Converting voltage to power in a receiver

• 3dB
In summary, the problem seems to be that the conversion factor, C1, is not a constant, and that the dynamic range of the input signal is decreased by a factor of 2 when calculated at the input of the APD.
3dB
Hi everyone. I'll describe my setup first, then state my question.

I have a receiver chain consisting of the following (where all conversion factors are assumed to be scalar and constant under the conditions in which I use them):
1) An APD (avalanche photodiode): Input: Optical power P1in [W]; Output: Current I1out [A]; Conversion factor: C1 [A/W].
2) A transimpedance amplifier: Input: Current I2in = I1out [A]; Output: Voltage U2out [V]; Conversion factor: C2 [V/A = ohm].
3) A monolithic high-frequency amplifier, 50 ohm differential input. (~50 ohm resistor connected across the amplifier inputs, negative input is AC-coupled to GND, positive input is AC-coupled to the transimpedance amplifier output.) Use Z0 = 50 ohm. Input: P3in [dBm]; Output: P3out [dBm]. Gain factor is C3 [dB].

My question is: How should I convert the input optical power to an electrical power, at the input of component 3) above? (Or more specifically, how do I convert U2out to P3in?)

I want to use power expressed as dBm when making the necessary calculations, since I will treat the receiver as a RF system. For example, the amplifier gain is expressed in dB.

Assume I have the input optical power and all conversion factors, no point in confusing the argument with a bunch of numbers.

**************************

My thoughts on the matter are as follows:

I am confident that the output from the transimpedance amplifier is U2out = P1in * C1 * C2, and the units seem to be correct.

But the only way I can think of to convert U2out to P3in is to take P3in = (U2out)^2 / Z0.

I have a gut feeling that this is incorrect, since (look at the formulas) this means that the original input power is squared. In my case this is a problem since the required dynamic range of the system, when calculated at the input (the APD), is _half_ of the dynamic range when calculated at the 3:rd component.

Since I do not care about the DC components of U2out, the Z0 resistor seems to be where the voltage drop occurs (but I suspect I should treat the differential input of the 3:rd component as a parallel connection of the external matching resistor and the input impedance, but that would only introduce a factor of 2 in the calculations).

I can't find the error. I suspect the root cause is in the C1 conversion factor, [A/W].

Can anyone please point me in the right direction? Or if I am already pointed in the right direction, tell me why? ;-)

P3in = (U2out)^2 / Z0 should work as long as you really have a 50 Ohm system.
Remember that you can always think of the amplifier as a VOLTAGE amplifier if you want to as long as all the components have the same impedance; e.g. a gain of 20 dB simply means that it amplifies the voltage 10 times.

Thanks for the reply, but I thought of that too. But I still have twice the dynamic range (in dB) compared to the dynamic range of the optical input signal.I think I might be on to an explanation:
The APD transforms a power to a current.
If one considers a grounded load, Zl, connected to a feed line carrying the current I, the power dissipated by the load is Pl=I^2*Zl, or conversely, I=sqrt(Pl/Zl).
But in my assumptions above, I used a linear transformation from power to current. It seems as it is here my "squaring" occurs.

I must investigate further if the linear transformation is valid or not (it is not a resistive load so it might be valid), but if I use the resistive load analogy, my formulas make sense.

It appears as if I did not do anything wrong. I just got it confirmed that power is actually squared in the APD, and that I will simply have to deal with the fact that the dynamic range is doubled inside my receiver than outside of it.

Perhaps my problem with the inital findings was mostly wishful thinking.

What is the relationship between voltage and power in a receiver?

In a receiver, voltage is the measure of the electrical potential difference between two points, while power is the rate at which energy is transferred or converted. The relationship between voltage and power in a receiver is that an increase in voltage usually results in an increase in power.

How is voltage converted to power in a receiver?

Voltage is converted to power in a receiver through a process called rectification. This involves using diodes to convert the alternating current (AC) voltage received from the antenna into direct current (DC) voltage, which can then be used to power the receiver's circuitry.

What factors affect the conversion of voltage to power in a receiver?

The efficiency of voltage to power conversion in a receiver can be affected by various factors such as the quality of the components used, the design of the circuit, and external interference. The type and strength of the antenna used can also play a role in the efficiency of voltage conversion.

Why is it important to accurately convert voltage to power in a receiver?

Accurate conversion of voltage to power in a receiver is important because it ensures that the receiver is receiving a strong and stable signal. This is crucial for proper functioning and optimal performance of the receiver. Inaccurate conversion can result in a weak or distorted signal, leading to poor reception or even complete failure of the receiver.

What are some methods for improving voltage to power conversion in a receiver?

One method for improving voltage to power conversion in a receiver is to use high-quality components and a well-designed circuit. Additionally, shielding the receiver from external interference and using a high-quality antenna can also improve the efficiency of voltage conversion. Regular maintenance and calibration of the receiver can also help to optimize voltage to power conversion.

• Electrical Engineering
Replies
11
Views
1K
• Electrical Engineering
Replies
12
Views
1K
• Electrical Engineering
Replies
14
Views
1K
• Electrical Engineering
Replies
10
Views
2K
• Electrical Engineering
Replies
8
Views
2K
• Electrical Engineering
Replies
61
Views
7K
• Electrical Engineering
Replies
10
Views
2K
• Electrical Engineering
Replies
10
Views
1K
• Electrical Engineering
Replies
47
Views
4K
• Electrical Engineering
Replies
4
Views
2K