1MHz output impedance query?

[mole42uk]

Thank you Jeremy, that's interesting. I'm doing financial stuff at work today :-( so it will be tomorrow before I can do any work to follow up.....

[G0HZU_JMR]

The reason it is possible to roughly predict the power level is that the HC gate should produce a fairly accurate 5Vpkpk squarewave when unloaded. Hidden inside that 1MHz 5Vpkpk squarewave is a 1MHz sinewave of amplitude approx 6.3Vpkpk. This may seem odd but the other sinewaves that make up the harmonics all combine together to keep the overall waveform square at just 5Vpkpk.

If you model this as a voltage source producing a 6.3Vpkpk sinewave at 1MHz then it is effectively being fed to a 600R load made up of the HC gate ESR of 30R plus 270R plus the 300R on the input side of the matching circuit.

6.3Vpkpk into 600R calculates to a power level of (6.3^2)/(8*600) = 0.0082W.

Half of this is wasted in the 300R total of the 30R ESR of the HC gate plus the 270R resistor. This leaves 4.1mW delivered to the 300R seen at the input of the pi match. This gets efficiently matched into the 50R load.

If you were to look backwards with a network analyser into the pi match with the gate stuck at either logic 0 or logic 1 then it would look like a very good (purely resistive) 50R source impedance because the gate itself is very close to looking like an ultra low impedance source in both states. See the simulation plot below. This shows the source impedance at 1MHz is almost a perfect resistive 50R. However, this only applies at 1MHz.

This is why it would be better to have an attenuator after the matching section. However, this might not be what you want.

If you still want to try with the emitter follower then I think you may have to be mindful of the power consumption. If it drew 50mA from 12V then this is 600mW. Most of this will be converted into heat and I'm not sure it would be a good thing to have a slowly generated thermal gradient creeping its way towards your GPSDO if it is warming up in the same enclosure. It might cause issues with the short term frequency accuracy if the GPSDO has to keep correcting for a nearby heat source.

The HC gate version will probably only draw 4mA from 5V when delivering 4mW to the 50R load and this is much more efficient. The AC gate version might draw 8mA when delivering +10dBm but this is still quite a low power consumption.

[mole42uk]

An oddity - I've tried to duplicate your results here and notice that I don't know how to make the Macintosh iteration of LTspice report power - I have to plot voltage across the load and current through it seperately.

The image here is the result - I have altered the LC components to off-the-shelf values. I notice that the output swings positive and negative around the ground rail - is this permitted and is it desirable? Should I decouple the output? That's what I was doing earlier with that impossibly large 100µF capacitor but smaller ones seemed to have a damaging effect on the output waveform.

[mole42uk]

Now I'm looking at this, which looks to me as though I have 42mW into the 50Ω:

Is that 336mA fom the 12V supply? I probably need to reduce that a bit?

[G0HZU_JMR]

The image here is the result - I have altered the LC components to off-the-shelf values. I notice that the output swings positive and negative around the ground rail - is this permitted and is it desirable? Should I decouple the output? That's what I was doing earlier with that impossibly large 100µF capacitor but smaller ones seemed to have a damaging effect on the output waveform.

It is normal (and desirable) to see the output swing either side of ground because the output is AC coupled via the 33nF cap. This is the best way to produce the sine wave because it means there is no DC offset riding along with the 1MHz sinewave. This DC offset could upset the piece of equipment you connect it up to so it is quite normal to remove any DC offset using a blocking capacitor like the 33nF cap.

To measure power you can hover the mouse cursor over the 50R load resistor and then press and hold the ALT key. This should show a thermometer. Then press the left mouse button to plot the power.

If you want the text box that shows the average power across the display, then hover the mouse over the relevant power trace logo at the top of the graph (a little pointy hand will appear) and then press and hold the CTRL button and then press the left mouse button again. The little text box should appear.

The power trace logo might say something like V(N005)*I(R1) at the top centre of the graph if this is the only thing being plotted.

[mole42uk]

Excellent! Thank you.

I've also done some more work with the emitter follower so now I have a better output which spice thinks is about 10mW into the 50Ω:

[G0HZU_JMR]

Ideally, the emitter follower should be able to produce an undistorted sinewave into lower impedances than 50R at 1MHz. Also, the source impedance of the BJT follower will only be a couple of ohms or so when biased at maybe 25-30mA. This means that if you want the source impedance to be close to 50R there has to be a series 47R resistor at the output and this is R4 in the diagram below.

