5mhz signal to DC converter circuit required.

[SUMNERE]

I'm looking for a non-complex way to convert a 5MHZ signal to a dc level with
little if any ripple. The signal is at 5MHz with amplitude of 0-5 volts. When
the signal is present a + 2.5 to +5 Vdc output is required. When the 5MHz
signal disappears a 0 volt output is required.

Any ideas? Thanks

 

[Spehro Pefhany]

I'm looking for a non-complex way to convert a 5MHZ signal to a dc level with
little if any ripple. The signal is at 5MHz with amplitude of 0-5 volts. When
the signal is present a + 2.5 to +5 Vdc output is required. When the 5MHz
signal disappears a 0 volt output is required.

In (0V/5V 5MHz square wave)

o-----[10K]---------x--> 2.5V = signal/0V = no signal
|
---
--- 0.1uF SMT ceramic
|
Gnd-----------------x



Best regards,
Spehro Pefhany

 

[Keith R. Williams]

I'm looking for a non-complex way to convert a 5MHZ signal to a dc level with
little if any ripple. The signal is at 5MHz with amplitude of 0-5 volts. When
the signal is present a + 2.5 to +5 Vdc output is required. When the 5MHz
signal disappears a 0 volt output is required.

Retriggerable single-shot with a time constant of slightly greater than
..2usec.

 

[Michael Black]

Retriggerable single-shot with a time constant of slightly greater than
.2usec.

But apart from multiple posting his question they guy really has not
defined his problem, and you can't come up with a solution unless the
problem is actually defined.

I read it that he merely needs to know that the signal is present or
not. He's not talking about a signal that may be on various frequencies.
Maybe that is the problem, but he has not said so.

So the solution could be as simple as a diode detector, and then a comparator
or schmitt trigger on it's output to square it up.

Or it could require detection of a specific frequency, 5MHz in this case.
But again since he hasn't said much, there is no indication of how much
selectivity would be required. If he is talking fairly large steps
between frequencies, the solution is much less complicated than if
he needs to differentiate between 5MHz and 5.1MHz.

Michael

 

[Keith R. Williams]

But apart from multiple posting his question they guy really has not
defined his problem, and you can't come up with a solution unless the
problem is actually defined.

1. I only see one posting.

2. The OP defined a problem and I gave a solution to that problem.
Indeed, there could be more to the problem (which may alter my answer
;-) that hasn't been stated, but a problem has been stated and *a*
solution offered.

I read it that he merely needs to know that the signal is present or
not. He's not talking about a signal that may be on various frequencies.
Maybe that is the problem, but he has not said so.

You're assuming facts not in evidence. The problem stated 5MHz. There
wasn't anything there about any other frequencies.

So the solution could be as simple as a diode detector, and then a comparator
or schmitt trigger on it's output to square it up.

Huh? How does that give a high level when the "clock" is present and a
low when it's not?

Or it could require detection of a specific frequency, 5MHz in this case.

Exactly. Is it there or is it not? That was the question.

But again since he hasn't said much, there is no indication of how much
selectivity would be required. If he is talking fairly large steps
between frequencies, the solution is much less complicated than if
he needs to differentiate between 5MHz and 5.1MHz.

Your inventing a problem, not a solution. ;-) Selectivity wasn't part
of the problem.

 

[mikem]

I'm looking for a non-complex way to convert a 5MHZ signal to a dc level with
little if any ripple. The signal is at 5MHz with amplitude of 0-5 volts. When
the signal is present a + 2.5 to +5 Vdc output is required. When the 5MHz
signal disappears a 0 volt output is required.

Any ideas? Thanks

Here is my favorite:

cut what follows into a text file called Detector.asc and open it using
LTSpice: You can then play around with optimising the turn on time
(determined primarily by the ratio of C2/C1), the turn off time
(determined primarily by the C1R2 time constant) and the amount of
ripple while it is on. My solution uses a series dc blocking capacitor
so that you dont have to know which state (high or low) your RF source
is in when no 5Mhz is present.

MikeM

___________________________________________________________________________

Version 4
SHEET 1 880 680
WIRE 192 304 192 240
WIRE 192 176 192 80
WIRE 192 80 144 80
WIRE 80 80 -32 80
WIRE -32 80 -32 176
WIRE -32 240 -32 304
WIRE -32 304 192 304
WIRE -32 368 -32 304
WIRE -512 112 -480 112
WIRE -512 112 -512 192
WIRE -512 352 -512 272
WIRE -160 80 -192 80
WIRE -96 80 -32 80
WIRE -480 112 -352 112
WIRE -192 80 -208 80
WIRE 304 160 304 80
WIRE 304 80 192 80
WIRE 304 240 304 304
WIRE 304 304 192 304
FLAG -32 368 0
FLAG -512 352 0
FLAG -480 112 i
FLAG -192 80 o
FLAG -32 80 c
FLAG 192 80 d
SYMBOL cap 176 176 R0
SYMATTR InstName C1
SYMATTR Value 1000p
SYMBOL SpecialFunctions\\modulate -352 48 R0
WINDOW 3 0 0 Invisible 0
SYMATTR InstName A1
SYMATTR Value mark=5meg space=5meg
SYMBOL voltage -512 176 R0
WINDOW 3 29 154 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value PULSE(0 2.5 0.5m 1u 1u 1m 2m)
SYMBOL cap -96 64 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C2
SYMATTR Value 100p
SYMBOL diode -16 240 R180
WINDOW 0 24 72 Left 0
WINDOW 3 24 0 Left 0
SYMATTR InstName D1
SYMATTR Value 1N4148
SYMBOL res 288 144 R0
SYMATTR InstName R2
SYMATTR Value 10k
SYMBOL diode 80 96 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D2
SYMATTR Value 1N4148
TEXT -176 384 Left 0 !.tran 3m
TEXT -352 176 Left 0 ;5Mhz 5Vp-p