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Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/08/2009 2:17 AM

hello all ,

I hope someone can help me with this.

My goal is to put out on the load 10W -20W sine wave.

so I need a " box " that have the:

1) Input of sine wave 0.5MHz >>2MHz and 1Vpp >>4Vpp Amplitude.

2) Output of 10W -20W sine wave on 50>> 500 Ohm load.

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#1

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/08/2009 2:35 AM

The box you desire is an amplifier.

If the input and output impedance is 50 Ohms, you can calculate the desired gain you need and find an off-the-shelf amplifier from any of a number of manufacturers. Let me know and I will provide names and urls. Most rf amplifiers nowadays are designed for 50 Ohm output but will operate into a 500 Ohm load without damage. However, an amplifier designed to provide 10-20 Watts into 50 Ohms definitely won't provide that kind of power to a 500 Ohm load. If you have different load impedances, you will need transformer coupling to match the relatively low impedance amplifier output to a higher potential capable of dissipating the required power into the higher impedance. Can provide source information on that kind of transformer as well.

If your input signal is high impedance it is a little more complicated but I can provide circuits/manufacturers of emitter follower type circuits that match a high impedance signal to 50 Ohms.

Finally, if it is high impedance input and output, that used to be more common, and we may need to look at some older technology.

A little more detail on the projected use of the amplifier will go a long way towards nailing down the design you need.

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#2
In reply to #1

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/08/2009 3:52 AM

Many thanks on fast response.

I will be clearer I need to heat a biological tissue to something around 40C so

If you apply sine wave at high freqwancy 0.5Mhz and higher (floating from earth from safety reasons)

It just heat the biological tissue without any damage.

This biological tissue resistance can begin with hundred of ohms and while heating

To tens of ohms.

As well as I need the ability to control the output power by increasing or decreasing.

For now I work with Howland current pump with the input of sine wave produce from DDS AD9833

That control the power via software, but I cant get the desired power I can get only 1-2W.

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#3
In reply to #2

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/08/2009 9:54 AM

This amplifier can deliver 25 W cw into any load impedance without oscillation or foldback - an important consideration given the varying impedance of your tissue samples. Bandwidth is 10 kHz to 250 Mhz. Here's the spec sheet (pdf).

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#4
In reply to #3

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/08/2009 10:31 AM

Manny thanks.

very interesting but Please let me know:

1. Is the output is sine wave.?

2. Can I control the output power?

3. The price is high ($6,700.00) for my product

Can you recommend on less expensive solution

since I need up to 2Mhz

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#5
In reply to #4

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/08/2009 11:10 AM

Europium beat me to it, but I will answer questions about this and other amps like it. Amplifier Research, IFI and ENI all build similar technology amps for similar prices. You can get these amp at some discount from merchants of used test equipment such as Tucker, but for really cheap you can't beat ebay.

They are amplifiers: what you get out is what you put in plus gain, as long as you stay in the linear range. Sine wave in, sine wave out.

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#6
In reply to #4

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/08/2009 11:35 AM

"... since I need up to 2Mhz"

Nearly all RF amplifiers in your frequency range have a much wider bandwidth than the one you specified. Generally, to get less bandwidth you have to artificially limit the bandwidth by means of filters. There are tuned amplifiers, of course, but these are not what you want given that they generally have less bandwidth than you spec'd.

If an amplifier works at 500 kHz (0.5 MHz), it's gonna work at 2 MHz, too - and much higher. You're not paying extra for this amplifier's bandwidth (re: my previous post). What you're paying for are the other qualities of the amplifier, such as its ability to handle a wide range of loads (including open and short circuit conditions), low distortion, lack of parasitic oscillation, excellent linearity, and so forth. These amplifiers are well-made and make a great addition to any laboratory.

One thing you don't want in an experimental setup are additional sources of error. Depending on how stringent your experimental controls are, you may wish to consider investing in good equipment that may cost a bit more. You get what you pay for, generally.

These amps are worth the money, IMO, but if your experimental setup is not particularly demanding you may opt for less expensive, used and/or reconditioned equipment. As emc_c pointed out, e-bay is a great place to start looking.

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#7
In reply to #6

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/08/2009 11:53 AM

Thanks both gentlemen's.

You absolutely right but I am not talking on a good laboratory equipment.

