How to Generate 10kHz–10MHz AC Signal at 0.6–2A: Signal Generator or DC-AC Conversion?

I want to get a ac signal(or a rippling signal) from around 10k Hz up to 10M Hz with current around 0.6-2A.
How can I do that?

A common signal generator may have such frequency but the current is low to around 200mA. How can I boost it up? If I boost it up successfully, would the large current damage me signal generator?

If signal generator cannot be used, what else can I try? I really wish to hear some simple way because I do not know much in this area.

Maybe can I convert to the required ac signal from a DC supply? Then the current magnitude can fulfill my need. But to to do such conversion to get such a high frequency?

Thanks a lot to reed all the words^^

Hello chunkit, I think you need to boost the current from 200mA or something to what is desired, you can acheive this in stages, genrate the signal, then boost the cuurent using a power transistor (e.g TIP32).

Hope this helps!
Cheers!

You need a wide band liner amplifier. 10MHz is not all that high and pretty much inexpensive components can be used to make one. What is the amplitude you are looking for, 5Vp, 10Vp ?

With your signal gen a buffer and a power stage , that would be pretty much all you need , of course a power supply is required.

And no worries about it damaging your signal generator. That's what an amplifier does. It boosts the signal using it's own power, not the power from the thing it is amplifying.

Just a simple transistor will work?? I am excited because this is my first work in ee. The transistor can cope with the ac current?? Can it handle backward current? or I need to set up a circuit to handle the different direction of current flow?

thx for answering.
but i have a silly question.
what is the word wide band means? Does it mean the amplifier can handle a wide range of frequency?

thanks for answering
this is my first task so i get nervous in damaging the device.

You got it. It's short for "wide bandwidth" ["band" being a "band of frequencies"]

True but it's generally means more than that. It also means, common understand that it has a flat response over some range, say 1KHz to 10MHz, where the lower and upper limits are the -3dB, & +3dB respectively.

In this case it could supply a load with minimal variation as a source impedance and minimal distortion over he same range as well. These are some of the basic requirements in a power amplifier.

Mentioned in the responses, a single transistor I am assuming it is an emitter follower with a buffer amplifier. which would be cap coupled to the load. I have done some simulations on such and I could recommend some components to start out with, although the PCB layout is another issue. One can buy kits that will do this for about $100, although not sure these kits will deliver 2A of load.

An emitter follower would only be a solution if the required output voltage is larger than around .7 volts less than the max voltage output of the signal generator. As far as I can tell, chunkit hasn't divulged that specification.

Edited [2012-10-22]: I meant, "...if the required output voltage is _lower_ than around .7 volts less than the max voltage...".

Here is the circuit. Is this correct?
Is there any method I can get the same sine shape signal as the input?

Help~~~~~~~~~
The signal generator only provide me the signal but no/low voltage or current.
I construct as the above graph but I got constant 10V in the output.
Then I tried to connect the two terminal of signal generator to a 3 ohm resistor, and set the peak to peak voltage to 5V(displayed on the signal generator). I expected to have a current of 5/3 A and heat up the resistor with power 25/3 W. But it failed. Although the CRO show that the peak to peak voltage across the resistor is 5V, but the resistor didn't heat up.
I have try to connect a 27 ohm resistor to 9V battery to have a power consumption of 81/27 W, and the resistor heat up quickly in this case.
So i suspect the signal generator only give me the signal but not a real voltage. Is that correct? If yes, the problem go back to the starting point---how can i get the current.

At 10 mhz you are talking microwave frequencies which require wideband amplifiers and your asking for 2 to 6amp at the ouput which is rf frequency You know full well that this is going to be very difficult to achieve but heres a link for you

http://www.guru.com/freelancers/Circuit-Desig...tion/California/Los-Angeles-Long-Beach/391584

I think you were thinking "GHz" when you saw "MHz" -- happens to the best of us ;)

10 MHz is actually a rather low frequency, and quite obtainable.

Nope, that's not an Emitter Follower (Google "Emitter Follower") circuit, and actually, that circuit is not biased correctly for class-A operation (more like class-C) -- unless the values you have chosen have to do with knowledge you have about the characteristics of the signal generator's output (since you do have it direct-coupled).

