Driving a Discrete Input Having a 50ohm Impedance

What logic family is that, "Wasteful" ?:-)

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |

It's actaully an input signal to a piece of equipment (1PPS GPS). Probably more a case of making a robust signal for going around a vehicle rather than a logic family per se.

Don

You're just full of buzzwordese today and that's a "per se." You're reading the specs wrong- there are no such drive requirements. Just because they have a BNC or some other coaxial input jack does not mean it's 50 ohms.

It's going to take over a watt to drive it with the logic level high (v^2 / r), and the frequencies are RF (7 mhz equivelent or so) but on the bright side it doesn't need to be linear.

I'd probably lean towards a half bridge design with two active devices... mosfet's might be tempting, basically just make a CMOS output on steroids?

If it's a high speed signal, then it *does* mean it's 50 ohms if the instrument is well designed... you want to match the characteristic impedance of the transmission line.

Hmm, question, how can scope inputs be high impedance and not cause reflections on the probe coax? It would seem like one end of the coax or the other would need 50 ohms of fairly resistive termination to ground, no?

well, not necessarily 50 ohms, could be 75 or 93 or whatever, but a lot less than a megaohm

the input:

INPUT IMPEDANCE: 50 OHMS +/-5% LOGIC 1: 8 to 11 VDC LOGIC 0: -1 to 2 VDC RISE TIME: < 50nsec FALL TIME: < 1usec

I guess I don't understand what you mean about "buzzwordese". If you know and work with the Global Positioning System (GPS), you probably know what 1PPS means (1 pulse per second - a pretty standard signal in GPS equipment).

For grins, I measured the resistance from the signal pin to ground, and guess what? 50 ohms. (Yeah, I know there is other stuff in there, but it does support the spec.) So. what I described is accurate and I did not mis-read anything.

I had to do one recently for a GPS emulator that needed 0 and 4.5v and wound up using a discrete transistor driver.

Norm

Don, If I understand your proposed circuit correctly, you are proposing shorting the "low" side resistor to ground in the "0" state. If so, the output impedance in the "1" state would be 25 Ohms, and in the "0" state, the output impedance would be "0" Ohms (neglecting Ron). I would suggest a discrete complementary (NMOS-PMOS) driver with a 50 Ohm back termination resistor. This way, the output impedance would be approximately 50 Ohms in both states. It would be a power waster, as Jim implies, but it should work. In the "1" state, the resistor would dissipate (20-10)^2/50 = 2 Watts. In the "0" state, the resistor dissipation would be 0. Regards, Jon

What do you mean by a 'back termination resistor'?

A resistor between the switched node and ground?

If so this does not set the output impedance, or even the termination impedance except when neither FET if on. Mostly what it does is waste power.

A resistor in series between the switched node and the driven cable?

This would set the output (source) impedance. However, it also functions as a voltage divider with the cable (initially) or load impedance (long term), meaning he'll need to switch 20 volts on the PMOS to make 10v show up across the receiver's 50 ohm impedance.

Unless you really need the 'softness' of a 50 ohm source impedance, or the receiver is high impedance, this just makes life complicated for little benefit.

Given the input specifications and measurement he posted, I wouldn't use a resistor on the order of 50 ohms at all. I guess you could consider one an order of magnitude smaller in series or an order of magnitude larger in shunt, but not closer. And I might use a fuse or some more sophisticated overload protection device...

Hi Don.

Your load looks like a 50 ohms resistor to ground. The 50 ohm coax has the property of "transporting" that 50 ohms to ground right into your driver circuit. So your job is to pull a 50 ohm resistor up to say 9 or 10 volts. You don't have to pull it down - the 50 ohm resistor can do that all by itself. In other words, you need a switch.

If you have a 10 V or 12 V or similar power rail, you can pull the 50 ohm load up to it with a switch - a transistor which turns on. In your circuit you have a logic output of some kind - say a 0 to 5 V logic level. You need something like this :

View in fixed pitch font

------------------ +10V, 12V etc | | R1 R2 | | | |/e |------| PNP R3 |\c----------out | |/c LogicIn --R4---| NPN |\e | ground

In fact, the spec you quote looks tailored to such a driver - fast on, not so fast off. Resistor and transistor values are going to depend upon your logic voltage and power supply voltage.

