# High Frequency amplifier

Discussion in 'The Projects Forum' started by Johnny1010, Jul 13, 2014.

1. ### Johnny1010 Thread Starter Member

Jul 13, 2014
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Hello everyone I am trying to built a discrete transistor amplifier using 2N3904 npn transistor. I am trying to achieve a 4Vpp output with 0.7Vpp (Sinusoid) input with a bandwidth of 120 MHz.

The amplifier that I have designed meets my requirement when simulated in Proteus but when implemented on PCB the gain starts to roll off after 1MHz.In my final design I used BFR93A (6 GHz UGB) still got upto 5MHz. Any help would be appreciated.

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3. ### Johnny1010 Thread Starter Member

Jul 13, 2014
89
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Thanks Bertus for your reply I will be looking into the document you gave.
I am quite new at pcb designing and used the proteus auto-routing. I have attached the pcb layout. Would be looking forward to more guidance.

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4. ### crutschow Expert

Mar 14, 2008
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What load are you adding to the output? Are you using a 10:1 oscilloscope probe?

Apr 5, 2008
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Hello,

When designing PCB's for RF, long traces are to be avoided.
Have a look at the trace from R7 to Q1.
I have changed the colors to make it more visible:

Bertus

• ###### johnny1010_PCB_layout_changed colors.PNG
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6. ### RichardO Well-Known Member

May 4, 2013
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8. ### Johnny1010 Thread Starter Member

Jul 13, 2014
89
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@crutschow yes I am using a 10x probe and a 100MHz bandwidth oscilloscope.

Last edited: Jul 13, 2014
9. ### Johnny1010 Thread Starter Member

Jul 13, 2014
89
2
Thanks Bertus,
I will be working on compacting the PCB as much as possible. I had used a simple single layer copper pcb. Do you think using a double sided one would make any difference. Also wanted to know if any other factor could be involved in ruining the gain and bandwidth?

10. ### RichardO Well-Known Member

May 4, 2013
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You should use a 2 layer PCB with one layer as ground. You only have to etch one layer if you allow yourself some jumpers.

Your scope and probe probably do not have wide enough bandwidth to accurately measure 120 MHz. Your 120 MHz signal will be down by more than 3 dB with a 100 MHz bandwidth scope.

The probe also has a bandwidth. What is the bandwidth of your probe? The bandwidth of your scope/probe system is less than that of either the scope or the probe:

$BW = 1/sqrt((1/BW_scope)^2 +(1/BW_probe)^2)$

For instance, a 100 MHz scope with a 100 MHz probe will only have a system bandwidth of about 70 MHz.

11. ### Johnny1010 Thread Starter Member

Jul 13, 2014
89
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@Richardo could you please elaborate what would be the significance of keeping one layer as ground.
Also I am using rigol rp1100 probe (250 MHz at 10x) and rigol ds1102e oscilloscope so using your formula I get a 92.8 MHz bandwidth which will be ok as far as I am concerned.
Instead of probes what if I use bnc connectors both at input and output wouldn't it be better?
http://www.icel-manaus.com.br/manual/RP1100-User%20Guide.pdf
http://www.rigolna.com/products/digital-oscilloscopes/ds1000e/ds1102e/

12. ### RichardO Well-Known Member

May 4, 2013
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The ground plane gives three advantages. The most obvious is that you have a large copper area to carry the power and to act as a solid ground reference for your signals. Second, it is used for high frequency power supply bypass capacitors. The final advantage is you can do controlled impedance traces such as microstrip.

To use the ground plane with a 2 layer PCB, I prefer to put surface mount components on one side of the board and use vias to connect to the ground on the other side. If you are making your own PCB then you have to drill a hole for each connection to ground and solder a wire to connect from the circuit side to the ground side.

92 MHz is marginal but you should be able to make things work if you allow for the errors in your measurements.

You _must_ use a coax cable from a signal source to your amplifier input. The characteristic impedance of the cable must match the output impedance of your source and the cable must be terminated with a resistor the same value as the impedance of the cable right at the input of your amplifier.

If you use a coax at the output, then the amplifier must have an output impedance matching the coax impedance and the coax must be terminated at the scope with a resistor that matches the cable.

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13. ### vk6zgo Active Member

Jul 21, 2012
677
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I think 5MHz, may be about the expected value for frequency response,considering the following:-

C1 (10nF) posesses inductance as well as capacitance.
At low frequencies,this is negligible,but at higher frequencies,it becomes more & more part of the circuit.

