Looking for 100 KHz - 10 MHz, 10W

Looking at the data sheet for the HP3314A, I notice that the normal output level is 10v p-p (250mw). With option

001 it goes to 30v p-p (2 watts). If you happen to have Option 001, you could probably get 10 watts out with just a broadband step up transformer, a simple emitter follower using a high voltage power xsister, and some ferrite beads.

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Jeff Liebermann     jeffl@cruzio.com 
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The Apex parts used to be real wire-bonded hybrids, plus a few monolithics done on a now-defunct (IBM?) high voltage cmos process. I think that they may be regular surface-mount nowadays, under all that epoxy.

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John Larkin                  Highland Technology Inc 
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Precision electronic instrumentation 
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This is what I was looking for -- I've been using my HF transceiver with the output power cranked down, but it won't go below 1.8 MHz for transmit. The matching networks are easy.

We'll try the LT1210!

Many thanks--

Unfortunately it's not the opt 001 model. Results with an emitter follower and beads weren't satisfactory. Ideally, the source plus amp fits in a small enclosure -- the DDS generator and the LT1210 should do the trick.

Thanks--

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Namaste--

Oh, And I was thinking the low end was easier. I'm a class A kinda guy, so just an opamp (or more for gain) into a transistor (with a heat sink)

To OP: There's a nice trick for driving non-resistive loads, you put the same R and other equivalent Z (L or C) in series, and that combo in parallel with the load, and it makes the load look resistive to the amp. (you should do the math at least once.)

What's the physics package? (or can't you tell?) I'm assuming the L is an air coil. What's L and R? (or if not that, what's the R/L time?)

George H.

Do your power calculations carefully, pay attention to overhead voltage when you set impedances (and when you do your power calculations), and when you think of the necessary heat sinking, think "humongous".

I can't remember the calculations for power dissipation from a class B stage (which is more or less what that thing is, or should be), but if you're getting 10W out you're probably going to be burning up at least

10W in the device. That means that you'll be getting at least a 50 degree rise from tab to junction in a TO220 case. That, in turn, means that (assuming a 125C junction temperature rating, which I'm too lazy to check for) you need to hold the tab temperature below 75C.

Consider the temperature around your heat sink, and size appropriately. And don't forget your heat sink goo.

--
Tim Wescott 
Control system and signal processing consulting 
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No. If he builds something in the style of a conventional 2-30MHz FET or bipolar RF amplifier, the matching transformers between stages and at the output rely on the ferrites to deliver enough inductance and coupling at the lower frequencies. It usually ends up looking like this: That's 1.6 to 54 MHz with 100 watts out.

At 17-30 MHz, there's probably enough wire in the transformers to produce usable inductance without any ferrites. Below about 8 MHz, it's either add much more wire, or add ferrites. The above amp probably works from 3 to 29 MHz (ham bands). To make it work down to

0.1 MHz, my guess(tm) is that two different ferrite materials will be needed and that the ferrites in the output section will be 4 times as large. The inductance requirements are less using FET devices, but still tend to be rather large.

There's also the problem of gain. If I double the bandwidth of the amplifier, the gain goes down by about 6 dBm. (There are resonant tricks to squeeze some more gain out at the high frequency end.) So, if I start with a 4 octave (2-30MHz) amplifier design, and want to extend it down to 0.1 Mhz or a total of 8.2 octaves, it will have about 12.6 dB less gain. 12.6 dB gain is fairly cheap at low power levels, but rather expensive at 10 watts and up. In this case, dropping the upper end from 30 Mhz to 10 MHz for 0.1 to 10 Mhz range results in about 7 octaves which only saves 3.6dB of gain.

I have no class, so that's not a problem for me. Class A is nice except for efficiency. If OP wants 10 watts of distortion and intermod free output, which is likely if he's feeding it with a function generator, he's likely to get only about 15% amplifier efficiency with class A. That means he has to burn 67 watts of heat in order to get 10 watts of clean output.

67 watts is going to require some efforts in heat disposal. For one time prototypes, I'm partial to a small aluminum (for heat transfer) can filled with mineral oil. The power amp gets immersed in the mineral oil. However, that's not enough for 67 watts, so the aluminum can is dumped into a much larger bucket of water. No need for gallons of mineral oil and the water can be cooled down with a garden hose.

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Jeff Liebermann     jeffl@cruzio.com 
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Hi Jeff, Yeah.. So like twenty years ago I did a 20 MHz pulse amp for an NMR system. It looked something like those pictures.. not at all easy to do. (Well at least it took me a while. I was learning RF at the same time.)

