Driving an inductor with a square current wave

Thread Starter

riftwarrior

Joined May 21, 2009
2
Hi everyone. I have a project which illustrates the driving of a linear induction motor (LIM). We are obligated to supply four square current waves with 90 degree phase difference, at least 300mArms amplitude and adjustable frequency from 20Hz to 200; to four 50mH inductor (with inner resistivity of 20 ohm). As a result, we will get a moving electromagnetic field between that four inductor. We can use only lm741 opamp, whole transistor types and other basic circuit elements with only a DC power supply. At this stage, i have obtained 4-phase voltage waveforms with 12 volt amplitude and adjustable frequency, but i have problems with the next stage. I have a well current wave on inductor with the following circuit (http://img196.imageshack.us/img196/911/circd.jpg) On 20 ohm, there is a good square wave with 500mV Vpp, which corresponds to a poor 25mA max value. This is why i use the base terminal of bjt. Hovewer, with any other variations of terminals, the current waveform collapsed. When i have removed the opamp from that configuration, i also can't obtain a square voltage on the resistor, it behaves as RLC circuits. I also try mosfet with similar circuits. Is there any idea for that problem?
 

mik3

Joined Feb 4, 2008
4,843
What is the value of Vin?

Disconnect the right side of the inductor from the transistor base and connect it to the transistor emitter.
 

Thread Starter

riftwarrior

Joined May 21, 2009
2
Vin is a square wave with 12 Vpp and 20-200Hz frequency adjust. And as i said, when the emitter i use it behaves like just an RL circuit. On 20 ohm, instead of a square voltage wave, i obtain that: http://img30.imageshack.us/img30/3619/rlc.jpg with emitter connection. And also, when i raise the resistance value, to make the RL circuit effect less, i also diverge from the desired 300mArms value. But with base connection, i am getting the desired shape: http://img198.imageshack.us/img198/2903/rlcs.jpg
 
Last edited:

THE_RB

Joined Feb 11, 2008
5,438
It sounds almost idential needs to a "unipolar stepper driver". There are a heap of google links to similar drivers. :)
 

mik3

Joined Feb 4, 2008
4,843
If you have the load on the base it is the same as not using the transistors. That is the season the current is limited to 25mA (the maximum output current of the op amp).

If you connect the load between the emitter and ground it will work.

Multisim simulations sometimes are not very good.
 

DickCappels

Joined Aug 21, 2008
10,171
Also (in addition to driving the inductor from the transistor's emitters), return the inductor (I assume the 20 Ohm resistor is really the resistance of the inductor, if not, eliminate it.) to ground through a small resistors, say 1 Ohm as a current sense resistor. Connect the inverting input of the opamp to that resistor. Now, a small voltage on the noninverting input of the opamp will result in a large current through the inductor.

Oh, and this will probably oscillate, so you will need to put a damping network - a resistor in series with a capacitor across the inductor. Find the values of these parts by experimentation.
 
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