I need this driver to be install on 100 – 300 machines so I need to take care

On cost price.

I hope you have still the patient to answer but what do say on paralleling some

Small custom current source like Howland current pump or something similar?

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#8
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Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/08/2009 1:36 PM

I would contact ENI about an OEM conversion of one of their off-the-shelf stand-alone amps. They are set up to do this kind of work.

http://www.industrialpartner.com/rf-microwave-amplifiers/amplifiers/eni-electronic-navigation-industries/

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#9
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Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/08/2009 8:50 PM

Many thank? you are caught with chaff. He must be getting large commission from the company.

haha,

Do you know what concept this 200Mhz is? 3db band...

find a microwave oven for heating your tissues. only 50usa.

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#10
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Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/08/2009 9:24 PM

I have no relationship, professional or otherwise, with any amplifier manufacturer. I suggested ENI because

1) I was aware they had done some medical applications, and

2) I have had no dealings at all with them, and have had problems, or heard of problems, with AR and IFI.

Aside from the unfounded and snide remark, cnpower is correct that commercial microwave ovens are a cheap source of rf power, but I assumed that the OP wanted 0.5 - 2 MHz because they want deep penetration. Microwaves do not penetrate deeply.

And it would be difficult to modify a microwave oven, which is a reverberant chamber, to do what the OP wants, which is to irradiate a body that wouldn't fit within the cavity.

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#15
In reply to #10

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/09/2009 1:40 AM

My god, Do you take it seriously? havnt you had humor among your colleagues?

Maybe it different between our two world.

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#18
In reply to #15

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/09/2009 2:22 AM

You might be right...

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#11
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Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/08/2009 10:20 PM

cnpower asks: "Do you know what concept this 200Mhz is?"

Why are you asking him this when in an earlier post he queried: "Can you recommend on less expensive solution since I need up to 2Mhz?"

The OP doesn't need the full bandwidth of the AR amplifier and he makes it clear that he understands at least this much by the mere fact of asking the question.

" 3db band..."

The guy is not an RF engineer, cnpower, so why clutter what is unfamiliar territory to him with additional unfamiliar terms? Does he really need to know about decibels right now? Really? Would such a discussion contribute anything solving his immediate problem? When you go hiking with friends, do you put rocks in their backpacks or do you try to share the load?

C'mon, cnpower, try to help this guy out.

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#16
In reply to #11

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/09/2009 2:19 AM

Is the C'mon = come on ?

I dont know if we could help him. he gave out a vague condition just as the front so-called Dr. Mr. from asia.

What you quoted him so higher price equpment is unneccessary.as well as so higher specification. thats why I make a jolke.

he needs no bandwide, only but a sine wave, single wave amplifier. up to 2Mhz.

There is only 50 times gain. about 34dbv enough.

I dont know what he will do with it? what we heat uniformly some thing like fiber or material use infrared or mocirowave.

Im not teaser at opposition. I will not do worst thing to anyone. I will not lurk to add rock in a sack of vegetable to cheat customers.

if he is located in your usa, you and mccc can suggest him an oscillator amolifier enough. either A or B class amplifier. not too expensive.

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#17
In reply to #9

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/09/2009 2:20 AM

He does not want to cook a dog, he wants to warm a tissue...

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#21
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Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/09/2009 2:12 PM

Given your stated parameters, depending on your engineering background, may I suggest looking into Apex Microtechnology's hybrid amplifiers, www.apex.cirrus.com. Something on the order of the model PA09/A can provide the needed stated power over the range of resistance you quoted. This TO-3 amp chip can supply in excess of 1 amp current at 2 MHz. I am assuming that you have a source of sine-waves. An adjustable resistive feedback network around the PA09/A can provide the adjustable output if needed if the sine-wave oscillator does not have an adjustable output level itself. The PA09/A operates with up to +/-40 volt power buses and 2 amps peak. This should more than supply 20 watts at your lowest expected load, "tens of ohms". There are additional models which can supply more output current with similar specifications if needed. These should provide the needed cost limits at the quantities you've mentioned.

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#22
In reply to #21

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/09/2009 4:00 PM

That doesn't look right at all.

The output Voltage is limited to VDD-8 Volts; if you set the supply to the maximum allowed 40-Volts you will only see 32-Volts on the output, which corresponds to 10.24 Watts for a sine-wave signal at 50 Ohm load - and only 1-Watt at 500-Ohms.