This means a regular 50R termination at the output will mean that the emitter will see the series total of its source impedance plus the added 47R plus the 50R load. This means it is looking into 100R.

To maintain (follow) a sinewave and produce 2Vpkpk into the output load this means two things. The minimum collector current for the BJT will have to be 2V/100R = 20mA for a 50R load at the output. Aldo the voltage swing requirement at the emitter doubles from 2Vpkpk to 4Vpkpk.

However, the emitter resistor R2 also loads the emitter and this means that the collector current would have to be even higher with a typical value for R2. The loading effect of R2 can be offset a bit by putting a choke in series with R2 as in the diagram below. The choke needs to be able to pass the collector current without any significant saturation effects in the choke. If the choke does its job then R2 becomes invisible and it doesn't load the emitter at 1MHz.

I'd still recommend running the collector current a bit higher than the theoretical minimum 20mA because the load impedance from your external test equipment might be lower than 50R at 1MHz.

In the example circuit below I've biased the BJT at about 33mA to produce a very good 2Vpkpk sinewave into a 50R load and it will also cope with a load impedance as low as 10R without too much distortion creeping in. The transistor would ideally be a metal can type and may need a heatsink if you start exploring highish collector currents like this.
The simulation below is just for the emitter follower and I've fed it with a sinewave at 1MHz. This assumes there is a pi filter ahead of this circuit that converts the square wave into a sine wave.

[mole42uk]

How does the choke unload the emitter? I've tried a model in spice but I don't get any notion of how it should work, and I can't make much difference in the collector current - I'm still getting a peak at 23mA with a 2.2V pk into the 100Ω load and biasing so that I get a cleanest sine wave.

[G0HZU_JMR]

Ideally the choke should prevent any AC current at 1MHz being wasted in the emitter resistor R2.

For example, R2 only needs to be 250R to see a 20mA collector current

With the choke included the emitter sees only the 47R + the 50R load in series = 97R. To put 2V peak into 97R requires that the BJT needs to be biased at at least 2V/97R = approx 20mA.

Without the choke the emitter sees the same 97R but it also sees the 250R of R2 in parallel with this. This means that the minimum current requirement to maintain a clean sinewave is 2V/70R = 28.5mA

Try changing R2 to 250R and repeat the simulation with the choke fitted and without the choke and aim for 2Vpkpk at the 50R load.

[mole42uk]

Right now I'm stuck. The attached circuit and chart seems to be the best I can do but I seem to have no voltage gain from the transistor - have I traded it all for impedance matching?
On the chart V(N008) is the base, V(N010) is the output and Ic(Q1) is, of course, the transistor collector current.
Right now I am preparing to build the thing on breadboard and put some real-time measurements on it.

[G0HZU_JMR]

If it helps I can take you through a few calculations that should deliver about 2Vpkpk at the 50R load resistor. First of all we know that if we do an FFT on a 5Vpkpk 1MHz square wave signal there is actually a 6.3Vpkpk sine wave at 1MHz contributing to the 1MHz square wave. We also know that the source ESR of a typical HC gate is about 35R.

We also know we have to add a wasteful series resistor here to limit the current taken from the gate. A typical value is 270R. This means the source impedance will be 35R + 270R = 305R.

An emitter follower follows voltage and so it doesn't provide voltage gain but it does provide current gain and this is exploited in the schematic below. We also know that the follower needs a series 47R resistor at the emitter to define a 50R source impedance and this is also wasteful. This means that to produce a 2Vpkpk sine wave at the 50R load we need to feed the base of the emitter follower with twice this voltage at 4Vpkpk.

So this helps define the design of the pi filter that is there to remove the harmonics and leave a sine wave. It needs to produce 4Vpkpk at its output if correctly terminated. If the 305R source is also correctly terminated (by the pi filter) then the 6.3Vpkpk at the logic gate will be halved to 3.15Vpkpk at the input to the pi filter. The pi filter therefore needs to produce some voltage gain to get 4Vpkpk from 3.15Vpkpk.

The termination impedance of the pi filter therefore needs to be higher and it needs to be approx 305 x ((4/3.15)^2 ) = 305 x 1.61 = 491R to get 4Vpkpk at its termination.

A pi network that matches between 305R and 491R needs to be designed and it should have sufficient Q to filter away the harmonic energy and deliver a nice sine wave.

I've included the results of all this in the schematic below. R7 provides the termination for the pi filter and R2 and C2 provide some noise cleanup from the 12V power supply in case the real version has any noise or ripple on it.

