Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

rf to dc converter schematic needed

Status
Not open for further replies.
Please explain more, are you saying there is DC present on the signal feeding the multiplier or you want to extract a DC output from the circuit. If you have DC superimposed on RF as your input signal, simply couple the signal through a capacitor, preferably one with good RF characteristics, the load impedance on the source will not change significantly.

Brian.
 
  • Like
Reactions: neazoi

    neazoi

    neazoi

    Points: 2
    Helpful Answer Positive Rating
hello, I gave the circuit as an example. The question is more general.
How one could take DC out of an RF AC signal of about 1W power and how many volts and mA one could expect?
 

neazoi said:
hello, I gave the circuit as an example. The question is more general.
How one could take DC out of an RF AC signal of about 1W power and how many volts and mA one could expect?
Click to expand...

Any type of RF detector is essentially a rectifier. The only problem is with diodes and high RF power and the RF frequency.
New DC/DC converters operate as switchers, and the frequency is growing from kHz to MHz range. Their rectifiers use fast Schottky diodes and in my opinion you can effectively rectify one watt or more at over ~10 MHz. With a growing frequency, however, the high-current Schottky diodes have too large capacitance, and the efficiency drops. For >100 MHz, detector diodes can barely handle ~10 mW with a reasonable efficiency.
 
  • Like
Reactions: neazoi

    neazoi

    neazoi

    Points: 2
    Helpful Answer Positive Rating
The original circuit is a frequency multiplier (harmonic generator) not a rectifier but Jiriipolivka is quite right, it's only the limitations of high speed power rectification that stops it being like any other power supply.

Brian.
 
  • Like
Reactions: neazoi

    neazoi

    neazoi

    Points: 2
    Helpful Answer Positive Rating
jiripolivka said:
Any type of RF detector is essentially a rectifier. The only problem is with diodes and high RF power and the RF frequency.
New DC/DC converters operate as switchers, and the frequency is growing from kHz to MHz range. Their rectifiers use fast Schottky diodes and in my opinion you can effectively rectify one watt or more at over ~10 MHz. With a growing frequency, however, the high-current Schottky diodes have too large capacitance, and the efficiency drops. For >100 MHz, detector diodes can barely handle ~10 mW with a reasonable efficiency.
Click to expand...

Ok thank you,
Have you got any idea about the valie of the shunt capacitor to try after the diode, or I just try and see?
 

It depends on the frequency and load current but basically works the same as in any half wave rectifier. In normal mains AC supplies we would typically use electrolytic capacitors of 1,000uF or more but the value needed drops as the rate of charging peaks get closer together. For frequencies higher than say 1MHz it would be better to use a combination of an elecrolytic and a ceramic capacitor in parallel. The electrolytic to hold the main charge and the ceramic to minimize the impedance.

May I ask why you are doing this? Is it an RF power meter or are you trying to recover power from a transmitted signal?

Brian.
 
  • Like
Reactions: neazoi

    neazoi

    neazoi

    Points: 2
    Helpful Answer Positive Rating
What do you call the "shunt capacitor"? After the detector, RF is shorted to ground with a capacitive filter. Depending on frequency and load current you can choose from several nF to several uF, or more.
 
  • Like
Reactions: neazoi

    neazoi

    neazoi

    Points: 2
    Helpful Answer Positive Rating
betwixt said:
It depends on the frequency and load current but basically works the same as in any half wave rectifier. In normal mains AC supplies we would typically use electrolytic capacitors of 1,000uF or more but the value needed drops as the rate of charging peaks get closer together. For frequencies higher than say 1MHz it would be better to use a combination of an elecrolytic and a ceramic capacitor in parallel. The electrolytic to hold the main charge and the ceramic to minimize the impedance.

May I ask why you are doing this? Is it an RF power meter or are you trying to recover power from a transmitted signal?

Brian.
Click to expand...

I am trying to obtain a voltage to bias the diode in the schematic I posted at the beginning of the thread.
Since current required is not too big (I think), and since the rf input power is 1W, I thought I could rectify this power in a parallel RF-to-DC circuit, to bias the diode, so I can have a DC power intependent circuit.

If you can think of any other ways, I will be glad to discuss it.
 

A simple way to achieve the same effect would be a variable resistor between Ux and ground.
 
  • Like
Reactions: neazoi

    neazoi

    neazoi

    Points: 2
    Helpful Answer Positive Rating
FvM said:
A simple way to achieve the same effect would be a variable resistor between Ux and ground.
Click to expand...

How does it work with no rectification? AC will appear to the diode bias point instead of dc ?
 

The voltage has to be high enough that the diode stays reverse biased for most of the input cycle. It should switch into conduction as fast as possible but only on positive peaks of the RF waveform. Remember that this is not a rectifier circuit but a harmonic generator, it intentionally produces a distorted output.