A bridge arrangement would be capable of 41-Watts at 50-Ohm (I'd probably place a 20-Ohm power-resistor in series to control the load seen by the amplifiers), but still only 4-Watt at the 500-Ohm extreme. The other issue is that we don't know what the load looks like at high frequencies, so the 150MHz gain-bandwidth could prove embarrassing.
Also, from the viewpoint of load tolerance, it is usually preferable to run the circuit with fixed feedback, and use an attenuator before the input when you need to reduce the gain.

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#23
In reply to #22

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/09/2009 4:52 PM

"... so the 150MHz gain-bandwidth could prove embarrassing."

I was once frequently embarrassed by 150 MHz gain/bandwidth products, but ever since I've been seeing a counsellor I've been fine. I really like the meds, too, and I'd recommend them, but they're pretty frequency-selective and roll off at nearly 48 dB/octave. My doc calls them "elliptical filters" and they taste like crap, but hey!

How frequently do you experience this problem?

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#25
In reply to #23

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/09/2009 9:20 PM

wow, you make the problem even more complex.

why use elliptical filter? its not a band signal, I said it before. only a double T or rc or lc selective frequency amplifier enough can do the trick.

next time you may throw out a higher order butterworth, bessel filters etc.?

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#28
In reply to #25

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/09/2009 10:13 PM

"wow, you make the problem even more complex."

I'm a complex person. What of it?

Look, some people are afraid of spiders. Some people don't like heights. Some people are sociopaths, like my ex-wife. I just happen to be embarrassed by 150 MHz gain-bandwidth products, like Fyz. But I take meds for it. He doesn't. Just look at his avatar.

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#39
In reply to #28

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/12/2009 8:32 PM

No what, you can do as you like. you can be of a gabber, none interfere.

We get used to businessman's brag on TV advertisement.

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#24
In reply to #22

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/09/2009 9:08 PM

Electronic Wiz is quit right at it.

some of your calculation is right , but others are not right.

The output limited at certain supply voltage depends on the output power devices. some of them can be trick to trick.

less of them have so large votage drop, like you said vdd-8v. most of them have 2-3v eought. you may misunderstand or unfamiliar with amplifier. this denpends on many factors.

your calculation about oupt power is also side. you calculate only constant voltge output rather than constant reisstance. thats why you worked out a 1w.

Im curious why put out so complex issue of gain banwidth concept? it s really no use at this place. what he needs is only a socalled frequency selcting amplifier, such rc, or lc selective amplifier with power output. its simple and cheaper.

dont make simple problem complex.

what the top poster does is speaking more clear what he wish to do or not?

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#30
In reply to #24

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/12/2009 6:39 AM

I checked the output Voltage drop on the amplifier performance graphs. I find I was only very slightly pessimistic - for a 50-Ohm load you should expect a peak output Voltage of about 33-Volts and a peak current of 0.66-Amps - or a peak power of 21.8-Watts. As the objective is known to be heating, we should assume that the requirement is average power, which is half the peak, or 10.7-Watts. The typical slew-rate limit (for 2-MHz operation) is also very close to the limits (220V/us corresponds to a 35-Volt sinusoid at 2-MHz).

But at least we now know that the power requirement is only at 50-Ohm load.

OK?

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#40
In reply to #30

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/12/2009 9:19 PM

ok, ok, yoiur calculation now is quiet right.

But in engineering, we usually use effective vaue for heating effect. so that the 33v will be effecitve value and peak value will be 1.414x . the power is 21.8 as you listed.

the power supply we can select 50-55v enough.

we can even use less 50, say 45v even down to 36v etc, depend on distortion of the output wave. He didnt give ouit this rate so far. for most of heating procedure, this can be more less than 10%.

we can also use transformer to match this lower voltage for the power. I know you well know this.

the pic the poster gave out is out of my depth. I rarely know which power IC has so higher frequency response with so higher ouput.

no problem for audio amplifier, but that has only less than 50khz at very most.

hoever there are many ways for us to implement. Im sure you are well know them.

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#43
In reply to #40

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/13/2009 6:21 AM

Your previous objection was invalid - my modification was trivial. Plus I got the slew-rate limit on amplitude wrong by a factor of two.

