mosfet RF amplifier with resistance

On a sunny day (Sat, 21 Apr 2007 11:14:15 +0200) it happened Matteo wrote in :

You must separate the AC characteristics from the DC settings. YOur idea about the impedances may be correct for RC (not checked), but at the same time you could use a different DC system by using decoupling:

Vb | | R1 [ ] Rd for DC | C3 |----------||----------------------

----| | | | |-- [ ]R2 === C1 < Z |-----| | | [ ] === | | | R3 | C2 | |

-----------------------------------------

Z sees R1 R2 and C1 parallel for high frequencies. But you are now free to select a value for R1 to match the Id max of the MOSFET, select a usable working point. C2 decouples for AC, and R3 sets Id. R1 could be a choke. C3 causes a low frequency cutoff.

If you wanted true DC coupling, then probably your only option would be to go differential.

The other issue is signal level. Very small AC signals in a low impedance are easier to realize.

If you wanted a voltage gain of 20 in say 5 ohm, S (ma/V) of the MOSFET need be very high. Even at 1A /V and 5 Ohm Zd the gain would only be 5!!!

Winfield will be able to tell you about FETS with low capacitance and high gain, but must it be a FET? At these low impedances know that transistors have a much higher 'voltage' gain......

So multistage differential transistor amp... Somebody is gone shoot at this......

Jan Panteltje ha scritto:

gain,

Thank you Jan! I'm not very skilled in FET amplifiers I missed to write that my target is to analise some design solution and to understand why they are not so good in order to understand why the differential amplifier is the best one for my purpouse.

I know that your circuit is better than mine about decoupling but I've to use mine :P

for example, I found that:

if I choose Rd = 625 ohm => Id = 2.6mA in order to have Av > 20 => Vov < 0.165 V => W = 412micron but Bw = 0.5Ghz instead of 1Ghz

if I choose Rd = 500 ohm => Id = 3.3mA in order to have Av > 20 => Vov < 0.165 V => W = 523micron but Bw = 0.53Ghz instead of 1Ghz

so I cannot reach my specifications!

Things that I don't understand very well:

- the replacement of the ideal current generator with a resistance makes the performances worse ..why?

- I supposed to work with R

On a sunny day (Sat, 21 Apr 2007 14:03:13 +0200) it happened Matteo wrote in :

Exactly.

We say: The MOSFET output is a current source. Ideal current sources have infinite high impedance. An infinite high impedance would give an infinite gain S x 00..... So if you put a resistor parallel to such a current source, you get the resistor value. Now the voltage gain is S . Rd, and this is a lower value.

Neither do I anymore.... There is plenty on the web.

But some remark, if you just want a RF amplifier at 1GHz, so not a _wide_band_ amplifier, then you could use a parallel tuned circuit in the drain, and in resonance that will have very impedance... and lots of real volatge gain (but a limited bandwidth).

Vb | --------------------------- |C4 | | ) === | R1 |C1 )L1 | [ ] Rd for DC === ) /// | C3 | ) |---------------------------------- out

----| | |-- |-----| [ ] === | R3 | C2

----------------------------------------- | /// C4 is for RF decoupling. Now The frequency is set by C1 L1, C1 inaprallel wit hthe MOSFET output capacitance etc. R1 sets the Q and with tha tthe bandwidth.

It is safe to assume the drain impedance is about R1, so gain is S . R1 R3 set Id for DC. Of cource you can add a second winding to L1, so make a transformer, to create any output impedance you want (impedance transformation is the square of the turns ratio. The transformed output load will appear parallel to R1 for AC. This is normally how _tuned_ RF amps are done, not with RC only. At 1GHz L1 could be a piece of stripline.

Good, I will remember that :-)

I did soome googling for 'transistor wide band amplifier diagram': In case you wanted a wide band _RF_ amplifier, here is a nice link:

A DC coupled differential 10GHz (note the inductors):

Just google a bit :-)

gain,

Hello Matteo,

For me a voltage gain of 20 up to 1 GHz with a single device seems very high. Did you check the specs of your active devices?

I agree with Jan that you need a device with very high gm to realize a voltage gain of 20. The High gm, may result in parasitic capacitance in a range that even with Rd = infinite, you cannot reach a voltage gain of 20.

I looked to your calculations. Your high voltage gain requirement will result in reasonable gate current (via Cgd). Are you sure that you can neglect the device internal gate resistance?

Your last steps, where you go to currents and resistor values, I cannot follow.

If your amplifier has to work from DC (with voltage gain of 20 from DC to 1 GHz), the balanced option is a good one. However, you cannot separate the AC design from de DC design. You also may add some inductance in series with the drain resistor(s) to improve the response at 1 GHz.

