Most of the receiver noise is created by the photodiode working resistor. If this resistor can be big in the night and small in the day, the receiver will have an excellent noise and dynamic range. For this purpose I will use a photoresistor coupled with LED and a controller regulating the LED current based on voltage measured on the photodiode.
Is here: PDF, PostScript.
SFH203 has power limit of 100mW. This will be limited by the zener diode resistor. How much does the converter give maximum? Say the drop on the transistors is 2*0.7V= 1.4V and the schottky drop is 0.2V. The maximum input voltage is 13.2V then we get 6*11.6=69.6V maximum.
Maximum energy transfer happens when the voltage is half, that's 34.8V and for 100mW we have 2.87mA. That's 12.125kOhm when divided into three resistors each is 4041 Ohm we have to round up so we take 2x3k9 and 1x4k7 and the total is 12.5kOhm.
Now the LED. The max. current is 20mA. The green LED dropout is 1.5V. The max. supply voltahe is 13.2V. That's 585 Ohm, round up to 680 Ohm.
An ordinary opamp would suffice to invert the characteristic of the coupler. However the coupler is slow and noise will be amplified greatly on high frequencies. For that a capacitor will be put across the LED. This will simulate the changing speed of the photoresistor - low current slow, high current fast. Furthermore I will measure the output of the opamp and if it makes too much noise, I will decrease the gain at higher frequencies.
The diode voltage will be picked up by a series of 10M resistors. This will minimize loading and parasitic capacity (200fF per SMD resister).
If there is a LF component to which the photoresistor cannot react quickly enough, for example 100Hz, it could cause saturation of the photodiode. With an ordinary resistor this wouldn't happen. We must make sure it doesn't happen with photoresistor. If the duty cycle of the interference is 50% and the ambient consists only of this interference, the voltage can be regulated from 25V to 50V without problem. Higher voltage gives more safety for signals that stay short time in light condition and long time in dark. The 100Hz or 120Hz ligting stays most of the time in light condition and short time in dark.
From noise perspective, if the photoresistor drop is 52mV, the photoresistor noise is equal to the photodiode shot noise (caused by ambient light). This would degrade the characteristic all the way from low to high illumination. Must be avoided. 5.2V drop makes 1% and 25V drop 1/480. Therefore I would suggest 30V voltage drop in general. In worst case where the Zener regulates at 45V, this makes 1/288 of the shot noise or 0.34%.
An opamp can't take 30V on the input if powered from 11V. Therefore I will use 5V for reference and divide by 6. 50/6=8.33 and 30/6=5.
How big should the divider be? I would use 3 10M resistors and 5.6M into the ground. That will make a parasitic of 66fF and 30M noise is 1.7nA. The resulting regulated voltage will be 31.79V.
The 36.8M will pull the voltage in case of high photoresistor voltage. Since we regulate at 30V, with 47V on the Zener, will result in 20.85M on photoresistor in case of no PIN diode current. This will create 2.5nA of noise.
The 2M2 in the gate is creating considerable noise 24nA. Therefore I would replce with 5.6M creating only 9nA. It has to be simulated and tested how the LF rejection looks like. For better performance with a better FET (HEMT possibly), the resistance can be further increased.
I hooked up the Perkin Elmer VT83N4 photoresistor (diameter 7mm) carefully to the smart tweezer and did 10 measurements with average of 3.169pF and sigma(n-1) of 11fF. The photoresistor was in darkness. Then the tweezer alone 1.12pF and sigma(n-1) of 10fF. The result is 2.05pF +/- 45fF. This will incerase the MOSFET noise but not to the point that it wouldn't pay off.
Similarly, Pelkin Elmer VT43N1(diameter 11mm) has 2.58pF +/- 37fF.
As the working resistor decreases the capacitive load of the PIN, photoresistor and MOSFET create a lowpass. How many k can we afford? At 31.7V the photodiode should havwe 1.63pF. The MOSFET input is 2.1pF and the photoresistor 2.05pF. Total 5.78pF. From the bandwidth simulation something like 200kHz cutoff is necessary otherwise the signal is distorted. That makes 137kOhm. That means with high illumination the jitter between 5 and 10MHz will degrade. I will not limit this at all. Better a degraded signal than none. 137kOhm is still 2.85x less than 390k and 390k has the same saturation lighting as the old Ronja receiver. So this is still a good improvement in resistance to ambient light.
To prevent noise from opamp output to couple into the input circuit through LED - photoresistor stray capacitance, I will put a capacitor across the LED. But how big? It should present a smaller but comparable slowdown than the photoresistor does. For this I need to calculate the dynamic resistance of the LED.
LED current is i=is*exp(v/(n*vt)). di/dv=i/(n*vt), dynamic resistance is dv/di=n*vt/i. What is the n for the green LED? I measured the voltage for current of 2mA, 2e mA and 2/e mA:
| Current [mA] | Voltage [mV] |
| 2/e | 1817 |
| 2 | 1868 |
| 2e | 1944 |
The second pair suggests n=2.92, the first n=1.96. Diodes allegedly have usually n between 1 and 2. The second n looks suspicious because it's bigger than 2 and differs from the first one. It's probably caused by internal resistance of the diode. Therefore I assume that n=1.96.