Now to your latest comments: the effective value for heating is the average power (you really don't care that the heating rate is double the average twice in each cycle, or that it is zero twice in each cycle. And the average power is what I calculated - 10.9 Watts - varying between 21.8 Watts (the peak that you can get into 50 Ohms from 33-Volts) and zero (when the amplitude of the sine-wave is zero).

Then we come to your comment about setting the supply to +/-55-Volts: the maximum supplies allowed for the amplifier are +/-40 Volts (80 Volts rail-to-rail specified). Given typical component and equipment margins, you would probably (no certainty) get away with a 10% over-Voltage on the supply, but the most likely effect of > 25% over-Voltage is an explosion.

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#46
In reply to #43

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/13/2009 8:49 AM

wow, I supprise at your insisting on the average power.

Do you forget your basic physics?

Could you explain what the avergae value is and what is effective value?

Do you know what adjust work is? what is adjust angle for heating an object?

as well as modify amplitude?

What is distortion? what is over voltage? what is rate of a transistor or power mosfet etc.?

ah, forget it, I think I'd better get away from this topic.

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#48
In reply to #46

Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/13/2009 10:34 AM

The objective is to heat a sample. So the user is interested in the total amount of energy in a given time. Energy=∫(power)dt≈∫((33.cos(2.∏.ω.t))2/50)dtIf we integrate over a full number of cycles, this comes out at 332/100.t.
This is both the effective and the average power.
It is also the reason why electrical engineers usually use the RMS values for current and Voltage - because calculations allow these to convert directly to average (or mean) power.
And of course the RMS Voltage of a sine wave is 1/√2 times the peak value.
We see that the standard electrical engineers' methods lead to the same answer as the mathematical physicists'.

So, a basic aide memoire for you: RMS Voltage, RMS current, but mean or average power.

This is (or at least used to be) high-school stuff - probably repeated in most year1 electrical engineering courses (I originally read maths and later lectured physics - in neither case was it considered necessary to repeat this at university level).
Maybe you need to review your understanding by working through some year1 exercises?

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#49
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Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/13/2009 10:38 PM

in one word. faint!

I was still impressed by your cloud shadow calculation. I know you can use some math operation like sine, cos. etc.

ok, ok, I shall back to review the baic average value meaning in electric engineering in middle school.

and you can go on with your slew rate of op.

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#27
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Re: put out on resistive load 10W -20W sine wave @ 0.5MHz >>2MHz

01/09/2009 9:38 PM

We may be both right, since the 'load' we're talking about hasn't been too well defined in definite terms, we're dealing with generalities. The PA09/A was used as a possible solution, depending on HIS real needs, there are other models available which can put out much more voltage/current.

He did not say that he needed 20 watts into a 500 ohm load, nor for that matter precisely where the 10-20 watt range was actually needed. It is generally concluded that more power is needed at the lower load resistance than the higher resistance, but we do not know that for sure.

If the amplifier is stable at unity gain, internally, then the use of a variable feedback network is permissible. Your suggestion of a variable input level (potentiometer) into a fixed gain arrangement is also permissible. If the amplifier has a minimum gain of more than unity, your variable input may be the better choice. However, a variable network can also be designed with a minimum gain if needed. We really need to know the specifics of his sine-wave source and load.

You are also correct in that if the load is not mostly parasitic free, i.e. swamped by the resistance of the load, then some additional compensation in the feedback loop may be required. Just because an amp as a 150MHz gain-bandwidth, doesn't mean you can't limit the bandwidth with feedback components.

Outside of the possible AC characteristics of the load, I do not think that these amplifiers are particularly fussy about reactive loading. The data sheet for the PA09/A specifies just how much capacitance and inductance can be directly loaded on the output and most likely, I do not think this particular load is going to exceed those values.

For all we know, this gentleman may not be able to go this route and have to rely on some kind of commercially available solution or he may not be able to purchase Apex where he is.

Thank you for your input.

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#12

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/08/2009 11:05 PM

ok go to circuit-test get one of the wave kits and a transister and a mosfit and you are all set simple as that

bout 50 bucks canadian just have to figure out the values for the transistor and the mosfit

Richard

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#13
In reply to #12

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/08/2009 11:26 PM

Would you like to post a link or a URL? That would be handy, especially for the OP.

