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How do I amplify a low Amp signal (12 v, 100 mA) to 12 v 3.0 Amps?

William P Hilmes

May 10, 2018
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I need to build a waveform amplifier to amplifier a sine wave generated by my DDS Signal Generator (Model JDS 6600). The input signal is 12 volts and 100 m Amp. The amplified signal needs to be 3.0 Amps at 12 volts to power a 33 watt electromagnet. I believe I need a non-inverting Op-Amp to do this job. But I am not sure which Op-Amp to use or how the other component of the circuit are designed. I would like to have a variable resistor at the at Rf position to have more control of the output Amp. An output Amp meter would give me more control to prevent overloading the circuit and destroying my electromagnet. Any ideas or suggestions would be appreciated. Thank you!
 

Harald Kapp

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I believe I need a non-inverting Op-Amp to do this job.
No op amp will deliver 33 W. You'll need a power amplifier. Which frequency? An off the shelf audio amplifier may be suitable.
At higher frequencies you'll neeed an RF amplifier. In that case, however, you'll have to ensure you comply with the regulations that aply at your location with regard to emission of high frequency signals.

Whether inverting or not is irrelevant: a sine is a sine, even when inverted.
 

William P Hilmes

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No op amp will deliver 33 W. You'll need a power amplifier. Which frequency? An off the shelf audio amplifier may be suitable.
At higher frequencies you'll neeed an RF amplifier. In that case, however, you'll have to ensure you comply with the regulations that aply at your location with regard to emission of high frequency signals.

Whether inverting or not is irrelevant: a sine is a sine, even when inverted.
I will be in the ultra low frequency range of 10 - 50 Hz.
 

William P Hilmes

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I will be operating in the ultra low frequency range of 10 - 50 Hz. I am building an underwater electromagnetic sound projector to test the effect of different frequencies on submersed aquatic plants and fish. I will be using an electromagnet to deflect a 4' in diameter 28 Ga galvanized sheet metal at a specific frequency to generate the waveform. Any ideas you may have would be appreciated. My field is not electronics. It is Chemistry. Thank you.
 

Chemelec

Jul 12, 2016
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I will be operating in the ultra low frequency range of 10 - 50 Hz. I am building an underwater electromagnetic sound projector to test the effect of different frequencies on submersed aquatic plants and fish. I will be using an electromagnet to deflect a 4' in diameter 28 Ga galvanized sheet metal at a specific frequency to generate the waveform. Any ideas you may have would be appreciated. My field is not electronics. It is Chemistry. Thank you.

What is this Electromagnetic?
Inductance and/or Resistance of it?
Does it have an Iron Core?
 

Harald Kapp

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I will be in the ultra low frequency range of 10 - 50 Hz.
Any audio amplifier with enough power should work, observe the low frequency limit (10 Hz may be lower than what a cheap amplifier can deliver).
 

William P Hilmes

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https://www.magnetoolinc.com/electromagnets/small-flat-faced-magnets.php

Go to the link above for a description of the electromagnet. It is Model EM-R4, Diameter 4", Height 3", watts 33 not 3.3 as indicated, 12 volts, 2.75 Amp, 4.4 ohms at 12 volts, it has 1000 pounds of pull on a flat surface in the air, it does have an iron core. I will make the unit water proof before I use it under water. This magnet will attract iron in air from a distance of 1.7". It is this attraction and deflection at a specific frequency that will create the waveform under water. A 18" aluminum speaker basket frame will be expanded to a 4' in diameter frame with the electromagnet at the center to attract and deflect 28 ga galvanized sheet metal. The electronics not part of the projector will not go underwater. The power supply will be a 12 volt boat battery. The projector will be deployed from a boat in different locations for testing. Any ideas on the design or logic would be appreciated. Thank you for your time.

Bill
 

Chemelec

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There is a Considerable Difference between powering them from 12 volts DC and a Low Frequency AC.
It is NOT so much about the DC Resistance as it is the AC Impedance.

Also are you using Sine Wave, Square Wave or some other Waveform?
 

William P Hilmes

May 10, 2018
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I am using a Sine wave generated by a DDS function signal generator at 12 v and 100 m Amp. How can I overcome the AC impedance issue? Thank you for your time and help with my project.
 

Chemelec

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I am using a Sine wave generated by a DDS function signal generator at 12 v and 100 m Amp. How can I overcome the AC impedance issue? Thank you for your time and help with my project.

100 m Amp?
Is that 100 Milliamps?

Possibly use a Power Amp and a Impedance Matching transformer.
But getting 33 watts with 12 volt supply is very unlikely.

You could use a Square Wave, Driving a MOSFET to get More Current flow.
 

William P Hilmes

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That is 100 Milliamps. The electromagnet is rated for 12 volts. Is there a ways around the limit without damaging the electromagnet?
 

Chemelec

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That is 100 Milliamps. The electromagnet is rated for 12 volts. Is there a ways around the limit without damaging the electromagnet?

You Said the Electromagnet was 4.4 Ohms and rated for 12 volt.
So 100 mA is FAR BELOW and Damage.
And the 100 mA is NOT Delivering 12 Volts. NO WHERE EVEN NEAR THAT.

If you Amplify that Signal with a Power Amp and use a Matching Transformer, you could bring it up to a MORE Suitable Power Level.

Or Using a Square Wave at a 50% Duty Cycle Driving a MOSFET with the Coil between the Dain and your 12 Volts will Bring you Closer to a Suitable Power level, WITHOUT ANY DAMAGE to the coil.

NEED MORE HELP on How to do this?
Email me at: chemelec at hotmail.com
 
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Alec_t

Jul 7, 2015
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As Chemelec mentioned, magnet inductance will be an issue.
Annoyingly the datasheet gives no info about the magnet inductance. For a 9lb 33W magnet I would expect the inductance to be quite significant. If we take it to be, say, 1H then the magnet reactance at 10Hz would be ~ 60Ω. You would need quite a high voltage to drive much current through that.
 

William P Hilmes

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Alec_t I will check with the company that I purchase the electromagnet from and find out what is the magnet inductance and the magnet reactance.
 

Audioguru

Sep 24, 2016
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How do I bridge the amplifier?
A bridged amplifier uses two amplifiers, each one drives one wire of the speaker. The voltage swing across the speaker is almost doubled then the current in the speaker is also almost doubled resulting in almost 4 times the power in the speaker with the same supply voltage, as a single amplifier. If a single supply voltage is used then output coupling capacitors are not needed because each amplifier has the same average DC output voltage.

I looked at all the amplifier ICs that I know and none provide enough output current to drive your 4.4 ohm load to 33W.
There are many car radio bridged amplifiers but their maximum supply is 16V which will not drive your 4.4 ohm load to 33W. If you bridge two separate amplifiers then each one has double the normal current so each one must be rated to drive a 2.2 ohm load. I could not find any.

A few audio amplifier ICs can drive your 4.4 ohm load to 33W if they use a 37V supply. Since your load has inductance then the output voltage swing and supply voltage must be higher than to drive a 4.4 ohm load.
 

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Externet

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A 12VDC powered automotive subwoofer power amplifier has a built-in voltage booster to be capable of delivering the output sought. For extra, most are 2 channel bridgeable.
 

Chemelec

Jul 12, 2016
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With a 16 VDC Supply, this can deliver 30 watts into a 4 Ohm load.
 

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