The transistor will get quite hot and may benefit from a heatsink or maybe a chunkier transistor. However, the circuit below and the equations above may help you home in on a circuit worth building. I've not used regular component values below but this is to show that you should get close to the expected 2Vpkpk at the output with the circuit below.

[G0HZU_JMR]

Note! In your first go at a spice simulation it might be worth it to reduce the value of C2 to 100nF or your circuit will take a long time to settle before you see the expected output.
Try setting C2 to 100nF and do a transient analysis for about 5ms.

In the simulation time settings of LTSpice you can set a time to start "recording", saves all the startup showing on the plots.

[mole42uk]

In the simulations I've done so far, the circuit settles after 0.6ms. I'm quite keen to see the start-up so check that I don't exceed any static values. I don't have an in-depth understanding of LTspice at all, so I'm leaving that to directives that seem to work for me.

I really want this to be finished, it's a real challenge but I do wonder if I'm trying to achive an impossibility - a sine-wave 1MHz nominally 50Ω output that is happy at almost any load that I throw at it. The output that I was originally designing for - a 1MHz into 50Ω reference to lock the Marconi 2955 onto seems to have been easy by comparison.

[G0HZU_JMR]

What can be a bit unrealistic is expecting it to keep delivering 2Vpkpk into any load. An extreme case might be a 5R load.

In this case the emitter follower would need to be biased at 2V/(50R + 5R) = 36mA to maintain a clean sinewave but the amplitude of the sine wave will fall a lot.
The circuit I recently posted up is biased at about 31mA so the sinewave would be just starting to show some distortion with a 5R load.

The voltage output will also fall because the 5R load forms a potential divider with the 50R source impedance of the emitter follower.

So instead of seeing 2Vpkpk at the 5R load you will see the 4Vpkpk at the emitter tapped down to just 4 * (5/(50+5)) = 0.364Vpkpk across the 5R load.

This is perfectly normal behaviour for a 50 ohm source impedance. A commercial 50R sig gen would do the same.

[mole42uk]

I've been breadboarding.

Using a 1MHz square-wave signal generator with a 50Ω output, I'm using the LC values as shown and find that the sine wave output peaks at about 850KHz. Obviously I want to adjust the LC network to bring the maximum output nearer 1MHz, I wonder if anyone knows the magic dot command to allow me to sweep the network in LTspice to view the output and get a closer approximation?

[Cruisin Marine]

Originally Posted by mole42uk

I've been breadboarding.

Using a 1MHz square-wave signal generator with a 50Ω output, I'm using the LC values as shown and find that the sine wave output peaks at about 850KHz. Obviously I want to adjust the LC network to bring the maximum output nearer 1MHz, I wonder if anyone knows the magic dot command to allow me to sweep the network in LTspice to view the output and get a closer approximation?

Without checking- this is all from memory:
You may need to understand the different types of filters.
You design the filter using tables for the type of filter (Chebyshev, Butterworth etc) and you use the tables which are normalised values you have to scale from 1 radian per second, i.e. 0.159Hz and an impedance of 1Ω. to suit your needs.

You design the LPF to have the flattest (ripple) response that is suitable for your design up to the cutoff frequency, if using a Butterworth, or if a desired bump at the "fundamental" you would use a Chebyshev.
The Butterworth response falls off more slowly than a Chebyshev, so has less harmonic rejection

The main factor though is usually making sure that your filter has the right spec. to attain the required harmonic rejection.
Remember, a flatter/Butterworth has less harmonic rejection than a Chebyshev.
Again, you would need to look up these figures in tables to see what they are.

In a quick answer (unverified cos I have no idea how you designed your filters) you need to use simple scaling to change from 0.850 to 1.000, I'll let you work that out, should be easy enough.

For a sweep in LTSpice, go to the top menu selection "simulate" then "edit simulation cmd" and click on the "AC Analysis" tab, there you are able to select the range etc. that you want. The source for this is a "voltage source" right click on it and put (I use 1V) into the AC Analysis slot. That generates the "dot" command for you. Run and you have a sweep.

LTSpice, days one to seven, baffled, weeks two to four enjoyment, weeks four to ... time disappears. Time to build it!

[mole42uk]

I’m using LTspice on a Macintosh. We don’t have the menu selection, there are various menus that can be invoked but I’m still in the baffled section of the training course.....

[mole42uk]

A long time ago I asked a question about providing a sinewave 1MHz output from my GPSDO. Today I have a spice simulation that seems to fit the bill, I have attached the screenshot below. A breadboard will follow to verify the simulation: