input impedance of preamp

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neazoi

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What is the input impedance of the 40673 preamp bjt in this circuit??
**broken link removed**
I want to add another identical 40673 stage at the 50ohm output to amplify the signal.
 

I think it is not a bjt input preamp. The 40673 is a MOSFET, connected as a source follower. The input impedance is capacitive, some pF, and it won't gain voltage, just matches impedances.
 
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The device is a dual-gate RF MOSFET not a BJT. It has a very high input impedance so it does not load the feedback path in the oscillator. The 47 Ohm resistor being to give some isolation of the gate(s) capacitance to further reduce loading.

As Frankrose points out, it is a source follower so it has no voltage gain. Adding an identical second stage will not give more output but will increase the current it needs. It looks like the MOSFET is configured to pass (Vled - Vbe)/33 = 30mA which I would guess without a datasheet to hand, is close to it's limit, they are really designed for VHF small signal amplification.

Brian.
 
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I meant dual gate mosfet ton bjt, this was a typo.
I see, it is an impedance matcher with a "constant" current drive due to the led.
It is weird but it barely gets hot, it does not actually need a heatsink, even with this 30mA current.

I am confused a bit about the output resistors, why he has put a potential divider at the output, just to have a high impedance output and at the same time a low one?

Another thing that confuses me is that: If I want to connect a 50 ohm stepped attenuator at the output of this generator, do I still need to have the 50R shunt resistor at the output? What happens if no attenuation is switched in and what happens if at least a switch is flipped in, regarding this shunt resistor?
 
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The output stage will work fine but the output voltage and impedances are not well controlled. It would be better to create a low impedance point then use series resistors to each output individually. To match a cable, which is what I assume the 50R output is intended to do, there should be 50R in series with the source and another 50 R across the load end of the cable. It matches the cable and also halves the voltage as the output point sits at the tap of two 50R series resistors.

I would guess the source pin of the MOSFET is at around half supply voltage so it should dissipate about 6 x 0.03 = 180mW so it wouldn't run hot. I was more concerned that it's tiny junction area might be overloaded rather than overheated.

The voltage regulator would be far better changed to an active one. There is no real advantage to using the 'capacitance multiplier' circuit and it will be less accurate and temperature stable than an ordinary IC regulator.

Brian.
 
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So you suggest a potential divider made out of two 50R, connected directly on the source?
The author says that because of the constant current, the output impedance of the fet is very close to 50R, so perhaps that is why he connects the 560R first, because he assumes the source impedance to be low.

How about the stepped attenuator I mentioned, how should I connect it to the mosfet?
 


Hi Brian,
You mean it is better to match it like this?
 

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That's correct except that a stepped attenuator usually has a constant 50 Ohm input impedance and 50 Ohm output impedance so you don't need the fixed 50 Ohm resistor at it's input. Good stepped attenuators have at least double pole switches so the resistor on the input side and the one on the output side are selected simultaneously. Usually the one in their junction to ground is also switched to maintain constant impedance regardless of the selected drop.

Brian.
 
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My stepped attenuator has double pole switches. It is similar to this one https://blog.novaeletronica.com.br/img/atenuator-de-RF-variavel-e1400940858682.gif although for other dB ranges. My consideration about the input 50R prior to the attenuator is that, what if attenuation is switched to zero dB? Then there will be no 50R anywhere in the output of the amplifier, apart from the 50R load resistor at the coaxial cable end (actually a 50R device). Please advise me on that. Maybe to include a small 3dB fixed 50R pad instead of the series 50R resistor prior to the attenuator?

Also, what if I also want to have the high impedance output available, is it enough to put a higher value resistor (560R) in series with the Source and take the output from there?
 

Now I am confused.
I had a better look at the PDF of the original article, and the output network is like that **broken link removed** notice that the 50R impedance is at the top, and the bottom output is used as a port for a frequency counter. I thought that this was a mistake but the PCB that the author has designed agrees to the schematic.

But to my view the output resistor network at the source, is a L-pad and despite attenuating the signal, it matches it accurately to 50R at the bottom output.

Is that a mistake of the author of am I loosing something?
 

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