Over Saturation ?

japer

hello

this post may be in a inappropriate forum this concerns
transistors.

in applied sensors an amplifying transistor is used to
run a pump. 300ma at approx. 3-volts. the problem is that on very long
pump on time over 5 sec. the transistor appears not to respond
to Base state change, on short runs the transistor responds correctly.
i bring this up because i have run into this type of occurrence
on a different type of circuit with an amplifying transistor.
Does this happen sometimes ? is the circuit flawed or overworked ?

any help or math explanation would be appreciated

japer

Beau Schwabe

Are you sure that your schematic is correct?

That 10K resistor is definitely in the wrong place.


These links may help...
www.parallax.com/dl/docs/cols/nv/vol1/col/nv6.pdf
www.parallax.com/dl/docs/cols/nv/vol1/col/nv23.pdf

http://forums.parallax.com/attachment.php?attachmentid=37701

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Beau Schwabe

IC Layout Engineer
Parallax, Inc.

Paul Baker

Well, I do note two issues, the first is that you are high side switching, ordinarily you use a pnp transistor to do such switching instead of npn. This is because npn's establish turn-on via the junction between the base and the terminal closer to ground.

What is happening is when the transistor is off is, the emitter of the npn is at Vss. Inductors (and motors are just like big inductors) resist changes in current by converting the change in current to an electromagnetic field. This field is what causes the motor to rotate. Once the current stabilizes (when the motor reaches speed), the inductive windings in the motor will try to keep the current constant. So when you attempt to shut of the npn, the windings in the motor will convert some of thier EM fields into current, this affects the voltage at the emitter so that the npn continues to conduct (ie the voltage at the emitter is < Vss, but the npn is not driven as strongly). This will continue to happen while there is stored EM fields in the motor, therefore it will take a very long time before the voltage at the emitter rises to the point that npn stops conducting. Also the rotataing motor when shut off becomes a generator, continuing to feed current into the system, exacerbating the situation even further.

So to solve this, first replace the npn with a pnp, doing this will reverse the control voltage input (Vdd for off, Vss for on), second you need to put a flyback protection diode across the motor, this is a reverse biased diode capable of carrying enough current to allow the field stored in the winding to collaspe safely to ground. By reverse bias I mean the line on the diode is facing the Vdd side of the motor. This diode will help protect the pnp transistor, because by using a pnp, it _will_ stop conducting (unlike your current setup), and the motor will desperately try to make the current continue to flow by droping the collector of the pnp to very a low voltage (< Vss), if this voltage exceeds the maximum Vce of the pnp, it will damge the pnp.·Having a flyback diode on the motor will prevent this from happening by conducting whenever it notices the·Vdd side of the motor·has a lower voltage than the Vss side of the motor (this only happens when the motor is turned off)·BTW pnp have thier collector and emitters reversed, the collector is connected to the Vss side and the emitter is connected to the Vdd side.

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·1+1=10

Post Edited (Paul Baker) : 3/30/2006 3:39:48 PM GMT

Tracy Allen

Hi Japer,

The circuit configuration is called an "emitter follower". The voltage at the motor motor follows the voltage at the Stamp pin, but reduced by 1) the voltage across the 100 ohm base resistor, 2) the base emitter voltage of around 0.6 volt, 3) the voltage across the 10 ohm resistor. The power comes from the 5 volt supply, not from the Stamp pin. Since that small motor is supposed to run off of around 1.5-3 volts, the reduction in voltage is acceptable. At 200ma current, the 10 ohms drops 2volts, the Vbe is .6 volts, plus <.1 volts across the 100 ohm resistor + Stamp output resistance. That leaves 2.3 volts across the motor. As I recall, the motor does not draw 300ma at those voltages.

The transistor called for specifically is a Zetex superbeta eline, which have remarkable power handling abilities and are very robust. Is that the transistor you are using?

I'm not sure what is going on with your circuit, to make it turn on but not off on long runs. That is why I ask about the transistor. If you are not using the one specified, it may be overheating.

While a catch diode in parallel with the motor could not hurt, this small motor does not generate much of a kick. Field energy tends to be dissipated in motor motion. You can see on an oscilloscope. Aristedes and I have talked about adding a diode there for instructive purposes.

The alternative circuit Paul mentioned is called "common emitter", a different thing. That circuit is the one you will see most commonlly for switching, and in the links that Beau posted. Another name for emitter follower is "common collector". The third configuration is "common base". All of those configurations (and others too, where those names don't apply) have their applications. In this case, emitter follower works because we have to reduce the voltage from the Stamp anyway. It could also work with a common emitter circuit using either a PNP or NPN transistor or a mosfet, with appropriate choice of a series resistor. One thing about electronics, there are many ways to go.

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Tracy Allen
www.emesystems.com

Tracy Allen

I checked it again, and you're right, the pump does fact draw around 330 ma under load at 3 volts. At 1.5 volts it draws about 230 ma. In the follower circuit with 10 ohms, it runs with at about 1.5 volts, but is still able to pump water over a 4" hill for the demo. If you decrease the resistor to 4.7 or 5 ohms, it will operate at oomph around 2.5 volts. But at those levels I never observed the lockup you mentioned.

However, if I left the resistor off entirely, I did see it lock up in a state where it would stay on and not turn off. The pump was then cranking along at over 4 volts. When it froze, I thought the transistor might be burnt out, but that was not the case. Instead, the Stamp was serving up a noisy 5 volts to the base of the transistor. It was the Stamp that had locked up. Cycling the power would restore operation. This was built on the BOE using the Vss and Vdd connections. There was a terrific level of noise induced on the power supply, that was getting back to Vdd on the Stamp. The voltage regulator couldn't keep up with the high frequency high current pulses. On the oscilloscope it looked particularly bad when the motor was first starting up, working to build up speed. The Vdd trace had terrible spikes that dropped ifrom 5 volts down to 4 volts, which I think would cause the Stamp to reset.

But I never saw that when there was 5 ohms or 10 ohms in series with the pump. The Vdd power supply then was as smooth as glass.

If you try this again and it locks up on you, please measure to see if the voltage from the Stamp pin is okay.

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Tracy Allen
www.emesystems.com