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#14
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Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/08/2009 11:37 PM

http://www.circuittest.com/English/Content/Items/CK307.asp

i have a supplier with 3 here already to ship if you need one fast

neutron electronics in guelph ontario canada

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#19

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/09/2009 7:15 AM

It isn't clear to me whether you require 10-Watts when the load is at the extreme values, or you require 20 Watts at 50-Ohm load and tolerance for 500-Ohms (which is what Europium's proposal would provide). If you need more than 10-Watts over the full impedance range, you will either need something like a 35-watt amplifier optimised for 160-Ohms or alternatively a 20-Watt amplifier plus a variable transformer between the amplifier and the load.

A further alternative, if it is not necessary to provide the required frequency as an input, would be to use a tunable power oscillator.

I can't give direct help with a commercial solution, but:

The most similar application I can think of is induction heating; they typically operate at much higher power, but use resonant loads to ease their task. A manufacturer of such equipment may be able to help.

Regarding the advice you have already received, it is correct that the cost of the proposed amplifiers is not for the bandwidth but for the combination of efficiency and robustness. If the use is intermittent, efficiency may not be so important. In addition, the robustness is much easier to achieve at lower bandwidths, and (given your volumes) it would be relatively simple to design an amplifier for the purpose*. The problem you will have with finding a suitable commercial amplifier is that there has not been an obvious large market.

*Again, it could be worth talking to specialists in other fields - these could include audio amplifiers, as the typical limit in the of 100-kHz is determined by the signals and the load (and not by the properties of the amplifier itself); also Amateur Radio power amplifiers, which start from 2-MHz

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#20

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/09/2009 9:42 AM

Aharon,

One small question that nobody has brought up. How do you intend to get the 1.25+/-0.75Mhz signal into the biological material? Do you intend to use needles stabbed into the flesh as electric probes with wires from the amplifier to the probes? This approach would only heat directly the flesh between the probes and by thermal conduction heat adjoining areas. Do you wish to use paddles that just touch the surface of the flesh to be heated? This will heat directly a larger section than the probes. Lastly do you wish the flesh to be placed inside an oven like box that requires no contact directly with the flesh to heat? This will heat all flesh inside the oven to a depth related to the frequency of the signal.

Each of these scenarios will change the impedance that the amplifier sees and will dramatically change the power requirements for the amplifier to impart the desired power to the target. Similarly each of these scenarios will require different topologies to impart a known amount of power.

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#26
In reply to #20

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/09/2009 9:23 PM

i will bet you that he is building a rife machine and wants this to heal his body ?

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#29
In reply to #20

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/10/2009 7:53 PM

Hello gentlemen's

Sorry, I miss a stormy discussion while I am on weekend vacation.

I will "edit " my needs while take into consideration the wise comments.

1) Input : adjustable (frequency & Amplitude) sine wave 0.5MHz up to 2MHz /

1Vpp up to 4Vpp Amplitude.

2) Output : 20 Watts at 50-Ohm load and tolerance for 500-Ohms.

3. The output (amplifier output and ground) will be connected to stainless steel

Electrodes That will touch the surface of the skin.(only touch non invasive).

If I want to summarize I need a controlled voltage to current converter with a power output Stage. I will control frequency & Amplitude of sine wave via software

And control the output power of the skin impedance (the skin is act as slightly capacitance load)

1. europium suggest a good solution but high price for my needs.

2. emc_c suggest contact ENI about an OEM conversion for lower price solution

3. Electronic Wiz suggest PA09/A power Amp from www.cirrus.com (apex).

4. Physicist? Comments regarding outpt voltage swing and suggest A bridge

Arrangement.(BTL amplifier) as well as talking to audio specialists.

5. redfred ask regarding probes arrangement (item 3 ).

On this stage my questions is as the following?

1. What topologies is the best?

2. PA09/A looks a good candidate (BTL arrangement) what do you think?

2. Can you send any starting point circuit?

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#31
In reply to #29

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/12/2009 7:14 AM

Other than in jest (Europium), this wasn't too much of a storm - at least by CR4 standards.

I've looked further at the PA09 recommended by Electronic Wiz. If you can tolerate a limit of 10-Watt sine-wave it would appear that you can use a single amplifier with a 33-pF compensation capacitor and the gain set (via feedback) toabout 8.5. Because the input capacitance is not negligible, it will be worth placing a 0.5-pF capacitor between the output and the negative input.

