:47 PM GMT