Home Community

Recent Posts

Recent Posts

Pages: [1] 2 3 ... 10
1
Electronic Projects Design/Ideas / Re: New Power Supply II
« Last post by redwire on December 21, 2014, 10:37:44 PM »
Hi Redwire,
On the original and fixed version your diode D2 was a resistor with almost NO voltage drop. Now the D2 has a voltage drop that changes when the temperature changes then the output voltage changes when the temperature changes.

Without a negative supply for the current control opamp the output current will not be regulated when it is set to less than about 0.8V/0.27 ohms= 2.96A. The output voltage of the current control opamp must be able to go to -0.7V so that regulation at low currents works. Then its negative supply must be at least -0.9V so use two diodes like I did.

Thanks for the reply.    I guess there is no getting around not having a negative charge pump with this type of design.   I'm curious to see how this guy gets around the problem.  Considering he is using a .1 ohm resistor, the theoretical load would be substantial before there would be stability.   I think from a practical standpoint if there is a heavy load and 0.47 ohm sense reisistor, the ground for the current op amp is pushed low enough down to provide some control.

I am also going to re-look at the design on the first post here.   I seemed to have the oscillation under control and I could change the voltage smoothly.   I was having some current control issues in that the opamps seemed to be fighting each other until the current setting was set to its lowest setting.   If I set the voltage at 14V and the current at 100mv, then connect a load of 1.5 A the current control would never kick in.  It was like in a runaway mode.





2
Electronic Projects Design/Ideas / Re: New Power Supply II
« Last post by audioguru on December 21, 2014, 08:43:03 PM »
Hi Redwire,
On the original and fixed version your diode D2 was a resistor with almost NO voltage drop. Now the D2 has a voltage drop that changes when the temperature changes then the output voltage changes when the temperature changes.

Without a negative supply for the current control opamp the output current will not be regulated when it is set to less than about 0.8V/0.27 ohms= 2.96A. The output voltage of the current control opamp must be able to go to -0.7V so that regulation at low currents works. Then its negative supply must be at least -0.9V so use two diodes like I did.

The value of your R9 is MUCH too low.

The current sertting pot should have a low value resistor in series with its lower end so that any input offset voltage of the current control opamp does not have a huge effect.
3
Electronic Projects Design/Ideas / Re: New Power Supply II
« Last post by redwire on December 21, 2014, 10:33:19 AM »
Here is the sketch with the diode on Pin 3 to compensate for the diode drop on the current op amp.  Would this work?
4
Electronic Projects Design/Ideas / Re: Can someone identify what these diodes are for?
« Last post by Hero999 on December 20, 2014, 07:16:46 AM »
Yes, it's an OR network. The diodes make it so the op-amps can only divert current away from the bases of the driver transistors.

A1 is the voltage error amplifier and A2 is the current error amplifier and the OR network stops them from fighting one another.

The drop-out voltage will be huge, made worse by the ORing diodes. All those transistors are connected together to form a lagre three stage Darlington so the over all voltage drop will be over 3 diode drops, plus the saturation voltage.
5
Electronic Projects Design/Ideas / Re: stepper motor output waveform
« Last post by Hero999 on December 20, 2014, 07:03:08 AM »
Yes, you need an oscilloscope.

Spin the motor at a constant speed, perhaps use a drill but don't go too fast as very high voltages will be produced, and measure the waveform using an oscilloscope.
6
Electronic Projects Design/Ideas / Can someone identify what these diodes are for?
« Last post by liquibyte on December 20, 2014, 12:57:34 AM »
I found this while searching out something else and I modelled it because I was curious.  I'm having a problem getting good results because of VD5 and VD6.  Regular diodes won't work here, or at least I don't think so.  Are these supposed to be references/zeners or something else altogether?

Edit: OK, figured out it's supposed to be an OR network.  I still can't get it to work right though.
8
Electronic Projects Design/Ideas / Re: Dual 1-15Vą tracking
« Last post by liquibyte on December 19, 2014, 10:05:53 PM »
How's this look?  I can't quite get the limiting fully linear but it seems to work for the middle range of the pot well enough.  I chose the 0.05 ohm sense resistor because I have a meter on the way that has that for its sense value, otherwise I'd have probably gone with 0.1 ohms.  I also kept the pots at values I have on hand at the moment.  It's hard designing around parts you have instead of what you should be using.
9
Electronic Projects Design/Ideas / Re: stepper motor output waveform
« Last post by hubble on December 19, 2014, 09:20:30 PM »
OK. Then how can the nature of this type of unknown waveform be found? Do we need some kind of oscilloscope?

Thanks.
10
Electronic Projects Design/Ideas / Re: New Power Supply II
« Last post by redwire on December 19, 2014, 07:37:13 PM »
The op-amp will need a negative supply to compensate for the diode voltage drop and that of the op-amp's output stage.

Yep, your right.   could you add another diode on Pin 3 of the Voltage op amp after  the other diode leading to the current op amp.  This way the voltage set for 0 V output is actually  0.7 on the line leading to the non inverting pin.   Then the current op amp would only need to bring pin 3  the line down to 0.7?   

.
Pages: [1] 2 3 ... 10