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Amplifier with led's and photodiodes.
This article is about an amplifier which uses led's and photodiodes as amplifying elements.
To start, I did a test to find out how much current is detected by a photodiode when it is illuminated by various types of led's.
Figure 1, test setup to measure detected current with various types of led's.
Six high brightness led's are tested.
- Two pieces HSDL4230 infrared led (datasheet_HSDL-4230.pdf).
One in original state, and one modified by removing the top of the led, see figure 4.
- One SFH4235 infrared led (datasheet_SFH4235.pdf).
- One LRW5SN red led (datasheet_LRW5SN.pdf).
- One TLCR5800 red led (datasheet_TLCR5800.pdf).
- One LXHL-MW-1D white led (datasheet_LXHL-MW1D.pdf).
The six led's under test are series connected, so all have the same current.
A BPW34 photodiode (datasheet_BPW34.pdf) on a piece of PCB board is held by hand above the led under test.
Figure 2 , the test setup with 6 led's.
The 3 led's at the left emit invisible infrared light.
Across all led's a 100 nF capacitor is placed, just to have an easy accessible point to put the leads of the multimeter on, to measure the led voltage.
For the rest these capacitors don't have much function.
Figure 3, for the HSDL4230 led I have tested two versions.
This is the original version.
Figure 4, this is the modified version of the HSDL4230, from which the top is removed.
Through this the photodiode can come closer to the light emitting chip, which gives a higher current in the photodiode.
The current through the led's in increased in 2 mA steps from 2 mA to 12 mA.
And the detected current through the photodiode is measured.
Figure 5, this diagram shows the detected current in the BPW34 photodiode for the various led's.
The highest current is detected in combination with the SFH4235 (infrared) and the LRW5SN (red) led.
These are the led's in a thin surface mount package, which allows you to place the photodiode very close to the light emitting chip.
From figure 5, the SFH4235 gives the highest detected current for a given led current.
But another interesting thing is the measured voltages across the led's.
|led type||led voltage
at 2 mA
at 12 mA
|change in voltage
(2 to 12 mA)
From this table we see the LRW5SN needs less change of voltage then the
SFH4235, to change the current from 2 to 12 mA.
This means the LRW5SN might need less change of input power, to change the detected current in the photodiode.
But anyhow, I choose the SFH4235 led to build my amplifier.
Also because the datasheet of the LRW5SN mentioned explicit (without giving the reason) not to use that led below 100 mA current.
As we have seen from figure 5, when you put 12 mA through the SFH4235 led, the detected current is slightly more then 4 mA, so there is a loss of current instead of gain.
How can we then make an amplifier with signal gain?
The answer is: driving the led's from a low impedance source, and loading the photodiode by a much higher impedance.
You then can get a high voltage gain, much higher then the loss of current, so the power gain stays higher then one.
Figure 6, circuit diagram of the amplifier.
The left and right audio line inputs are combined to one mono signal by resistors R1 and R2.
Then the impedance is stepped down with audio-transformer TR1 to drive the two led's (D1 and D3).
The used transformer is a 100 Volt loudspeaker transformer made by Visaton, model TR10.16.
It has on the primary side power taps of 0.625Watt , 1.25W , 2.5W , 5W and 10W.
These corresponds to nominal input impedances of 16 kΩ , 8kΩ , 4kΩ , 2kΩ and 1kΩ.
But the actual input impedance depends on the value of the load resistor.
In this design the load caused by the led's is higher in impedance then the 4Ω of the transformer winding they are connected to.
This means, the transformer input impedance is also higher then the 2kΩ you might expect from the 5Watt winding.
The both led's are biased to 9.5 mA DC by resistor R5 and R6.
Via capacitor C1 and C2, the audio signal is added to the led voltages, causing the led current to vary with the audio.
Both led's are driven in anti phase, so when for one led the current goes up, for the other led the current goes down.
This reduces audio distortion.
Resistor R3 and R4 keep the minus pole of the elco's (C1 and C2) at 0 Volt DC, while the plus poles are at the led voltage of about 2.5 Volt.
Maybe elco's C1 and C2 and resistors R3 and R4 can be omitted, and connect the transformer directly to the led's.
When the led's have exactly the same voltage drop, this should be possible I think, but I did it as shown in the circuit diagram.
The two photodiodes are mounted directly above the led's, so they pick up maximum current.
By measuring the voltage across resistors R7 and R8, we can determine the DC current through the photodiodes, which should be about 3 mA (0.3 Volt across R7 and R8)
Output transformer TR2 has between the 0 and 2.5 Watt winding an impedance of 4000 Ω.
Between the 2.5 Watt and 0.625 Watt winding is also 4000 Ω impedance.
So, both photodiodes are loaded with 4000 Ω.
The transformer converts this to 16 Ω at the output.
Figure 7, amplifier with led's and photodiodes.
Figure 8, detail of the led and photodiode.
The photodiode is mounted on a small piece of PCB board, which is mounted above the led.
Figure 9, for this picture, the board with the photodiode is removed.
This picture is taken with another camera which has some sensitivity for infrared light.
So you can see the light from the led which is invisible for the human eye.
Figure 10, the amplifier in action, with a headphone connected to the output, and line level signals to the input.
The headphone has two 32 Ω speakers which are parallel connected, to give an impedance of 16 Ω.
The input impedance of the amplifier is 5.2 kΩ.
The maximum input level is 700 mV RMS, with higher input levels the output signal is clipping.
I connected the input of this amplifier to a CD player, which has a maximum output level of about 700 mV RMS.
And it sounds very good, and loud in the headphone.
A volume control at the input might be a good idea, but in this project I omitted this.
It's not my aim to use this amplifier in practice, I just wanted to proof that the concept with a led and photodiode can work, and has power gain.
I did a test with 1 kHz 0.47 V RMS at the input.
And as the input impedance is 5.2 kΩ, the input power was (0.47V)² / 5200Ω = 42.5 μW.
Then at the output I got 0.25 V RMS across a 16 Ohm load, which is a power of 3.9 mW.
This means the power gain of the circuit is about 92 times.
Making an oscillator.
In the next test, the output of one photodiode is via a 150 nF capacitor fed back to the amplifier input.
This causes the amplifier to oscillate.
Figure 11, by feeding the output back to the input, the amplifier is turned into an oscillator.
Figure 12, the output signal (measured at the photodiode).
Horizontal: 1 ms / division, the frequency of the signal is about 590 Hz.
Vertical: 5 Volt / division, so about 25 Volt peak-peak.
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