Disclaimer: I'm more of an MPU/digital guy, and I'm not as good with analog, but I'll give you what I've got:

As for constant 10V on the output of your amplifier, it might have to do with direct-coupling it to the signal generator output (because, otherwise, the way you have that transistor biased, I would expect it to be in "saturation" or "on" mode and the output to be near zero. But, if the signal generator's output impedance is 50ohm, then, that would explain the 10V output, as the low [50ohm] impedance would bias the transistor off.

I suggest using a capacitor between the sig. gen. output and your amp's input. For that frequency range, though, you need a large enough cap to pass the low frequencies, and yet you need to consider parasitic inductance which will limit it's ability to pass higher frequencies (not sure how to solve that, but perhaps with parallel caps, one a high value and one with a low value (and resonate frequency well above 10 MHz).

If you put a 3ohm resistor across the output of your sig gen and read 5V P-P across it (3.5V RMS), and it didn't get hot, it is either a VERY high wattage resistor (one that can dissipate around 4 watts [(5*.71)**2/3] and not get hot), or your testing device (not sure what a "CRO" is) is giving you an incorrect reading (it needs to be able to measure RMS at whatever frequency you have the sig gen set to). Try setting the sig gen to it's lowest freq setting. If the resistor is, indeed, 3 ohms, and there is 3.5 volts RMS across it, then it will dissipate around 4 watts -- that's the physics and nothing can change that. In other words, your statement "the signal generator only give me the signal but not a real voltage" is nonsensical. What?... as opposed to "imaginary voltage"?

When you measure the "signal" what are you using to measure it? If it's a scope, and that scope measures voltage in the time domain (or even frequency domain), then the voltage is there--the signal IS the voltage. So, something else must be wrong. I would suspect your instruments (broken or out of calibration) or an error in interpreting them. It's easy to be mislead by a reading, if you've made incorrect assumptions about the setup or device(s) that you are using to get that reading.

Ah-ha! I just reread your post and it says that you "set the peak to peak voltage to 5V(displayed on the signal generator)". When you say "displayed" do you mean, what is "written" on the generator? If so, then that is the output _expected across whatever output impedance the generator is designed to deliver to_. Never trust what is _written_ on a device. That is merely an assumed result. The real world might intercede with conditions that were not intended -- like sticking a 3 ohm resistor on the output of a generator that is probably designed for higher impedances (like 50 ohms).

What you need to do is look at what the _actual_ voltage is across the 3 ohm resistor, preferably with a scope. You will probably find that it is a LOT lower than 5 V peak-peak.

Addendum to the suggestion of looking at the voltage across the 3 ohm resistor: consider that putting a 3 ohm resistor across the output is probably stressing the generator and perhaps not a good idea (unless the generator's output circuitry is well designed and can handle boneheaded maneuvers like applying 3 ohms to the output ;)

Perhaps, instead, using a 50ohm resistor (but only if the generator is spec'ed to handle such a load). In other words, find out what the output is designed to take and use that for your little test

Microwave frequency? Do you know what a microwave is, no it isn't the thing that heats up the hot dog.

Wow!! Wow!! Wow!!!!!

A microwave is 1/1-million of a meter or C*f = wavelength ( speed/time = distance). I will let you Google that. 10MHz isn't going to harm anyone, cook you or make you go crazy. Look up ,"C," if you don't know and no it isn't after B and before D.

10MHz is nor that difficult, not at all. What is more challenging is a wide band amplifier up to 10MHz that has good flat response/characteristics.

You need a new friend, Google or Wilipedia.

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CRO means cathode ray oscilloscope
The scope have nothing wrong when measuring other potential such as a 9V battery. When I put the scope across the 3 ohm resistor, it give a sine wave of exactly 5V peak to peak voltage, same as the setting on the signal generator. That's the most strange thing......

Calm down, Earl -- and ask yourself if you ever make mistakes -- for instance, what is "Wilipedia"?

And isn't it _C/f_, not _C*f_ that wavelength is equal to?

And isn't it _speed/frequency_, not _speed/time_?

And, I believe a microwave's wavelength can be anywhere from 1 millimeter to 1 meter. That's "milli" not "micro" -- i.e. "1/1-thousand", not "1/1-million".

If you don't believe me, check Google, or Wikipedia

That is puzzling. Are you sure it's a 3ohm resistor? Try measuring it with an ohmmeter.

Actually it is called micro because it uses these frq to heat up your hot dog.....
Wow! Wow!!