Now since I don't have datasheet for the photoresistor I will try to take a similar one. According to the datasheet I downloaded a photoresistor with 200k @10lx (the biggest one) hast 35 rise and 5ns fall time. Let's take geometric average of 13ms. 200k in our case corresponds to LED current 700uA. That's LED dynamic resistance 72.38 Ohm. At this resistance a capacitor should have RC constant of 13ms. That's 179uF. Since 100nF is enough to suppress the noise from the Zener, let's take 100nF. It has lower input resistance. The time constant of the LED will then be 7.6us.
Since the opamp presumably also creates some noise on the input and that could couple into the input through the parasitic capacitance about 80fF (200fF/2.5), I will block the input. But how to make sure it isn't slow? The total resistance on the input is 1/(1/2.47M+1/0.47M)=395k. That's 32n. Since I want to suppress 100Hz the time constant should be smaller say 1ms. For that I need 2n2.
To prevent unnecessary noise, I will reduce the gain of the opamp. But how much do I need? Let's say the output swing is 12V and we want the PIN diode voltage swing only 0.5V. The megaohm divider divides by 6, therefore the gain necessary is 144. That doesn't seem big. According to the datasheet the gain is 79000 and UGB is 480kHz. That means 3.3kHz bandwidth with this gain. The voltage noise is 100nV/sqrt(Hz) at 1kHz. How much noise will that make in a 3.3kHz band with 144 times gain? 827uV RMS. I guess that's not enough to induce anything adverse into the PIN through the optocoupler.
To get 144 gain I will use 10M and 68k resistor. This will be tied to the 5V and will inject current into it. However the divider for the differential limitters has total resistance on order of 5k so a 10M resistor from 12V cannot screw this up.
Since both the opamp and the feedback look like a RC lowpass the loop phase will eventually approach 0 for higher frequencies. So it needs to be tested with an oscilloscope that it doesn't oscillate for any illumination level.
I have tested the Rhodopsin module and it works fine, but it goes to only about 35uA of the PD current. Then the opamp ICL7621 cannot give enough current - the output of the opamp hangs at 10.3V. Therefore I will replace the LED with a high intensity green LED. I have measured the characteristic of the optocoupler with this LED and it's excellent: PDF, PostScript.
Unfortunately, the ICL7621 has temperature range only 0 to 70 degC and military version is not normally available (Pusterla, Distrelec). Firtunately CA3140 has always range -55 to 125 degC! It comes even in 2-pack as CA3240 and both Pusterla and Distrelec have it! Output short circuit is indefinite so we can leave the 680 Ohm for the LED. However care has to be taken with the output higher voltage. The datasheet guarantees 4V from the upper rail at 25 degC. So I will hook the LED simply to the 5V (now Vcc-3V) and it will be without problem. Then I can decrease the resistor to 560 Ohm (20mA at 13.2-35V).
Now it regulates up to 5V luxmeter (500uA PD current) but the opamp output oscillates at 15Hz (according to multimeter). It makes triangular needle impulses. Long stays at full voltage (13V) then goes slower towards 4.4V then goes quickly back uo.
The ICL7621 has quiescent current 10uA and GBP only 44kHz. With 144x gain it makes 306Hz bandwidth! Another slow element is the photoresistor, then there are some poles (less dominant) at the LED and at the feedback! This is the probable cause why it oscillates. I will replace the opamp for CA3240.
First the PIN diode voltage regulation works until the HV supply is overloaded. No oscillation is shown by multimeter. However the opamp output is at 9V which doesn't make sense. I realized the glass diode in series with the LED is light sensitive and therefore may cause false behaviours. This is experimentally verified with a lamp and black plastic plate. I decided to replace it with BAS140. This will have additional benefit of moving the opamp operating point 400mV away from the lower rail.
CA3240 cannot go to the upper rail - but at least 3V from it. This could could cause the LED to stay lit. I hooked up the LED first to 5V supply but then realized this would overload the RC filtering network, causing serious offset in the RSSI circuit. Therefore I will put a 5.1V Zener in series with the LED. For 1N5338, the Zener voltage will drop between 4V and 3V for 1N5338. That's why I take such a high zener voltage. The Zener has to be biased into the 3V operation otherwise the LED could leak and the photoresistor couldn't reach it's maximum potential. Already 100nA influence the tens-of-megaohm range!
From the 1N5338 graph it looks like 10uA could be the right bias for 3V. The min. power supply is 10.8V, filter cascade max. 0.5V drop, that's 10.3V, minus 3V 7.3V, divided by 10uA is 730k, let's take 680k.