R1 posseses inductance as well as resistance.
Again this becomes a large part of the circuit as frequency increases..
Both components have stray capacitance to earth.

The junction of R6 & R8 is not bypassed,so U1 is part of the base circuit for both Q1 & Q2.
At HF & VHF U1 will not look like zero impedance to earth.

R5,the emitter load for the Common Collector stage Q2 will have series inductance & shunt stray capacitance,making it a complex load at HF & VHF.

C6 & C7 will also tend to look like Inductors as frequency increases,so that the base of the Common Base stage Q1 will no longer be at earth potential.

R7,the Collector load for Q1will have series inductance & shunt stray capacitance,making it a complex load at HF & VHF.

Wideband Amplifiers are not easy to design-----Common bandwidths for the cheaper Oscilloscopes were around 10-20MHz,well into the semiconductor era.
OK,'scopes are much more complex than your small circuit,but,again,if it was easy,Instrument manufacturers with their much greater resources would have loved to turn out cheap 100MHz 'scopes, back in the 1960s.

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14. ### Johnny1010 Thread Starter Member

Jul 13, 2014
89
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@vk6gzo I am a bit confused here because from what I have read and what other forum members have pointed out is that all these complex impedances being formed at high frequencies are a result of bad pcb design. So do you think that the amplifier design is ok and I need to concentrate on pcb design to get the most out of the circuit or is it the amplifier's configuration that's faulty.
Cheers.

15. ### Bhabrooo New Member

Jul 15, 2014
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@vk6gzo i'm working with johnny1010 on this project
you were right about C1 changing it's value even in simulation is giving better results but i fail to undestand that why C6 and C7 would fail to act as ground reference....?

Nevertheless following the suggestion of others we've deigned a "2 sided PCB with one side as a power(GND) plane"
here's the design screen shot and a separated layer PDF... as well as a 3D image of both the bottom and top sides...

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16. ### vk6zgo Active Member

Jul 21, 2012
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No,the PCB design does contribute,but these effects are an unavoidable characteristic of Electronic components & circuitry.

If you build it as a "spider web" with no PCB,you will reduce the stray capacitance,but not totally eliminate it.
Apart from any inductance due to the construction of the components,the leads on "through hole" devices look like small inductors,having substantial reactance at VHF.

Amplifier configuration?----Remember what I said about the lack of bypassing of U1?

Wideband amplifiers use various types of "peaking" circuits to try to correct for the drop in response.

17. ### vk6zgo Active Member

Jul 21, 2012
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C6 & C7 can only act as a ground reference if their impedance is low at the frequency of interest.
Fairly large value capacitors cease to look like pure capacitors as frequency increases.
Stray series inductance can increase the impedance of the "ground reference"

You will probably get better results with SMD components,& a redesigned PCB,but I am still very doubtful about 100 MHz.

18. ### Bhabrooo New Member

Jul 15, 2014
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so if the C6 and C7 develop impedance how're we supposed to bring it back (compensate for it)?... i mean will the impedance developed be capacitive or inductive in nature...
regarding components, they are SMD unlike last time..
Also i dont think johnny or i mentioned this before our PCB is smaller now.
It's 24.8mm by 22mm as compared to 36.6mm by 48.8mm
We took following steps to reduce stray capacities:
1-reduce trace size
Power: 0.5mm to 0.2mm
Signal: 0.76mm to 0.25mm
2-make smaller pad areas for the SMD components
3-use bottom copper power place (ground)
So after these improvements should we get better results ?
high freq. designs are tough !

19. ### RichardO Well-Known Member

May 4, 2013
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You never want to deviate from the standard SMD pad sizes -- either to small or too large. The pad size helps determine the amount of solder paste applied to the pad for reflow soldering.

I was involved in a project where the PCB designer thought it was a good idea to make the pads a little bit larger than usual. He assumed that it would make the assembly easier.

The PCB assembly house asked that the pads be made smaller because the resistors and capacitors were tombstoning. Tomstoning is where the part is on end. soldered to only one pad. In our case it was caused by too much surface tension from the excess solder on the pad.

20. ### ronv AAC Fanatic!

Nov 12, 2008
3,599
2,537
Try to make the board flow from input to output on the board. Start by placing the two transistors right next to each other and tying the 2 emitters together. Forget about the auto router and make all traces as short as possible.