frequency counters

[mneary]

True. I missed the rms requirement. Hopefully the counter is more sensitive than its spec.

 

[Hero999]

In his first post, Patroclus says the sensitivity is 50mV rms. This means he needs a 100mV p-p square wave (AC coupled - DC doesn't count). I doubt the spec applies to any waveform other than 50% duty cycle.

Don't you mean 141mV p-p?

If you are going to use a series resistor, it needs to be applied at the source, to minimize reflections in the coaxial cable.

That's true, the cable if already matched to the input impedance of the counter and adding a resistor at the counter end would mess that up.

 

[patroclus]

I'll take advantage of the opportunity to get something clear.
When I studied microwaves, the teacher told us that at high frequencies, when the wavelength of the signal approaches the size of the components in the circuit, the reflexions have to be taken in account. So, impedances should be matched to achive the maximun power transfer to the load. Ok, this also applies in any circuit at any frequency, even in DC. Impedances should be equal. But, no reflexions happend if frequency is not high enough.

From that moment, I started to see a microwave circuits as a different kind of circuit than "standard" AC circuits, where reflexions don't happend (or are negligible). Now, relating this to what have been said, I suppose I need to use a 50 ohm coaxial cable for frequency counter 50 ohm input, and a 1Mohm oscilloscope probe for the high impedance input.

Now, as Roff talked about reducing the reflexions in coax cable, should I suppose that they do happend at 100 MHz - 300 MHz frequencies of an hipotetical CMOS bus / signal ?
If they do, then a impedance far away from 50 ohm would cause a terrible standing wave ratio,... maybe I'm confused. To be honest, I didn't like microwaves too much

 

[Roff]

Don't you mean 141mV p-p?

A 141mV p-p sine wave has an RMS value of 50mV. A 100mV p-p square wave (which I specified) has an RMS value of 50mV.

 

[Hero999]

A 30MHz square-wave will have harmonics extending to 300MHz and beyond.

For the transmission line to be critically damped (to remove reflections, resonance and standing waves) only one end needs to be terminated with the characteristic impedance.

This has got nothing to do with maximum power transfer theorem. If the load is matched to the source impedance, the efficiency will be 50% at the most. You can get CW transmitters that are much more efficient than this, especially at lower frequencies so their actual output impedance will be much lower than the load they're designed to drive.

The reason why standing waves can ruin a transmitter is because: If the class C output stage is left open circuit a large voltage can build up which will destroy the switching device. If the class C output stage is short circuited, a large current will flow which will also destroy the switching device. In practise a transmitter is designed to drive a certain load impedance and will be damaged if a significantly different impedance is connected.

 

[Roff]

I'll take advantage of the opportunity to get something clear.
When I studied microwaves, the teacher told us that at high frequencies, when the wavelength of the signal approaches the size of the components in the circuit, the reflexions have to be taken in account. So, impedances should be matched to achive the maximun power transfer to the load. Ok, this also applies in any circuit at any frequency, even in DC. Impedances should be equal. But, no reflexions happend if frequency is not high enough.

From that moment, I started to see a microwave circuits as a different kind of circuit than "standard" AC circuits, where reflexions don't happend (or are negligible). Now, relating this to what have been said, I suppose I need to use a 50 ohm coaxial cable for frequency counter 50 ohm input, and a 1Mohm oscilloscope probe for the high impedance input.

Now, as Roff talked about reducing the reflexions in coax cable, should I suppose that they do happend at 100 MHz - 300 MHz frequencies of an hipotetical CMOS bus / signal ?
If they do, then a impedance far away from 50 ohm would cause a terrible standing wave ratio,... maybe I'm confused. To be honest, I didn't like microwaves too much

OK, here's the deal. Waveform distortion due to reflections only occur if the source termination and the load termination do not each match the characteristic impedance (Z0) of the coax. Let's assume the coax has Z0=50 ohms. If the load is exactly 50 ohms, with no reactive (capacitive or inductive) component, then the load will completely absorb the incoming wave, and no reflection will occur, regardless of the source impedance. However, if the load is mismatched, then a portion of the signal reflects back to the source. If the source is exactly 50 ohms, it will absorb the reflection. So, even though there has been a reflection, there is no waveform distortion at the load (where it matters). However, if the source is also mismatched, a portion of the reflected wave will travel back to the load, distorting the waveform. For these reasons, a well-designed coaxial transmission system will attempt to match Z0 at both source and load.
Another point is that, in digital systems, you will not get overshoot and ringing in an unterminated transmission line so long as the round-trip delay time of the line is less than the risetime of the signal. For example, if your signal has a 10ns risetime, the line can be up to 5ns long. As a rough rule of thumb, the propagation velocity of coax is around 1.5ns/foot. If you use less than ~3ft (1m) of coax, you will not get overshoot and ringing.

