The only kind of transformer that has a hope of offering that kind of bandwidth is a transmission line transformer.
Impedance goes as the square of the step-ratio, so you are asking for step-up ratio of 4.47, which is difficult to realise in a transmission- line transformer - impossible in a low-order transformer
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If you can live with a more accessible impedance transformation ratio, a transmission line transformer could be perfectly practicable, and possibly even practical. Some off-the-shelf small pulse transformers are bifilar wound, which makes them transmission line transformers, but the characteristic impedance of a twisted pair is around 110R, which won't suit your set-up.
It should be easy enough to wind something with miniature coax on a biggish ring core or two.
If you want 20 ma at 100 volts, I don't think you generator will supply the power you need. If the generator supplies 10V into 50 ohms that is 20 ma. With a 1 to 10 voltage step up you generator would need to supply
As Mike hinted, the math doesn't compute here. 50ohms into 1k is not a
10x voltage step-up.
The transformer has to be of multi-filar winding technique in order to provide the required bandwidth. Not sure if two of these would work:
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One winding, both transformers parallel, for the input. Five plus five in series on the output side. Should result in 10:1. Three of them could give 15:1. I've never tried to use them for that but have used them in switch-mode converters and they behaved very nicely.
You could also check Mini-Ciruits. Rule of thumb: The cumulative inductance on each side should represent a Z of four times the presented impedance. It's ok if the "stated impeance" in the datasheet is lower, then the transformer just has to be listed for much lower frequencies than 200kHz.
If Paul's generator is too weak to drive loads under 10ohms then he'd need to find or build a staunch buffer.
I'll add to that, some signal generators have an amplifier with a very low output impedance and a series 50 ohm resistor. If you have this type and want to load it at 10 ohms, make sure the 50 ohm resistor is rated for the wattage needed, especially if you expect continuous use. I have seen blacked 50 ohm output resistors. I won't name any HP names. Mikek
Essentially you can do this using a PNP/NPN follower pair driven by an opamp, and then a resistor from bases to emitters for the crossover region. Just make sure the transistors are sufficiently SOA rated. One of those situation where the old American saying holds, if the transistors ain't big enough get bigger ones :-)
Much cheaper than trying to obtain some extinct buffer.
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Voltage (turns) ratio goes as the square root of the impedance ratio,
so instead of a step-up of 20, you'll only get a step-up of 4.47.
That means that if you want 100 volts out you'll need to input about
24 VAC.
I once had a Wavetek generator die on me when I connected a 50 ohm termination to its "50 ohm" output. Fried an output transistor.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net
Tell me about it.. I fried mine driving a B-E junction... same idea. How can something not be made to drive the rated load? Beats me. They used a weird output stage too, it has an op-amp around it for DC; the fast stuff is AC coupled, a peculiar complementary buffer.
I replaced the transistors with some TO-126 parts I had on hand, not nearly as much fT so it hardly does 5MHz from the output proper; I tapped a jack off a testpoint so I can still use a full-bandwidth signal, though it doesn't have attenuation or offset on it.
Tim
--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
I don't recall. I needed the generator, so I opened it up and put in a beefier transistor. It slowed down the edges a bit, but didn't blow up.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net
Why do you care about matching the load? The two normal reasons for wanting to match impedances are (1) to get maximum power transfer, and especially (2) to avoid standing waves in a transmission line, which give you weird variations with drive impedance vs frequency.
You won't have reflection issues at those frequencies, and if you build an amp, you can have any power level you need, so maximum power transfer isn't an issue. (Maximum power transfer is usually a red herring anyway--as Joerg never tires of pointing out, in an impedance-matched system, half the power is being dissipated in the transmitter, which usually isn't what you want.)
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net
In my Wavetek 23 it looks like the EPROM died partially. Those things become very uncomfortably hot on the inside. After finding that the EPROM was the problem I took the chicken exit and bought a Chinese 20MHz arb generator.
Believe it or not but I never owned an EPROM programmer.
Agree. And if you really wanted to source terminate for some reason and have a 10:1 step-up ratio just hang a resistor in series with the source. 10ohms minus its native output impedance.
There is nothing that beat drive power. Except for more drive power.
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