# Current Amplifier

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#### Marco203

##### New Member
Hello,
I'm using LTSpice IV. I have a sinusoidal voltage (+5V -5V) and I need to have a sinusoidal current flowing in a coil with amplitude of 1 ampere. How do I do this? I tried with an Inverting amplifier followed by a transistor but I have problems..

Marc

#### alec_t

##### Well-Known Member
Welcome to ETO!
Is this a Homework question?
What are the values of the AC frequency, the coil inductance, the coil resistance and the voltage source impedance?
Why do you need an amplifier or transistor?

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#### MikeMl

##### Well-Known Member
I=E/Xl
1=10/Xl
Xl = 10

Xl=2πfL
10=2πfL
L = 10/2πf = 5/πf = 1.59155/f

Tell us f, calculate L, and then connect L in parallel with the voltage source, no amplifer required. If you cannot do it this way, then you will need a voltage-controlled current source. Last edited:

#### Marco203

##### New Member
I'd rather do that with a transistor because I need a precise value of L and a variable frequency for the input voltage. Actually the Alternating Current flowing through the coil should have variable amplitude and frequency

#### MikeMl

##### Well-Known Member
I'd rather do that with a transistor because I need a precise value of L and a variable frequency for the input voltage. ...
Then you need to use an active device, opamp or transistor circuit configured as a voltage-to-current converter. Does one end of the inductor have to be grounded?

#### Marco203

##### New Member
No, it has not to be grounded. What kind of solution are you thinking of?

#### MikeMl

##### Well-Known Member
Simplistic voltage-to-current converter using an ideal power opamp: Note that I(L1) = I(R1). However, note how much voltage it takes to drive 1A through a 10mH inductor at 1kHz. That is not your average opamp... A garden variety opamp can only drive 25mA at +-15V.

You will have to carefully calculate the opamp specs from the highest frequency and inductance values...

Also, be aware of the stability requirements for a real circuit... Even the ideal circuit is only marginally stable...

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#### spec

##### Well-Known Member
Hi Marco203,

Here is an outline circuit that is commonly used to define a current in an inductor.

The current flowing through the inductor is defined by V input/R8

spec

#### MikeMl

##### Well-Known Member
...
Here is an outline circuit that is commonly used to define a current in an inductor....
Just like in the previous example, the amplifier must be capable of sourcing/sinking > 1A. The +- power supply voltages are dependent on the highest frequency at which the inductor is driven.

#### crutschow

##### Well-Known Member
For simulation purposes you can just use a current source instead of a voltage source.

#### MikeMl

##### Well-Known Member
For simulation purposes you can just use a current source instead of a voltage source.
Or the G-source I showed in post #4.

#### spec

##### Well-Known Member
Just like in the previous example, the amplifier must be capable of sourcing/sinking > 1A. The +- power supply voltages are dependent on the highest frequency at which the inductor is driven.
In your circuit there is no current amplification; the input current is the same as the output current. But I did not mention that.

There are many considerations about any amplifier- are you going to give a dissertation on them all?

My post clearly states that it is an OUTLINE CIRCUIT.   Last edited:

#### MikeMl

##### Well-Known Member
...
My posts clearly states that it is an OUTLINE CIRCUIT.   Then why did you show +-9V power supplies? How can you possibly know that those specific voltages are sufficient to establish the required current in the inductor?

#### spec

##### Well-Known Member
Then why did you show +-9V power supplies? How can you possibly know that those specific voltages are sufficient to establish the required current in the inductor?
I do not understand your point of view. 9V is just a voltage. Like I said it is just an outline circuit to illustrate the principle. Besides which 9V supply lines may be adequate as the OP has not specified the frequency of the signal or the characteristics of the inductor.
(in case you think that I posted the circuit to counter your circuit, that was unfortunate: our posts crossed)

#### crutschow

##### Well-Known Member
Or the G-source I showed in post #4.
Quite true.
I didn't notice that you had posted a voltage-controlled current source.
Does that have any advantage over just using a current source?

#### spec

##### Well-Known Member
View attachment 100678

I understand how it works, but I don't know how to get currents with magnitudes of amperes in the inductance.
what should I change?
As I said, that is just an outline circuit. The opamp needs to be able to sink and source the current that you want to flow through the inductor and the opamp needs to be able to produce the voltage required to get that current flowing through the inductor.

Thus, in order to specify/design a suitable opamp, the following parameters will need to be specified.
(1) Voltage across the inductor: already specified as 5V peak sine wave.
(2) Current required to flow through the inductor: already specified as 1A (1A peak assumed)
(3) Inductance of inductor.
(4) Frequency or frequency range of sine wave.

In general, a direct coupled audio amplifier with an opamp input (traditionally called a servo amplifier) is the sort of topology that would be used. Alternatively, you could use a high current, high voltage opamp like these:

spec

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#### MikeMl

##### Well-Known Member
I understand how it works, but I don't know how to get currents with magnitudes of amperes in the inductance.
what should I change?
You need to know three things:
First is the inductance (L)
Second is the peak current (I) you want to drive through the inductor.
Third is the highest frequency (f) with which you want to drive the inductor.

for example:
Suppose you want to drive a 1A Sinusoid through a 50mH inductor at 100Hz. This is how to calculate how much voltage you will have to apply to the inductor to make that happen.

X is the reactance of the inductor at 100Hz
X=2*pi*f*L = 6.28*100*0.05 = 31.45Ω

Ohm's law says that E=I*X, so E = 1A*31.45Ω = 31.45V

Here is a simulation of this example. Note that it confirms a zero-to-peak voltage V(a) of >31V is required to force 1A of current through a 50mH inductor at 100Hz: That would require a power amplifier (with feedback around it to make it a voltage-controlled current-source) that can produce a Voltage of +-31.45V, at a current of +-1A. The LM675 is such a power opamp that will tolerate up to +-60V power supplies, and is capable of sourcing and sinking up to 3A.

Note that if you double the desired inductor current, you will have to double the applied voltage... If you double the frequency, you will have to double the applied voltage... If you double the inductance, you will have to double the applied voltage...

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#### spec

##### Well-Known Member
Here is an outline schematic showing the topology of a typical current driver amplifier. This amplifier would be capable of sinking and sourcing around 3 Amps and would be capable of an output voltage of around plus and minus 15V.

spec

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