5mhz signal to DC converter circuit required.

[SUMNERE]

Sorry guys if I did not provide enough details for my question. Below is more
details.

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.

I'm running a LC Colpitts osc. running around 5MHz (the Fq is not critical in
this applications. The Oscillator is constantly running until I touch a "touch
plate" which dampens the signal to below .6V.

I need a simple circuit that will produce a positive or negative DC ouput when
the 5Mhz is dampend. The rise or fall time must be fairly quick in the order
of time 100 usec. Recovery of the oscillator is quick and thus the
detector/converter circuit must also follow.

From there It runs into digital logic which I can handle.

 

[Jan Panteltje]

Sorry guys if I did not provide enough details for my question. Below is more
details.

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.

I'm running a LC Colpitts osc. running around 5MHz (the Fq is not critical in
this applications. The Oscillator is constantly running until I touch a "touch
plate" which dampens the signal to below .6V.

I need a simple circuit that will produce a positive or negative DC ouput when
the 5Mhz is dampend. The rise or fall time must be fairly quick in the order
of time 100 usec. Recovery of the oscillator is quick and thus the
detector/converter circuit must also follow.

From there It runs into digital logic which I can handle.

OK, assuming no other signals present:

------------------------- +5V
|
[ ] 10k
| R1
------------- out, 0V is signal present
/ | 5V is no signal
|| C1 |/ |
5MHz ---||-------------| BC 547 ===
in || | |> | 10n
100p --- \ | C2
/ \ D1 | |
| 1n4148 | |
5MHz ------------------------------------------------ GND
GND


Rise time is set by R1 x C2 = 10E4 x 10E-8 = 10E-4 = 100 us.
Fall time is set by the gain of the transistor, just s a few 5 MHz periods.
The BC547 can be replacesd by almost any NPN low power si transistor.

The way it works is that if HF is present, a current flows in teh base of
the transistor, and a beta x bigger one in the collector.
That current will discharge C2 almost immediatly.
D1 creates an actual AC path, else C1 would just charge up via the base
emitter juntion of the transistor, and not much would happen.
You can reduce C2 for faster, but if the time constant gets close to 200 nS
(5MHz) you get RF in the output and the thing no longer works right.
JP
JP

 

[budgie]

Sorry guys if I did not provide enough details for my question. Below is more
details.

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.

I'm running a LC Colpitts osc. running around 5MHz (the Fq is not critical in
this applications. The Oscillator is constantly running until I touch a "touch
plate" which dampens the signal to below .6V.

I need a simple circuit that will produce a positive or negative DC ouput when
the 5Mhz is dampend. The rise or fall time must be fairly quick in the order
of time 100 usec. Recovery of the oscillator is quick and thus the
detector/converter circuit must also follow.

From there It runs into digital logic which I can handle.

One of the "standard" 555 circuits in the original Signetics AppNote was for a
missing pulse detector. That will do what you ask in <2 cycles.