Driving a Discrete Input Having a 50ohm Impedance

[Don]

Any one know of a good reference design for a driver for an input to
unit with a 50ohm impedance and that's looking for 8 to 11VDC as a
"high" and "-1 to 2VDC for a logic 0? Needs to have a fast (50nsec)
rise time.

I have been pondering just driving a transistor cable of handling about
500mA with the collector ties to the mid-junction of two 50ohm
resistors hooked in series. The high side resistor connected to 20VDC
and the other resistor connected to ground.

Any references or suggestions will be greatly appreciated!

Thanks!

Don

 

[Jim Thompson]

Any one know of a good reference design for a driver for an input to
unit with a 50ohm impedance and that's looking for 8 to 11VDC as a
"high" and "-1 to 2VDC for a logic 0? Needs to have a fast (50nsec)
rise time.

[snip]

What logic family is that, "Wasteful" ?

...Jim Thompson

 

[Don]

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

 

[[email protected]]

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.

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?

 

[[email protected]]

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.

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?

 

[[email protected]]

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

 

[Don]

From the spec for the piece of equipment, here is the description of
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.

 

[Norm Dresner]

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

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

 

[Jon]

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

 

[[email protected]]

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.

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...

 

[Roger]

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.

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

 

[Jon]

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

 

[[email protected]]

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.

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.

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.

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.

 

[Tony Williams]

+5V
|
+----+--[10K]--|>|---+----------------------+---|<|-----.
| | 1N6263 | | 1N5817 |
| | | | |
| | | ---74AC240 |
| | | | | |
| | | | |\ | |
| | | | +--| o-+ | |
=== | === | | |/ | | |
0.47u | 0.47u | | . | | |
| | | | | . | | |
| | | | | |\ | | |
| +---------|>|---|---------+------------+--| 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.

 

[Winfield Hill]

Tony Williams wrote...

+5V
|
+----+--[10K]--|>|---+----------------------+---|<|-----.
| | 1N6263 | | 1N5817 |
| | | | |
| | | ---74AC240 |
| | | | | |
| | | | |\ | |
| | | | +--| o-+ | |
=== | === | | |/ | | |
0.47u | 0.47u | | . | | |
| | | | | . | | |
| | | | | |\ | | |
| +---------|>|---|---------+------------+--| 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!

 

[Tony Williams]

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!

A novel circuit that is interesting to look at though.
I'd never thought about a single 5V pump of the rail
and a simultaneous 5V swing of the output.

Will the output reliably swing from 0 to 10v in about
50nS? A possible killer might be load capacitance and
the effective transient current load on the 74AC241.

What might be other ambushes?

 

[Tony Williams]

Maybe even change the 74AC240 for a mosfet.

[/QUOTE]

Can you find one that has 20pF input capacitance, handles 200mA
at low threshold, and costs 27 cents? I assume it would be a
P-channel.

No.

 

[Winfield Hill]

Tony Williams wrote...

No.

What about ON Semi's NTA4151P that was just suggested? Only
0.49 ohms with -1.8V drive, and 16 cents each. Oops, 156pF of
input capacitance, well OK, then a similar but smaller MOSFET.
The NTA4151 can handle 760mA, well over the 200mA requirement.