According to ENG-TUPS FORUMS, LEDs have a reverse breakdown of 5V. THe 1N5338 comes in 10% worst precision that makes 5.61V. This extra 0.61V has to be dissipated on the 680k resistor, resulting in about 900nA reverse current. I can't imagine damage by such a low current, even if it were permanent.
The 9V is not caused by the diode because it happens with BAS160 too. It actually oscillates, can be measured with the multimeter, changes from 5Hz in darkness to 53Hz in bright light. The feedback strength is too much because the LED was changed to high brightness (100x more light output!) and the gain remained the same.
I have put a potentiometer into the feedback and determined the boundary of oscillation is at Kc=5x gain and Pc=45ms period with 0.8uA photocurrent. With 50uA it's Kc=4.2 and Pc=13.5ms. Not much difference in gain!
Therefore I will use gain of 1. The PID controller theory says I should reduce the gain to half of the oscillation which is apparently 2x. I will rather take safe side and since the LED shines 100x more than the LED and for LED the gain should have been 144x, for this one it should be 1.44x. It also saves 2 resistors ;-)
Unfortunately, this will create large error. To fix this I will use PI controller.
Using Ziegler-Nichols method, for 800nA Kp=2.25 and Ki=60 and for 50uA Kp=1.8 and Ki=160. I will use the one with slower photoresistor because doing the other way will lead to overshoot when the photoresistor is excessively slow.
To give an extra margin I will redce Kp to 1. However this would increase Ki relatively to Kp and as it looks from the Wikipedia article graph, this would leed to excess overshoots. Therefore I will reduce Ki proportionally too to Ki=27. This corresponds to lower resistor 10M and feedback cap 3.7nF. I will use 3.3nF.
With 3nF it's pretty stable. Tested with a flashing light, when the light goes on it makes a ring. I will try 10nF in the feedback instead of 3n3. It didn't help so I'll put 3n3 back.
The diode regulator wastes energy even in darkness. In my test circuit there is 463uA idle current through the zener. Since the multiplier multiplies 6 times, it is a draw of 2.7mA from the 12V source.
I will use a transistor regulator. First I will reduce the Zener current to 50uA since this will make only 0.3uA idle draw from the 12V. Second I could employ a collector resistor to limit the max. current from the source. However the max. current would depend heavily on the source voltage. Therefore I wil employ a resistor between the emitter and the photodiode. The PIN has max. dissipation 0.1W, the source will give 50V max., max. transfer occurs at half of the voltage or 25V which is 4mA. This resistor is then 6.25kOhm we have to round up not to exceed the power limit, since we have 3 resistors in series in the RF filter, we take 3x2.2k making to 6.6kOhm
The multiplier output should be 69.6V max. and the Zener nominal has to be max. 50V + transistor drop=50.7V. The upper tolerance of BZX84-C from NXP is either 54V (too much, 51V nominal) or 50V (47V nominal). Let's take the supply max. voltage and diode nominal voltage: 69.6V and 47V. That makes 452 kOhm, round to 470 kOhm.
What is the maximum voltage on the transistor? In zero current draw, there will be min. 44V (let's say 42V because of low current through the Zener). That's 25.6V Vcb and 26.3V Vce. For this BC547 is OK.
The highest voltage on the transistor is when the regulator is overloaded. the measurement diodes have a drop of 1.4V, the limitting resistor is 6.6kOhm and the BC547 has min. 110 beta. With 470kOhm for the Zener it makes the regulator behave approx. as a resistor 4272 Ohm. It will form a divider with the 6.6kOhm fed by 69.6-1.4=68.2V. Max voltage on the transistor is then approx. 27V. BC547 has a great reserve here.
Should we specify BC547 or BC547C? If the beta is to low the regulator will droop at max. current of 4mA. What is the multiplier minimum voltage? Say the drop on the transistors is 2*0.7V= 1.4V and the schottky drop is 0.2V. THe minimum 12V power supply voltage is 10.8V this makes 9.2 before multiplication by 6, which is 55.2V. 4272 Ohm will have a drop of 17.2V @4mA which will droop the regulator down to 38 Volt. If I use BC547C, the min. gain is 420, the regulator internal resistance 1.12 kOhm, 4mA drop 4.47V and the regulation will be maintained. Therefore I will specify BC547C. In this calculation I omitted the diode drops on the transistor and effect of the Zener diode.
The circuit with Zener diode doesn't seem to work properly. The drop on the Zener isn't min. 3V as designed for, but 2.5V. After disconnecting the branch and leaving the Zener and 680k alone, the drop is 2.6V. The diode type is BZV85C 5V1. Unfortunately the graph in it's datasheet is too rough to predict behaviour at current less than 5mA. I will place a pot in series with the zener and find the resistance that produces 3V drop on the zener.
189 kOhm make exactly 3.00V on the Zener diode. I will put twice as less there to have some reserve for unpredictable variation in those diodes. So I am using 100k. The drop is now 3.23V in low light and increases to 3.24 when illuminated by table lamp.
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