one transistor amplifier help

maybe

reduce R9, R10 to 39R reduce R8 to something like 680 - your aimimg to get the collector to 13V

Fundamental novice error... amplifier outputs should be LOW impedance for maximum power transfer.

What are you actually trying to accomplish? Input amplitude? Source impedance? Gain (X3)? Load impedance? ...Jim Thompson

Sorry Jim, I responded to the wrong address.

Drive a Diode detector with a Maximum amplitude of 5Vrms

Max amplitude 1.67Vrms into the amp (gain stage)

Not sure yet, I want to drive a diode detector, haven't figured out which one yet.

Yes, I have 5vrms source. 3 to 1 capacitive divider, just to limit swing on FET gate to 1.67VRMS. FET amp has a gain of 1X. Gain of 3X brings me back up to 5Vrms. It looks like diode detector is very linear between 3 Vrms and 5 Vrms input.

What ever I need to drive the diode detector.

I'm thinking I may need high input Resistance DC amp to drive the meter. I have only tested one so far and the meter pulls it away from the linear line.

Here's a graph of the detector input/output. >

Mikek

Drivel below that may help you understand what I'm doing.

Hey I now have two working detectors. One is here, >

and the other is this, from a Boonton Probe. It is high impedance and may need an amp to drive a meter. I need to check that. Ok checked it, when driving the meter it deviates further from a linear line. That is a 109K Load when driving the meter. The Boonton probe has better linearity on the low end, But it has BAT62 diodes where the upper circuit has 1n34A.

BOONTON PROBE 1/8W CF 1/4W CC Low noise double 100 ohm 330 ohm shielded cable 5% 10% +---| out - | | 2 1n5F | diode --- 600V? | 1n0F --- Cable End | Cer | S GND Probe | Cer | 1n0F --- | diode --- | | 1 +---|>|----/\/\/----#----/\/\/-----------> out + 5% 10% 100 ohm 330 ohm 1/8W CF 1/4W CC

A casc-ode amplifier might improve swing, but it has high output impedance and uses 2 transistors.

The grounded base section gives no current gain, so that has to be done by the common emitter - but the grounded base section has to be able to carry that current.

actually for maximum power transfer the amp output impedance should be equal to next stages input impedance. If the amplifier has a low impedance it results im maximum efficiency not power transfer

dl=0

So, I have a 50 ohm load, a +12V supply, and I build an amplifier that can do 12Vp-p open-circuit, and has a 50 ohm output impedance. Working into 50 ohms, it has a 6Vp-p output. For a sine-wave signal, that works out to 90mW.

Contrast that with the same load and p-p limits, but an amplifier with zero ohms output impedance. That one can put 12Vp-p into the load, or

You're saying that 90mW is bigger than 360mW?

See...

It's a common error with the school-book-only crowd >:-} ...Jim Thompson

I think the argument (problem) usually starts with some output impedance. (Physics types always start with a battery and the internal battery resistance.) And then asks for which load is there a maximum of power transferred from the device to the load. So if you had 50 ohm Z-out for which load do you get maximum power? It's a silly argument. (And sets me off, every time I hear it.)

George H.

There are times when impedance matching is important -- but for the most part impedance matching isn't the best thing to do. Even in a radio receiver, the best SNR is often achieved at some input impedance that does _not_ match the impedance of the line seen by the receiver.

I'm not sure about radio receivers. Somewhere in there I think the antenna is doing some sort of impedance matching.

Going out on a limb, I think you need impedance matching when you are interested in energy (power) and not just the signal (amplitude) and don't have any gain available.

So; microwave horns, trumpet horns, ultra sound transmitters/receivers, AR coating on optics... there's probably some mechanical examples...? George H.

Of course, the antenna impedance is involved. But, keep in mind, there are best impedance for best SNR and best impedance for best power transfer... and quite often they are NOT equal.

...Jim Thompson

Sure, gears and levers, the middle ear.

In the case of SNR optimization, it often happens that the noise and signal have different source impedances, and the best SNR match would then be off w.r.t. the maximum power match.

Maximum power matching is useful for antennas and various transducers, but it mostly doesn't matter after the first amplifier stage. Of course, inserting resistors to equal source and load impedances is just plain silly, although there may be other reasons to do so anyway.

Jeroen Belleman

For receiver only, the noise figure and SNR and impedance matching can get tricky. While the NF of the receiver is almost never where the impedance is matched to the transmission line. Then if the impedance is not matched closely you can loose more signal in the transmission line if there is very much loss in the transmission line.

Yup. That's what I was trying to say.

There are techniques to _make_ them equal, using transformers in the feedback path of your LNA, but AFAIK that measure doesn't improve SNR -- it just reduces the reflection back from the receiver.

Yes, (duh) and pulleys.. (my mind went to the capstan (for pulling up anchors)). What's the electrical equivalent of the ratchet and pawl? (I've got my answer.) George H.

Sample and hold? Mikek

TRACK and hold. ...Jim Thompson

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Thinking outside the box... producing elegant solutions.

I came up with 33 ohm resistors for R9 and R 10, and 549 ohms for R8. I got 20Vpp before distortion, however my gain is about 13. The value of R8 is critical, small change make big differences in the collector voltage.

Thanks for the input Mikek

Acoustic transducers, shock absorbers,...

Cheers

Phil Hobbs