Regarding the possibility of using a bridge, I would need more information on what you are driving. A bridge will be fine if you are driving a load that can be balanced; however, if you are driving a large body and the second electrode serves as a pseudo-ground, we would consider whether dissipation in the bulk of the body is acceptable.

BTW, I looked through the other Cirrus/Apex amplifiers. I didn't find any that were any that were more suitable than the PA09A - i.e. capable of greater output Voltage slew rate with 1-Amp output current.

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#32
In reply to #31

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/12/2009 8:54 AM

Hi Physicist,

Thanks on response I really appreciate it!.

Please let me know how I can load /send a fast schematic to know if I understand what you mean.

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#33
In reply to #32

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/12/2009 9:07 AM

I usually convert my drawings to GIF and then use the "camera" in the message creation toolbar to load it into CR4. Experience suggests that multiple simple drawings work best.

Good luck

Fyz

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#34
In reply to #33

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/12/2009 9:16 AM

Thanks .

Attached Gif file.

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#35
In reply to #34

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/12/2009 10:23 AM

Starting with the output stage, the combination of the feedback network and the 6pF input capacitance of the negative input will create an extra pole in the loop at about 22-kHz. It probably doesn't matter, but I would still be inclined to provide a compensating zero using a capacitor of about 0.5pF in parallel with R51. (For breadboarding, a 5-mm section of 50-Ohm transmission line will do the job).

I haven't used the THS3001, but you really don't need that sort of bandwidth or slew rate here - and excess speed may mean you will need to be unnecessarily careful about layout. Basically, you are looking for a gain of 10 at 2-MHz, so a gain-bandwidth product of about 50-MHz would probably be sufficient (coupled with a linear slew-rate of at least 50 V/us). If you are going to use this amplifier, you need to keep the pole in the feedback network at quite a high frequency - greater than about 100-MHz. As I've no experience of this amplifier, I'm probably being overcautious; however, I would be tempted to reduce the feedback resistor to 1-kOhm (and of course the value of the digital potentiometer accordingly)

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#36

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/12/2009 12:28 PM

Warning (and apologies) - you need to reduce the feedback capacitor (your C1) to 6.2-pF in the PA09A, and even then you will only get the 10W you desire at frequencies below 700kHz.

The reasons are:
a) I missed a factor of two in my calculations of 2*; and
b) When I checked the graph** of "POWER RESPONSE", it became apparent that the data-sheet slew rate is not usable - it obviously requires that the input be overdriven, which is not helpful.

If you are happy to take a risk, there is a good chance that you could get the bandwidth you need by using an unconventional topology. The potential problems are that we don't know whether the output stage is capable (though I expect it will be) and that the amplifier designers would not have taken care of any implications. What you would need to do is reconnect the PA09 in an inverting configuration, and then connect an 82-pF capacitor in series with 150-Ohms between the output of THS3001 and pin7 of PA09. [The way this might work is that it drives the AC signal directly into the inverting output stage, and uses the rest of the opamp to provide fine control]

*I should stick to spread-sheets and never use a basic calculator, as the latter cannot be checked. I only noticed when because there was an inconsistency between my numbers for the PA09 and those for the THS3001.
**Page 3 of data sheet, right-hand column, one from top

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#37
In reply to #36

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/12/2009 1:39 PM

Hi Aharon and Physicist?:

I generally concur with the Physicist?'s statements and schematic, the PA09/A is about as good as it gets from Apex and it would not provide 20 watts at 2 MHz into a 50Ω load (takes a little digging into the data sheet for that bit from first appearances) but it will not be bothered by a 500Ω load either. This amp was designed to work with large values of capacitance, 0.1uF and inductance, 11mH so I do not think it will have a problem with your load. There are self-limiting features in the chip.

If you find that you do need 20 watts at 50Ω (and the 'trick' outlined above doesn't work), a possible solution would be to add an outboard current/power booster stage on the PA09 using either bipolars or power MOSFETs. This would complicate the design a little but by backing the power demands on the PA09/A down, you would allow it to reach the 2MHz output you need. Using the PA09/A to basically drive the booster stage instead of the load would solve some of your requirements and still keep it fairly simple. This would add three small power resistors and two transistors with appropriate supply bypassing (needed anyway) and of course, heat-sinking. Current limiting can be added to the output stage if necessary easily.

