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 The Projects Forum Working on an electronics project and would like some suggestions, help or critiques? If you would like to comment or assist others with their projects, this is the place to do it.

#21
01-23-2014, 11:48 PM
 ronv Senior Member Join Date: Nov 2008 Location: Tucson AZ. Posts: 696

Five volts won't be enough. You FETs have a threshold voltage between 3 and 4 volts. That is the voltage from gate to source. So you can see for example if you were drawing 3 amps from the load the source would rise to 3 volts - the gate can only go to 5 so it won't turn on. The second is that the FETs share the same resistor. So imagine one turns on at 3 volts and the other at 4. The one that turns on at 3 will steal all the current. You can add a 1 ohm resistor in each FET. That's a good start. Now one will steal the current until it's source rises to where the other on starts to turn on. One amp thru 1 and 2 amps thru the other. The better solution is to give each one it's own op amp and resistor like you post 8 or 14. Then the gate voltage will go to where it needs to go for each to share equally. That's a pain cause you only have 4 op amps. But you don't need to buffer the pot because the + input of the op amps have very high input impedance. So just tie the wiper on the pot to the + inputs where the buffer goes now. Assuming you change the +5 to +12, add a 150k resistor to the top of the pot to +12 so the highest voltage you can get from the wiper is about 3 volts. This will be 3 volts for each 1 ohm resistor for a total of 6 amps maximum. Or a 10k pot and 27k resistor would be ok too.
You will need to change the gain going to the meter since it's only measuring 1/2 the current now.
Make the 10 nf cap smaller and put it between the output and the - input.
#22
01-26-2014, 03:12 PM
 Dr.killjoy Senior Member Join Date: Apr 2013 Location: Southern New Jersey Posts: 467

Quote:
 Originally Posted by R!f@@ Any value would work there if you know what I mean. But higher values would not load the PSU... Wait...Ur regulator is 78L05...means it can supply 100mA safely. So lower pots will load it too much
I really have no idea what's going on or what your saying ... Can you explain it simply for me ???
I just started learn the math to figure out everything ..

Thanks
Jason Sr
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#23
01-30-2014, 01:54 AM
 Veracohr Senior Member Join Date: Jan 2011 Posts: 192

As R!f@@ said, the regulator can only supply 100mA. The 50k pot at the input of IC2A will draw 5V/50kΩ = 0.1mA from the regulator. The current going into the opamp is so small you can call it 0. If you replaced the 50kΩ pot with a 100Ω pot, it would draw 5V/100Ω = 50mA, which is half the available supply current. If the rest of the circuit required more than 50mA from the supply, the regulator output voltage would drop. Using a high-value pot means minimal current will be used for the reference voltage IC2A is producing.

Theoretically any value of pot will work to get the proper voltage at the input of IC2A, because that's determined by the pot ratio, not the overall resistance. But the overall resistance determines the current draw from the source, which is an important consideration with a low-current supply.
 The Following User Says Thank You to Veracohr For This Useful Post: R!f@@ (01-30-2014)
#24
01-30-2014, 08:52 AM
 R!f@@ Senior Member Join Date: Apr 2009 Location: Euphoria.., The One & Only {GMT +5} Posts: 6,828 Blog Entries: 3

Quote:
 Originally Posted by Veracohr As R!f@@ said, the regulator can only supply 100mA. The 50k pot at the input of IC2A will draw 5V/50kΩ = 0.1mA from the regulator. The current going into the opamp is so small you can call it 0. If you replaced the 50kΩ pot with a 100Ω pot, it would draw 5V/100Ω = 50mA, which is half the available supply current. If the rest of the circuit required more than 50mA from the supply, the regulator output voltage would drop. Using a high-value pot means minimal current will be used for the reference voltage IC2A is producing. Theoretically any value of pot will work to get the proper voltage at the input of IC2A, because that's determined by the pot ratio, not the overall resistance. But the overall resistance determines the current draw from the source, which is an important consideration with a low-current supply.
Well Put .
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