Hmm is it that bad? (the efficiency) I was thinking about just biasing the transistor at the mid current point. And then the sine wave turns it on or off.. but the average power loss still looks to be about 1/2. So ideally 50% efficiency.. but of course it will be a bit worse than that in practice. (I may be making some huge blunder here, but then (hopefully) I'll be corrected and learn something.)

Upon further reflection, (sorry for thinking out loud) I was picturing an opamp->transistor, voltage to current source for driving the inductor, since that's what I often use for driving coils. Hey, if the OP's load is a coil then he may want a current source. (no change in the B-field as the coil warms up.)

George H.

We may have been competitors. I did an NMR driver and power amp system using then new VMOS devices. My guess(tm) was about 1979. Getting ultra low distortion of -60dB 3rd order IM at rated power gave me headaches and nightmares.

It's always worse, never better.

The problem is linearity. If I setup the amplifier for best efficiency, I end up driving the output from rail to rail. That's fine for getting the maximum output from the device, but not so fine for getting the best linearity, lowest intermod, and minimum distortion. Generally, the output swing needs to stay away from the rails. A safe guess(tm) would be about 50% voltage swing, which is

1/4th of the maximum output power. For example, for 10 watts of clean output, I would need an amp capable of producing 40 watts. Of course, if one doesn't care about distortion, intermod, and such, a 10 watt amp would do just fine.

As a side note, the bandwidth should be at least 3 times the maximum usable frequency. If the function generator is expected to be usable to 10 MHz, the amp has to go up to at least 30 Mhz in order to amplify the 3rd harmonic necessary for something resembling a square wave.

I dunno about using the LT1210. From the data sheet: the maximum output swing is +/- 11.5V. Into a 50 ohm load, that's: (22.5v / 2 * 0.707)^2 / 50 = 0.4 watts rms So much for direct drive. 0.4 watts seems a bit less than the required 10 watts output power.

However, the LT1210 can drive a 10 ohm load with a 22.5v swing. Using a 4:1 impedance ratio output transformer, the maximum output would be: (22.5 / 2 * 0.707)^2 / 12.5 ohms = 1.6 watts rms which is still shy of 10 watts by a large amount.

In my never humble opinion, getting rid the transformers and ferrites is a necessity. To deliver 10 watts into 50 ohms, one needs: (10 * 50)^0.5 = 22.4 v rms * 2.828 = 63.3 volts p-p which means a +/- 50v power supply and output swing.

All I know about the load is that it's a "physics package" which is the common term for rubidium and cesium fountains used in frequency standards. I don't recall the excitation frequency (and am too lazy to Google for the numbers).

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Jeff Liebermann     jeffl@cruzio.com 
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As was pointed out elsewhere in this thread, see page 16 of the data sheet you reference above. There is the schematic of the amplifier requested.

Memory fault from 30 years ago. The 1979 version used bipolar devices, which were difficult to work with and difficult to produce low IM levels. The 1981 version used Siliconix VMOS-FETs, which was much easier.

Oops, arithmetic error.

With a rail to rail output voltage swing, the output power of a Class-A state will be the same its quiescent DC power dissipation resulting in the theoretical maximum efficiency of 50%. If I cut the output swing in half, the output power is reduced to 25% but the full DC power dissipation remains the same. The overall efficiency is now

12.5%. So, in order to get 10 watts of squeaky clean output, one needs to build an 80 watt amplifier.

To be fair, this is probably overkill for the OP's purpose. A reduction to 70% of rail to rail output voltage swing will result in

50% of the output power, 25% efficiency, and requiring only a 40 watt amp.

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Jeff Liebermann     jeffl@cruzio.com 
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Oops. That should be +/- 35 volt power supply.

Pg 6 thru Pg 8.

The schematic is almost exactly what the OP wants, except for one little problem. There wasn't enough output voltage swing necessary to produce the required 10 watts of RF output power. That's what I calculated above. Otherwise, it's a fine circuit for producing 400 milliwatts of RF maximum.

There's also the not so minor problem of the bandwidth, which seems to literally die at 10 MHz. If the OP is producing sine waves with his function generator, that will work fine. However, if he's doing ramps, square waves, pulses, and complex waveforms, he'll need at least 30 MHz of bandwidth.

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Jeff Liebermann     jeffl@cruzio.com 
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Hi, Jeff -

I don't know exactly what the problem is with your analysis and I really don't want to chase it down, but I can supply an LTSpice simulation that closely matches their spec sheet. It performs in LTSpice as advertised. If you want it, let me know.

My analysis is fine. However, my arithmetic is atrocious (as usual). Slight change in numbers below.

