Comment summary #47
The comments are shown without formatting, links and images.
Comment 461 ... 470
| date | article | author | comment |
| 02 aug 2014 17:02:44 | Fast Lux meter | Freddy | Only R3 has to be changed, so the virtual ground will be approximately 1 volt. If the frequency isn't an issue, then the feedback C1 may be increased more to reduce noise. Say a hundred times greater value. |
| 02 aug 2014 15:37:48 | Fast Lux meter | Josh | Thanks Freddy, I'll try the LTC1050 and switch out R3 (and R2 as well?) for 1kΩ. Frequency actually isn't an issue for me, I was drawn to this article because of it's specific use of the BPW21R, which I desired for it's spectral response similar to the human eye, and not so much for the "fast" aspect of the lux meter. The highest frequency I'll ever need is to take 1 measurement per second. |
| 02 aug 2014 15:15:22 | Fast Lux meter | Freddy | The problem is the limited output voltage swing of the OP37 opamp. The maximum output voltage is 3 volts (or 4.8 volts in worse case) lower than the supply voltage. So when the circuit is powered with 12 V, the virtual ground (set with R2, R3) is at 6 volts. Subtracting from this the 3 volts reserve, than the maximum output voltage is only 3 volts (or 1.2 V worse case). You could increase the supply voltage (max 22V for the OP37) but there's a risk of destroying the BPW21 due to it's maximum reverse voltage of 10 volt. Also the maximum frequency range will decrease at higher output levels. It is better, as you suggested, to make selectable ranges by switching the feedback R1, C1. Keep the leads to the switch as short as possible. Another solution is to take a rail-to-rail opamp like the LTC1050 and set the virtual ground at a lower voltage lower by decreasing the value of R3 to 1kΩ. You should than have a possible swing of 10 volts, but the frequency range will be smaller. |
| 02 aug 2014 14:23:46 | Fast Lux meter | Josh | Thanks for the quick response Freddy! I did read that section, and I hesitate to decrease the value of R1 much because I would then loose my lower range of measurement, which is just about perfect right now. If I'm doing the calculations right, with R1=50kΩ I can measure a 0.1mV reading for illumination levels of about 0.2 lux (assuming 10nA/lux) up to a 4.88V reading at just short of 10 klux. For my application (an integrating sphere) I need a very wide measurement range. Looking at the BPW21R datasheet, it looks like it should have a linear current response well beyond 10 klux, so it's not the photodiode that is causing the clipping. Do you know what part of the circuit causes the clipping at 4.88V? Ideally I would like to raise that upper voltage limit, but if that's not possible I'll build in a switch to be able to be able to easily go between two R1 & C1 values for a low sensitivity range and a high sensitivity range. |
| 02 aug 2014 14:02:58 | Theory and Definitions | Freddy | Simply calculate the area (P*t) under the curve. Because the curve is exponential it is hard to do. You can ease this by dividing the total time in smaller time chunks, and calculate the the energy in each chunk (Pav*dt). After that sum all the chunks (Etotal = A+B+C+D...). The example above shows how to do this. (only the first five chunks are calculated.) A = -11.3*10^6 W * 0.001 s = -11300 J B = -6.25*10^6 W * 0.001 s = -6250 J C = -4.0*10^6 W * 0.001 s = -4000 J D = -2.5*10^6 W * 0.001 s = -2500 J E = -0.5*10^6 W * 0.001 s = -500 J Total energy Etot = A+B+C+D+E = -24550 J If you want a more accurate calculation divide the total time in more chunks. |
| 02 aug 2014 12:45:12 | Fast Lux meter | Freddy | The circuit draws only 4mA, so, a 1A supply is a bit overkill but should be fine. The maximum reading of 4.88 volts is because the circuit is clipping caused by a high light exposure. You can make the circuit less sensitive by lowering the value of R1. See chapter "Feed-back R1 & C1". |
| 02 aug 2014 10:25:35 | Theory and Definitions | Raghavendra H R | Good afternoon Sir , just I red your reply. I used BLDC motor which is AC type of motor. now let i explain the voltage graph of BLDC motor. During 0s to 1s the bldc motor is running i.e called motoring mode of BLDC motor in that time period power is zero. After 1s to 1.0097s means between fraction of seconds i got a POWER CURVE in braking mode of BLDC Motor . see At t=1s, power is peak value =-14*10^6 in negative direction of a graph ( the power curve coming downwards at 1s which is the peak value ) then the power curve is dacaying up to the time period t= 1.0097s . Finally the time period t= 1.0097s the power bacame zero means coming back to original zero position. in this curve i dont know how to calculate enery please solve out my problem quickly. if its graph not understand clear to you then write your email ID in this massage site i will take one photo of that graph and then mailling to you today. since the power curve is exponentially decal i.e at t= 1 p=-14*10^6 which is high value and at t=1.0097 p=0 which is low vaue of power. hence power is decaying from the period 1s to 1.0097s . THANK YOU SIR |
| 02 aug 2014 03:24:49 | Fast Lux meter | Josh | Hi, I've put together this circuit as shown in the post above by Freddy "At 16 feb 2014, 23:47:08", with the multimeter and power source connected as drawn in upper version. I'm fairly certain I've built it correctly, except for a few alterations: my power supply is a plug that converts AC house current to 12V DC with a max amperage of 1A; C2 and C3 are not polarized (I was told this should be fine); and the IC is OP37GPZ-ND (I couldn't find anything just labeled OP37). When I fire it up, it seems to respond correct at low illumination levels, but then at higher illumination levels seems to max out at about 4.88V reading on the multimeter. Is this a limitation inherent in the circuit design, or have I made a mistake? My best guess would be the power supply's 1A limit is causing trouble, but I don't know for sure. Any ideas? |
| 02 aug 2014 01:49:03 | Theory and Definitions | Freddy | I assume that you're speaking about breaking an electric motor. Normally the breaking energy is calculated by integrating the product of the measured voltage and current during the breaking action. Is it an AC or DC type and can you describe the curve shape? |
| 01 aug 2014 12:39:36 | Theory and Definitions | Raghavendra H R | sir, during braking mode i got power curve in that curve how to calculate energy I knew formula E=P*t . my power curve is between 1s to 1.0097s, that power curve is delay and its peak value is -14*10^6. then reply in that situatiobn how to calculate energy. |