Take care to carefully read the PA09/A data sheet about using and mounting this amplifier, while it is not made of glass it does require a little more careful handling than a regular TO-3 type transistor case. There is more data at Apex on the use and application of this amplifier, definitely go there and check it out before proceeding to save yourself some trouble later on.

I am not familiar with the THS3001 amp either but it is definitely best to keep gain-bandwidth to a minimal figure as too much can easily cause instability from parasitics.

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#38
In reply to #37

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/12/2009 4:27 PM

Assuming that the slew rate limit is internal (see below), the only effect of adding followers buffers will be to increase the phase lag through the circuit.

The slew rate appears to correspond to a current in the input stage of about 9-mA. I.e 9mA driven into a 340pF feedback capacitor would give a slew rate of about 28-Volt/us; unfortunately the topology is a little odd, and it isn't clear whether the limit on the speed is capacitance from Q4_base to +VS, or it's the current in Q4. If it's the former, feedforward from th input source to pin7 would do what is necessary - but if it's shortage of current in Q4, we are well and truly stuffed.
(If we needed to do anything more complex with the PA09, it would be easier to design the ****** thing from scratch).

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#42
In reply to #37

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/13/2009 3:12 AM

Hi Electronic Wiz

1. This what you mean?

2. What is value of R61 R62 .

3. Q1 , Q2 should be very fast.

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#45
In reply to #42

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/13/2009 7:36 AM

I think it's what Electronic Wiz meant - but in this particular case* I don't believe it will help, as the load we are using is already much lighter than the loads for which the amplifier is designed (<0.7Amp vs >2-Amp).

So, even without the buffer we would already obtain any slew-rate benefit that could be provided by lighter loading. However, the most likely slew-limits are internal to the amplifier - i.e. internally limited currents driving device input currents.

*On the other hand, if I were to design an amplifier specifically for this purpose, I would use something rather similar for the output stage

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#41
In reply to #36

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/13/2009 2:58 AM

Hi Physicist

1. This what you mean?

2. Is PA119CE is a better solution? (900V/Us , 4A peak current ,100Mhz)

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#44
In reply to #41

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/13/2009 7:24 AM

Well done finding the PA119CE - that looks a much better solution.

It should not really be relevant now, but the circuit is mostly what I meant. However, as further examination of the internals of the circuit suggests that it only has a 30% chance of working I wouldn't bother.

I looked at the data sheet for the PA119CE. Definitely a better solution. No need for any clever intervention. (In addition, the power response graphs look like measurements, whereas the PA09 looked to be estimates).

I think I would use Cc=15pF to allow a little margin on the slew rate. Also, I would use the amplifier in an inverting configuration; the reason is that you will not be short of small-signal bandwidth, and this gives marginally more phase margin than the same gain in a non-inverting arrangement. (Even at 500-kHz, I think the peak output signal may also be slightly greater than for the PA09 - so you could probably use a slightly higher gain - though I would not exceed x-10).

I note that the data sheet has a 500 Ohm feedback resistor; as you are operating at fixed gain you can increase it. However, if you want to go beyond 1.8-kOhm it will be safest to place a 0.5-pF capacitor in parallel with the feedback resistor (note that the feedback resistor should be rated for 40-Volts - so a 1k8 resistor would need to be capable of dissipating almost a watt).

A note on safety: if there is any possibility of the user handling the plates when the circuit is powered, I would use a capacitor in series with the output - this will protect against most forms of internal failure. A good value (combining lowish sognal loss and adequate protection) would be about 33-nF. Ideally, it would be capable of withstanding mains Voltages, but if there is protection elsewhere, a 50V rating might be adequate.

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#47

Re: Resistive Load 10W - 20W and Sine Wave @ 0.5MHz >> 2MHz

01/13/2009 10:30 AM

Hello to Aharon and Physicist?:

Yes, the PA119 is a better fit, sorry I didn't have the time to drill down through Apex's listings further and find it earlier. I agree with Physicist?'s suggestions for the components. Just follow my earlier suggestions on using these power op amps.

No, that buffer isn't quite what I had in mind, a little bit different configuration, but that doesn't matter at this point.

Well done, both of you.

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