EDIT: I was composing this as Hero was posting, so some of what I said duplicates his remarks.

 

[patroclus]

Thank you, I understand now.
So, I suppose the only way to go safe, and get the required voltage, would be using emitter follower or equivalent stage, with an output impedance of 50 ohm, right?

I suppose frequency counters count sine wave frequency, don't they?
maybe what I'm really looking for is a pulse counter or something like this

 

[Hero999]

The load on the circuit side doesn't have to have a 50hm: impedance because all the energy would be absorbed at the counter end.

Frequency counters just count where the waveform crosses 0V. It doesn't matter whether it's a sinewave, trianglewave or squarewave, a frequency counter will measure the fundermental with no problems.

 

[patroclus]

then it shouldn't work on TTL or CMOS, as they don't cross 0V ?

 

[Hero999]

The input of the frequency counter has a DC blocking capacitor which removes the DC component. Therefore a 5V clock waveform which only goes from 0 to 5V will be converted to a waveform that goes from -2.5V to 2.5V.

 

[patroclus]

Oh, yes, it's AC coupled, like you can do with an oscilloscope. Now I think I get the idea.

The load on the circuit side doesn't have to have a 50 impedance because all the energy would be absorbed at the counter end.

well, it won't load the source, just the emitter follower. The signal to be measured would be loaded by a high impedance input seen towards the transistor's base. So, what's wrong with it? you get a clear and strong signal at the counter.
Maybe I'm letting something out... maybe I forgot about the 50 ohm coax cable?

 

[Hero999]

There's nothing wrong it using an emitter follower, it's just not normally necessary, most of the time a 1k to 2k2 resistor will do.

 

[patroclus]

instead of a 2k2 resistor why not just directly couple it?
the source will see 50 ohm.
if a resistor is used, the source will see 2k2 + 50 ohm, for example.

if I understood correctly, coax cable is 50 ohm, and matches the counter input, there would be no reflexion. So, why use an extra resistor. Yes, the source will output a higher voltage because of the higher impedance (2k2+50 instead of 50), but again it will drop on the resistor's voltage divider.

 

[Hero999]

Because a TTL/CMOS output can't drive a 50hm: load.

 

[patroclus]

If a common value for TTL's IOH is 400 uA, driving a 50 ohm load would cause a mere 20 mV voltage drop.

If you use a 1950 ohm resistor in series, as you proposed, you'd get a 2k ohm equivalent load, and 0.8V drop (400uA * 2000). But the voltage divider 1950 + 50 puts no more than 20mV in the 50 ohm input.

 

[Roff]

If a common value for TTL's IOH is 400 uA, driving a 50 ohm load would cause a mere 20 mV voltage drop.

If you use a 1950 ohm resistor in series, as you proposed, you'd get a 2k ohm equivalent load, and 0.8V drop (400uA * 2000). But the voltage divider 1950 + 50 puts no more than 20mV in the 50 ohm input.

I explained this once already:

The output of a TTL gate does not look like a 12.5k resistor. I realize you inferred this from the 400ua output sourcing current spec, but this is the worst case, and is specified with the output at 2.8V. I tested a 74LS04 on my bench, and at 25MHz, it drove about 3V p-p into 560 ohms, and over 1V p-p into 47 ohms. So, if you put a 510 ohm resistor in series with your output (put it on the end of the coax that goes to your TTL gate, not on the freq meter end), you will still have marginally valid logic levels on a typical gate.

Remember that 400uA is at 2.8V, not 400uA at 0V, and it is worst case.

 

[Hero999]

If a common value for TTL's IOH is 400 uA, driving a 50 ohm load would cause a mere 20 mV voltage drop.

If you use a 1950 ohm resistor in series, as you proposed, you'd get a 2k ohm equivalent load, and 0.8V drop (400uA * 2000). But the voltage divider 1950 + 50 puts no more than 20mV in the 50 ohm input.

I just gave a range of values, you should adjust them to suit your meter, generally higher values are better than lower values as there's less chance they will interfere with the operation of your circuit.

 

[patroclus]

I might have not explain myself very well. Sorry.
I'm just trying to understand why it works that way, not how.
For example, Roff, you said that you were able to get 1 V p-p into 47 ohm resistor, then why not just use the 50 ohm coax cable directly coupled to the source ??
Thank you for being patient

 

[Hero999]

1V is not a valid logic level, if there are other gates connected to the output of the gate then they will not work correctly.

Also it will be drawing more current than the IC is designen for which could cause it to overheat and die or at least cause unreliable operation.

 

[patroclus]

Now it's loud and clear. thank you