Sure I'm interested. I want to see if it can produce 10 watts of output with only a 23v P-P output swing. No need for LTSpice for that calculation. 23 v p-p / 2 * 0.707 = 8.2 v rms With a 50 ohm load (i.e. no xformer): Power = Volts^2 / Load = (8.2)^2 / 50 = 1.3 watts rms

However, the LT1210 can handle loads below 50 ohms, so with a transformer, more power can be delivered. The data sheet suggest a 10 ohm load, which might be convinced to deliver 6.6 watts rms with an impossible broadband power xformer or 5.3 watts, with a 4:1 impedance transformer.

Another line in the data sheet suggests that it can deliver 1.1 amps peak output current. Converting to rms, that's: 1.1 * 0.707 = 0.78 amps rms Power = E * I = 8.2v rms * 0.78 amps rms = 6.4 watts rms output. That's maximum voltage and maximum current simultaneously, which is generally considered a bad idea.

I'm sure the model closely matches the data sheet and applications schematic. However, that's not the problem. I'm questioning whether it meets the OP's vague requirements. Methinks not.

My abilities with LTSpice are similar to my arithmetic and spelling abilities, but I still would like to see the model. I could use the practice. My email in the signature below works. However, I won't have time to play with it for a few days as I just found myself with a very strange intermod problem on a local ham radio repeater and need to build and install an FM broadcast trap filter to eliminate the garbage.

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Jeff Liebermann     jeffl@cruzio.com 
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On 7/27/2013 4:18 PM, Jeff Liebermann wrote:

It is a bridged amp configuration. The 23V is _peak_. Do not divide by two. That means your power calculation is 1/4 the actual.

The transformer shown is configured as 2:3 ratio.

It does. Again, see page 16 of the data sheet.

Version 4 SHEET 1 880 680 WIRE -176 -224 -208 -224 WIRE -32 -224 -112 -224 WIRE -208 -208 -208 -224 WIRE -288 -160 -608 -160 WIRE -32 -128 -32 -224 WIRE -32 -128 -144 -128 WIRE 144 -128 -32 -128 WIRE 528 -128 384 -128 WIRE -288 -96 -320 -96 WIRE -32 -64 -32 -128 WIRE 144 -64 144 -128 WIRE 384 -64 384 -128 WIRE -608 0 -608 -160 WIRE -320 48 -320 -96 WIRE -32 48 -32 16 WIRE -32 48 -320 48 WIRE -800 64 -800 32 WIRE 528 64 528 -128 WIRE -32 80 -32 48 WIRE 384 80 384 16 WIRE -800 96 -800 64 WIRE -608 96 -608 80 WIRE 144 96 144 16 WIRE 528 160 528 144 WIRE -32 208 -32 160 WIRE -32 208 -320 208 WIRE -800 224 -800 176 WIRE -800 224 -880 224 WIRE -32 240 -32 208 WIRE 144 240 144 160 WIRE 384 240 384 160 WIRE -880 256 -880 224 WIRE -800 272 -800 224 WIRE 384 336 384 320 WIRE -320 352 -320 208 WIRE -288 352 -320 352 WIRE -32 384 -32 320 WIRE -32 384 -144 384 WIRE 144 384 144 320 WIRE 144 384 -32 384 WIRE -800 416 -800 352 WIRE -288 416 -320 416 WIRE -800 432 -800 416 WIRE -320 448 -320 416 WIRE -208 480 -208 464 WIRE -144 480 -208 480 WIRE -32 480 -32 384 WIRE -32 480 -80 480 FLAG -320 448 0 FLAG -608 96 0 FLAG 528 160 0 FLAG 384 336 0 FLAG -880 256 0 FLAG -800 64 +15 FLAG -256 304 +15 FLAG -800 416 -15 FLAG -256 464 -15 FLAG -256 -48 +15 FLAG -256 -208 -15 SYMBOL res -48 -80 R0 SYMATTR InstName R1 SYMATTR Value 680 SYMBOL res -48 64 R0 SYMATTR InstName R2 SYMATTR Value 220 SYMBOL res -48 224 R0 SYMATTR InstName R3 SYMATTR Value 910 SYMBOL voltage -608 -16 R0 WINDOW 123 24 124 Left 2 WINDOW 39 0 0 Left 2 SYMATTR Value2 AC 1 SYMATTR InstName V1 SYMATTR Value SINE(0 2.5 100k) SYMBOL cap 128 96 R0 SYMATTR InstName C1 SYMATTR Value 100n SYMBOL res 512 48 R0 SYMATTR InstName R4 SYMATTR Value 50 SYMBOL ind2 128 -80 R0 SYMATTR InstName L1 SYMATTR Value {L} SYMATTR Type ind SYMBOL ind2 128 224 R0 SYMATTR InstName L2 SYMATTR Value {L} SYMATTR Type ind SYMBOL ind2 368 224 R0 SYMATTR InstName L3 SYMATTR Value {L} SYMATTR Type ind SYMBOL ind2 368 64 R0 SYMATTR InstName L4 SYMATTR Value {L} SYMATTR Type ind SYMBOL ind2 368 -80 R0 SYMATTR InstName L5 SYMATTR Value {L} SYMATTR Type ind SYMBOL Opamps\\LT1210 -256 -128 M180 WINDOW 3 -125 69 Left 2 SYMATTR InstName U1 SYMBOL Opamps\\LT1210 -256 384 R0 WINDOW 3 -209 -1 Left 2 SYMATTR InstName U2 SYMBOL voltage -800 80 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V2 SYMATTR Value 15 SYMBOL voltage -800 256 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V3 SYMATTR Value 15 SYMBOL cap -112 -240 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName C2 SYMATTR Value 10n SYMBOL cap -80 464 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName C3 SYMATTR Value 10n TEXT -800 -224 Left 2 !;tran 0 100u 20u TEXT 616 0 Left 2 !k1 l1 l2 l3 l4 l5 .996 TEXT -800 -264 Left 2 !.ac dec 10000 10k 100e6 TEXT 656 104 Left 2 !.param L=33u

Oops(tm). My mistake. I was looking at a single amplifier section and missed the bridge amp which is further down the page. Pg 6 thru 8 and 13.

I was trying to eliminate the transformer.

The single LT1210 version does not. The bridged amp comes very close.

9 watts instead of 10. Your LTspice model shows the output -3dB down at 10MHz. No word from the OP as to what waveforms he's going to be feeding this amplifier but if it's a 10 MHz square wave, the amp will need much more bandwidth.

Looks like Coiltronics no longer makes the VERSA-PAC CTX-01-13033-X2 xformer. I couldn't find a data sheet. Plenty of possible substitutes, but I can't determine which one might work best: I'm worried about the comment "Frequency range to over 1MHz".

At full output (9 watts), the voltage swing across the 50 ohm load will be 21 V rms or 0.42 A rms. With a 2:3 turns ratio, that's 0.42 a rms * 1.5 = 0.63 a rms Peak current in the xformer primary is: 0.63 * 1.414 = .89 A peak Looking at the Mouser catalog page for the transfomer, anything with more than about 1A peak saturation current should work.

In order to have gain at 0.1 KHz, the transformer primary reactance must be high enough so that it doesn't reduce the load amplifier impedance very much. Generally, that means 4 to 10 times the amplifier output impedance (10 ohms) or about 50 ohms reactance. XL = 50 = 2*Pi*f*L = 2.828 * 100*10^3 * L L = 200 uH Since there are two windings in the primary, that's 100 uH per winding. Looking at the Mouser catalog page, none of the transformers qualify with about 100 uH winding inductance and >1A Isat.

Assuming I haven't screwed up the math again (an all too common occurrence), my guess(tm) is that LT simply ignored the saturation current, which only affect amplifier linearity on power peaks, and used whatever had an acceptable inductance. Even so, there's still nothing on the Mouser page that has enough inductance and that will handle the current.

Where did you find the 33uH winding inductance shown on the LTspice model?

Thanks for doing the model. I'll play with it for a day or two and see how it works.

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Jeff Liebermann     jeffl@cruzio.com 
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Skype: JeffLiebermann     AE6KS    831-336-2558

(...)

The closest approximation on the Mouser and Cooper pages is the VPH5-0155-R. That's 22.3 uH and Isat = 1.6 A. k = (1 - Leak_Induct / Inductance)^0.5 k = (1 - 0.235/22.3)^0.5 = 0.946

At 100 KHz, 2 * 22.3 uH = 28 ohms which should be good enough with only a slight drop in gain at 100KHz.

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Jeff Liebermann     jeffl@cruzio.com 
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Sorry, type. That should be k = 0.995

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Jeff Liebermann     jeffl@cruzio.com 
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Skype: JeffLiebermann     AE6KS    831-336-2558

Here's were I am now with the amplifier.

I tweaked the transformer to resemble a Coilcraft VPH5-0155-R and added the 6th winding.

I managed to expand the -3dB bandwidth to 50 KHz to 25 MHz. The way it works is that C4 resonates with the primary winding inductances to provide a parallel resonant peak at about 20 MHz.

After I do what I'm suppose to be doing today, instead of playing on the computer, I'll see if I figure out how to model the linearity, power handling, etc.

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Jeff Liebermann     jeffl@cruzio.com 
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Skype: JeffLiebermann     AE6KS    831-336-2558