R2 can drop any unwanted voltage if your supply voltage is too high. R1, R2 and R3 can be chosen to stop the PNP saturating, which speeds up turn off. This kind of circuit can easily meet your spec.

Roger Lascelles

No- the high impedance probe cable is lossy and ~350 ohm eliminates reflections.

Okay- well if it was ME doing this I would find the minimum acceptable pulsewidth worst case- not necessarily 50% on the 1PPS, it is a trigger after all, and then go with cheap parallel logic buffer drive into pulse transformer and onto the line of that pulsewidth. But if you want to play dumb, then you can go with something like this or similar, use 50 ohm series termination:

Don, A back termination resistor is just a resistor in series with the source signal. It is used to provide impedance matching with the characteristic impedance of a cable. This minimizes reflections and ringing. ~ I was assuming that you were concerned with ringing at the load input, and would, therefore be using cable with a 50 Ohm characteristic impedance. If this is not the case, then there is no need to provide the impedance matching provided by the 50 Ohm resistor. ~ By the way, it is very common in video systems (which commonly use 75 Ohm cable) to use video drivers with a voltage gain of 2. This compensates for the gain of 1/2 that results from the 75 Ohm back termination resistor that is commonly used. Regards, Jon

Yes, but it wastes half the power and is not the only way to solve the problem.

If you match the line impedance at the receiving end, source termination is unecessary as there won't be a return wavefront arriving at the source end to reflect off of a lower source impedance. Terminate one end, and you are limited to at most a single-round-trip reflection, which may not even be in a direction of concern.

If the signalling level is only a volt or two, this is easy to do, and lets you kill off rather than send back any reflections returning from bad cables or connectors or mismatched loads (or multiple 75 ohm loads), not to mention providing short circuit protection. But at 10 volts desired across the load, it's a substantial penalty to up the driver to 20 v - now you have to handle 4 watts instead of 2, and have a more limited choice of parts capble of that voltage range, especially if you are looking for a PMOS device.

Also, even with just that load resistor I'd look at the duty cycle requirements - would be nice if you could use just a brief 10v pulse and then return it low for the rest of the second, that could get the average power much more reasonable.

Look at the TC4421 and TC4422 gate driver chips, and try to limit the 'high' time to a microsecond or less, so that the duty cycle is very small, and therefore you will not need to waste a lot of power.

As long as the load is 50 Ohms, which you say it is, then you don't need a resistor at the driving end, only at the load end. Those who say otherwise are mistaken. You can test this all quite nicely if you have a fast scope and a few tens of metres of 50 Ohm coax cable, this length being desirable so that you can clearly discern the reflections or absence thereof in the waveforms. You need to have a low capacitance scope probe or you need to use the scope in 50 Ohm mode in place of the load termination for testing. Be careful not to put too much DC for too long a time into a 50 Ohms scope input as this will fry it.

Chris

This one might be too fast, but it's a cheap way of driving that 1PPS trigger pulse train from something like a PIC. I am assuming that up to

. . . . . +5V . | . +----+--[10K]--|>|---+----------------------+---||---|---------+------------+--| o-+----+->

. | | 1N6263 | | | | |/. | | | . | | | | | | . | | | . | ----74AC241 | | | | |\ | | | . | | | | | | +--| o-+ | | . | | |\ | | [560] | |/ | | . | | +--| >-+ | | | | | | . | | | |/ | | | | ---------- | . | | | . | | | | | | . | | | . | | | | | | . | | | |\ | | | | | | . IN>--------+--| >-+------+---------+------------+---|>|-----' . | | | |/. | | 1N5817 . | | | . | | . | | | |\ | | . | | +--| >-+ | . | | |/ | . | | | . | ----------- . | | . | | . '----+--------------------------------------------------+ . | | . --- --- . /// /// . . . . . . . . . .

Perhaps add a Schottky to stop the gates of the 74AC240 from going above it's supply rail when the input goes down. Maybe even change the 74AC240 for a mosfet.

Tell you the truth Fred I'm sitting here, buttocks well- -clenched, waiting for the next edition of this bloody hairy circuit.

This is the kind of circuit we think about from time to time, and quickly reject. For example, a flying opamp with supplies pinned to the rail of another opamp, to extend ordinary cheap +/-15V capable opamps to +/-30V operation, rather than using a HV opamp to begin with. In the case of some steady-state AC signal applications, the flying opamp's entire power supply can also be obtained from the first opamp's AC output. Arrggghhh!