# Power Supply Sees Electromagnet as a Short Circuit

I'm designing a circuit in which current periodically passes through a copper coil (with an air core), which acts as an electromagnet. The resultant magnetic field then repels a neodymium magnet attached to a rod, such that when the current stops, the magnet+rod returns to its original position.

In the circuit I'm presently trying (below), the periodic output from a 4060 counter is amplified by a BC547 NPN transistor, which then charges a relay that switches 9 volts through the coil. The power supply is a Mastech benchtop HY3005f-3.

I'm presently using quite thick wire (24 SWG, 0.56 mm), and so the resistance of the coil is only 0.5 ohms, which gives a large enough current to do the job (the magnetic field of an electromagnet is a function of both the current in the coil and the number of turns in it). However, the problem is that the power supply sees the coil as a short circuit, and so the short circuit protection in the supply is activated whenever the coil is charged. One way around this is would be to wind the coil with thinner wire and more turns, so that it has a sufficiently high resistance to not appear as a short circuit to the supply.

My question is, is there a way to design the circuit so that I can keep using my present, low resistance, coil without the supply seeing it as a short circuit? Perhaps by discharging a capacitor through it? Or is the only way to go with thinner wire and more turns?

• Compare the current the power supply can provide with the current you are trying to draw from it Aug 10, 2015 at 16:01
• Simply put: You are trying to draw a current higher than your PS is able to provide. Now answer this. Aug 10, 2015 at 16:01
• @JakaQuan: you need to limit the base current (put a resistor between Q3 and the base) otherwise the 4060 and/or the BC547 may be damaged
– Curd
Aug 10, 2015 at 16:09
• If the schematic (relay) is drawn correctly, the coil is on continuously, except for when pin Q3 is high. If the 74 or BC fail, the coil will be stuck "on." And generally, BJT arrows should point down or right for maximum readability. Aug 10, 2015 at 18:03

Like many 'output' devices, a coil for producing a magnetic field has a (DC) impedance. This impdance, in combination, with the applied voltage, determines the current, and current and voltage determine the power.

In your case, a 9V DC power and a 0.5 Ohm coil would result in 9/0.5 = 18A current, and 9 * 18 = 162 W power (if your PSU could supply this current, which it can't, as it is rated for 5A). I assume you did not intend to use 162W of power, in which case you have an impedance mismatch: you have a (relatively) high-voltage power, and a (very) low impedance coil. You'll have to change one or the other:

• you could use a lower voltage: assuming a safe 4A, your voltage should be 2V.

• you could increase the impedance of your coil, by using a smaller diameter wire (with more turns, to compensate for the lower current).

• You could use PWM to effectively lower the voltage or increase the impedance (depends on how you look at it), but I would not recommend this: it is much more complex than the transistor-and-relay level electronis that you seem to be more-or-less comfortable with.

PS check Curd's comment, you realy need a base resistor. And you probably need something to absorb the energy in your coil when you switch it off.

Depending on the amount of time that you want the coil energized, is it a flash or is it sustained, you might try rewiring your relay to charge one or more super caps upon closing then discharging it / them through your coil when it opens. The only downside to this is burning your relay contacts. You might try a 5 or 10 uF cap across the relay contacts to reduce arcing but at the power levels you are asking them to to switch, this will only slow the burning unless you are using a heavy duty contactor. This capacitor setup was how they extended the life of ignition points in the older cars that used point contact distributor ignitions. Good luck.