Bridged-T attenuator circuit using PIN diodes as variable resistances

Thread Starter

sindre jacobsen

Joined Apr 1, 2019
40
Hello,

I am trying to analyze a bridged-T attenuator circuit with the variable resistances being PIN diodes. I am trying with KCL to figure out a formula for the input voltage and current required to get a certain resistance value on the diodes from a table. Will it work to assume the diodes are ideal current sources when analyzing it this way? Or do I require both the forward current and voltage to find the appropriate input values?

Best regards,
Sindre
 
The resistance of a PIN diode is inversely proportional to the current through the diode. You will need to look at the specification sheet for the diode to determine the resistance vs current. The diode is not a current source, you must supply the proper current to get the desired resistance.
Here is an example from a Microsemi spec sheet:
upload_2019-7-16_7-26-9.png
 

Thread Starter

sindre jacobsen

Joined Apr 1, 2019
40
Thanks for replying.
What I mean is that I have set up a table of certain resistance values with its own current from a graph similar to the one you just showed. What I have done next is set the diode as a current source so that I can make a function that tells me the required input to get the certain current over the diode. What I'm wondering, is if I can do it this way, or if there is a proper way for how I should analyze the circuit with PIN diodes to get a function that tells the required input signal level.
 
You need to create an external current source to supply the diode. The table you desire would be determined by the current source control input vs output current of the external current source. If you want the table to reflect control input vs resistance you will need to use the diode specification sheet to determine the relationship.
 

AnalogKid

Joined Aug 1, 2013
8,306
1. Is this homework or a schol project?

2. Your circuit has no clearly defined output or purpose. What is it you are trying to achieve? If you assume ideal diodes (Vf = 0.6 V, independent of If), then the voltage across R3 is independent of any diode current.

ak
 

Thread Starter

sindre jacobsen

Joined Apr 1, 2019
40
I am supposed to build a Bridged T attenuator where the output (voltage across the R3 resistor) is attenuated based on the resistor values that are apparent in the PIN diodes. To do so, I need to find a way by using PIN diodes to create a variable resistor based on the input level that gives out the necessary resistor values to attenuate the signal a certain amount. From my limited experience/knowledge, I believe I require more than just the diodes in order to achieve this.
 

Thread Starter

sindre jacobsen

Joined Apr 1, 2019
40
1. Is this homework or a schol project?

2. Your circuit has no clearly defined output or purpose. What is it you are trying to achieve? If you assume ideal diodes (Vf = 0.6 V, independent of If), then the voltage across R3 is independent of any diode current.

ak
1. It is part of a project.
2. I am supposed to build a Bridged T attenuator where the output (voltage across the R3 resistor) is attenuated based on the resistor values that are apparent in the PIN diodes. To do so, I need to find a way by using PIN diodes to create a variable resistor based on the input level that gives out the necessary resistor values to attenuate the signal a certain amount. From my limited experience/knowledge, I believe I require more than just the diodes in order to achieve this.

Thank you for taking the time to help.
 

Thread Starter

sindre jacobsen

Joined Apr 1, 2019
40
Your design won't maintain a 50Ω impedance except at one point. What are your impedance requirements?
Using the formulas R1 = R/(A-1) and R2 = R(A-1) I found the required impedance values for attenuation range of 1 to 15 dB in 1 dB increments. By doing so I got the following resistor values (1dB left, 15 dB right):
R1 = [410 193.1 121.2 85.5 64.2 50.2 40.4 33.1 27.5 23.1 19.6 16.8 14.4 12.5 10.8]
R2 = [6.1 12.9 20.6 29.2 38.9 49.8 61.9 75.6 90.9 108.1 127.4 149.1 173.3 200.6 231.2]
In this region I require 50 ohm impedance.
 

Thread Starter

sindre jacobsen

Joined Apr 1, 2019
40
Then you are using the PIN diode incorrectly, your circuit will not work as you expect.
The attenuator would need to go before the detector.
So I either have to change the configuration on the variable resistances to make it work, or move it before the detector? Cause when I calculated it with variable resistors and didn't change them with PIN diodes, I got out correct values, with the only difference being that I look at voltages rather than RF power.
 

AnalogKid

Joined Aug 1, 2013
8,306
What is the input signal you are trying to attenuate? DC, audio, video, RF, etc? Voltage range? Frequency range? Diode attenuators often are used in RF circuits, but you have not said that that is what you are doing.

ak
 

Thread Starter

sindre jacobsen

Joined Apr 1, 2019
40
What is the input signal you are trying to attenuate? DC, audio, video, RF, etc? Voltage range? Frequency range? Diode attenuators often are used in RF circuits, but you have not said that that is what you are doing.

ak
The Vrms output of the detector (ADL5511) will be used, so a DC signal will be the input to the attenuator. Just read that the PIN diodes are seen as resistances for RF signals that are varied due to its forward current. Do this mean that I either have to move the attenuator before the detector and create a variable current source for the diodes that decides the resistor values or is it possible to make do with them after the detector?
 
The signal after the detector will not be RF, it will be the demodulated signal. If you are trying to set the resistance with the demodulated signal then the resistance will vary with that signal. Your circuit is using the demodulated signal as both the signal to be attenuated as well as the source for control current. That won't work! Or I should say that you won't get what you are expecting.

Your demodulated signal is probably at audio frequencies but you are trying to use RF components to attenuate the signal. You need to rethink your project.
 

Thread Starter

sindre jacobsen

Joined Apr 1, 2019
40
The signal after the detector will not be RF, it will be the demodulated signal. If you are trying to set the resistance with the demodulated signal then the resistance will vary with that signal. Your circuit is using the demodulated signal as both the signal to be attenuated as well as the source for control current. That won't work! Or I should say that you won't get what you are expecting.

Your demodulated signal is probably at audio frequencies but you are trying to use RF components to attenuate the signal. You need to rethink your project.
The signals that enter the RF detector have a constant magnitude. The signals will be attenuated if they reach a certain value that brings the amplifier into the nonlinear region. So the task of the attenuator is to limit the magnitude to stay within the boundary of the transfer function as to where the amplifier has a linear characteristic function.
Will the PIN diodes work for this application after the detector if I make a control current that is dependent on the input signal? I have read that the PIN diodes are meant for use in RF applications and that the whole point is that the bias current will give the intrinsic layer carriers such that it looks like a resistor, where the RF signal requires a frequency large enough so that it won't make the intrinsic layer discharge completely, making it turn off. Will the PIN diodes still work for this sort of application or won't it work unless it is RF signal input?
 

Thread Starter

sindre jacobsen

Joined Apr 1, 2019
40
The signal after the detector will not be RF, it will be the demodulated signal. If you are trying to set the resistance with the demodulated signal then the resistance will vary with that signal. Your circuit is using the demodulated signal as both the signal to be attenuated as well as the source for control current. That won't work! Or I should say that you won't get what you are expecting.

Your demodulated signal is probably at audio frequencies but you are trying to use RF components to attenuate the signal. You need to rethink your project.
Figured out that I have to do it before the RF detector as you said.
Thanks for all the help!
 
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