Best regards,

Wim PA3DJS

Your schematic shows a Cout of 250 pF plus the mosfet output capacitance. The impedance of 250 pF at 1 GHz is only about .6 ohms. To get a gain of

The 250 pF load will make it very difficult to get your required gain without some kind of impedance transformation.

I think that 20 GHz gain-bandwidth product is difficult, maybe impossible for a simple 1-stage mosfet amplifier like this. You just won't find a fet technology that will give you the required Gm without a huge amount of capacitance, at least in silicon. Every semiconductor technology has its limits.

Adding some inductances ("peaking") in various places can roughly double GBW, but that's not an option if this is an all-IC design.

A compound-semiconductor phemt might be able to pull this off. They have phenomenal transconductance and tiny capacitances. 20 GHz gbw is easy in a single-stage bipolar MMIC, in InGaAs or SiGe.

John

Matteo,

as many others mentioned , it seems that the requirements ( GBW = 20 GHz , Cload = 250 pF) are too hard for a single transistor.

you may look at something like a cascode plus a follower- buffer, or something like a Cherry-Hopper amplifier.

Jure Z.

Hi Jure,

I'm having trouble finding information on the Cherry-Hopper circuit configuration on the web. Some links require subscriptions, so they are of little value.

There is a schematic in Figure 2 of

(sorry for the wrap)

If you have time to look at it, is it the feedback from the MN3 source to the MN2 gate?

The rest of the circuit seems very conventional. But its not clear why that arrangement would be much more effective than other feedback configurations. There's a bit of explanation in the text - do you know of any other links that might explain it in more detail?

Regards,

Mike Monett

Never mind. I found it.

Feedback to the base instead of the emitter. Cherry, Hopper, 1963 paper.

And here I thought it was some new exotic configuration discovered by a computer running genetic software:) Regards,

Mike Monett

Hi Matteo and Mike ,

The trick here is to load a common source MOSFET with a TIA ( basically the two following devices ), providing a very low load impedance for the CS input device. In a cascode, the load of the first common source device is a Common Gate device , with potentially Higher input Z.

Here are some nice references (beyond the original Cherry-et-al paper that you have already found ):

there are many more papers available on the net, try google with Cherry, VCSEL , transimpedance, etc.

the old papers mention Hopper, newer ones refer to Hooper

Thanks , Jure Z.

Thanks for the help and nice links. Downloading them now.

I saw the reference to Hooper but thought it was a spelling mistake, since your post spelled it Hopper:

Negative feedback to the base is also useful in high gain amplifiers running on low voltage. One example is a simple 3 stage amplifier. The collector of one transistor is connected directly to the base of the next, and a simple pullup to VCC supplies current. The last stage supplies negative feedback through a large resistor to the base of the first stage. A series resistor and coupling capacitor at the input set the AC gain. The DC feedback is much higher due to the blocking capacitor so the circuit is quite stable with changes in beta, temperature and supply voltage.

These amplifiers can easily have bandwidths of a hundred MHz or so.

Regards,

Mike Monett

Here is the C-H simplified AC circuit with NMOS

^ | |-----------------| |-------VWV-----------|----------------| | | | |-- >-------VWV----| /// | Zm | | | |----------| | |---------| | | | |--

| /// | | | |--------|_____________________________________| IN ------| | |-- >--------| ///

Note: BTW, Jan Panteltje directly pointed to the same C-H concept earlier in this thread, referring to the Tao paper.

Thanks, Jure Z.

The diagram seems a bit garbled:)

I checked the source - it is the same. Regards,

Mike Monett

Sorry Mike , Google Groups ate my whitespace... I drew the three fets one after the other emphasizing the location of the feedback resistor Zm, from the Source of the last to the junction of the first Drain and the second Gate.

Thanks, Jure Z.

If you like, email it to me and I'll post it for you. Here is my address:

Mike Monett

gain,

Thank you all! I tried to come to a conclusion about the Rd limits

my gain is Av = gm . Rd = Vdd / Veff

in order to increase the gain if I decrease Veff I'll increase gm but I'll have to use a bigger W FET; moreover if I take Veff < 200mV my equations will not be true any more!

if I increase Rd I'll not be able to have Vds = Vdd/2

therefore I'd like to have a infinite Rd an ideal current generator

now I'll try to solve this problem using a cascode configuration

hope it's allright :-?

Negative feedback doesn't increase gain-bandwidth product.

John

completely agree !

Jure Z

Well, my question no longer applies since I found the feedback arrangement is very old, 1963, and well-known, even though it is often misspelled:)

You can apply feedback to the base of the input stage, which lowers the input impedance and requires a known and fixed source impedance, or you can apply it to the emitter, which raises the input impedance and requires a resistor in the emtter. Plus a lot of other details. But I don't understand the point of your comment. It is obvious, but what does it have to do with the feedback configuration? Regards,

Mike Monett