If you want to self bias it, you might be able to do it by tapping into the input signal and using a voltage doubling rectifier then drop it to the required level using a potentiometer. The current will be virtually zero except at signal peaks and even then quite low. Normally, attaching a rectifier to a transmitter output results in serious interference being caused but that's what this circuit is supposed to do anyway!

Brian.
 
  • Like
Reactions: neazoi

    neazoi

    neazoi

    Points: 2
    Helpful Answer Positive Rating
How does it work with no rectification? AC will appear to the diode bias point instead of dc ?
Click to expand...
You noticed the 1 nF bypass capacitor? So how to get AC at the bias point?

My simple thinking says, that the conduction angle will drop to almost zero with a DC-wise floating Ux terminal. Applying a higher DC-voltage than this self-biasing level (e.g. by using an auxilary voltage doubler) will only sligthly change the output signal magnitude and waveform. Things are different of course if the diode has a considerable reverse leakage current, but the suggested BA244 hasn't.
 
  • Like
Reactions: neazoi

    neazoi

    neazoi

    Points: 2
    Helpful Answer Positive Rating
I will try these solutions to see how it goes.
Thank you very much all.
 

Ok I have ended up to this schematic, not tested yet. Any suggestions?
 

Attachments

  • harmgen_w_rtx.png
    harmgen_w_rtx.png
    11.9 KB · Views: 105

Yes.... it won't work!

In truth, it will work if you have DC positive on the Tx signal but not more than 0.6V and your frequency is less than about 100KHz.

You have not mentioned the frequency you intend to use but 10uF as the input capacitor, especially a polarized type probably isn't what you want. If your frequency is > 1MHz or so, I would go for 1nF ceramic instead. If you are using above say 100MHz, I would drop it to 100pF and also look at using Shottky barrier diodes instead of 1N4148s.

Incidentally it's self powered but not self adjusting!

Brian.
 
  • Like
Reactions: neazoi

    neazoi

    neazoi

    Points: 2
    Helpful Answer Positive Rating
betwixt said:
Yes.... it won't work!

In truth, it will work if you have DC positive on the Tx signal but not more than 0.6V and your frequency is less than about 100KHz.

You have not mentioned the frequency you intend to use but 10uF as the input capacitor, especially a polarized type probably isn't what you want. If your frequency is > 1MHz or so, I would go for 1nF ceramic instead. If you are using above say 100MHz, I would drop it to 100pF and also look at using Shottky barrier diodes instead of 1N4148s.

Incidentally it's self powered but not self adjusting!

Brian.
Click to expand...

Thank you Brian,

I have made the corrections. The input frequency will be 144MHz/432MHz and maybe 50MHZ as well. How about the attached schematic?

I have also found on the net an RC filter shown at the lower end. Simulation showed good results at 100K I/O impedance and a cut-off of about 80KHz..
Would this be of any good to include it between the 100K pot and the 820K (to reduce doubler harmonics)?
 

Attachments

  • myharmgen.png
    myharmgen.png
    17.9 KB · Views: 96

I wouldn't bother with the RC filter but I would increase the capacitor across the potentiometer to say 100nF so it removes any residual modulation from the TX signal. The impedance at the potentiometer is so high that a simple capacitor should be sufficient to filter it.

The circuit is similar to one I used many years ago to multiply x10 to produce a signal on 10.315GHz as part of an amateur TV link. Be careful to select the correct harmonic and only one at a time when you use the output because it will wreak havok on the radio spectrum if you let it reach an antenna!


Brian.
 
  • Like
Reactions: neazoi

    neazoi

    neazoi

    Points: 2
    Helpful Answer Positive Rating
betwixt said:
I wouldn't bother with the RC filter but I would increase the capacitor across the potentiometer to say 100nF so it removes any residual modulation from the TX signal. The impedance at the potentiometer is so high that a simple capacitor should be sufficient to filter it.

The circuit is similar to one I used many years ago to multiply x10 to produce a signal on 10.315GHz as part of an amateur TV link. Be careful to select the correct harmonic and only one at a time when you use the output because it will wreak havok on the radio spectrum if you let it reach an antenna!


Brian.
Click to expand...

Ok here it is.
The purpose I wonder about the filter use, is because if not using a filter, harmonics from the doubler will present at the RF output port. But there will be no means of controling the level of these harmonics (at least not through the multiplier diode process), so better not to have them?

I wonder what will happen if I use a 1N23 as a multiplier diode instead... better for X-band harmonics?
 

Attachments

  • myharmgen.png
    myharmgen.png
    18.4 KB · Views: 89

More harmonics are better! the whole purpose of the device is to produce as many of them as possible. If you insert the filter all you will do is halve the DC bias available so you may then have to resort to a voltage quadrupler to get back where you started!

I doubt a 1N23 will work any better and it may be significantly worse. The property you need in the diode isn't it's microwave performance but it's rapid switching action. The harmonics are generated by the rapid turn-on and turn-off as it moves to and from conduction, that's why fast and step recovery diodes work best.

Brian.
 
  • Like
Reactions: neazoi

    neazoi

    neazoi

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top