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Kindly provide me the circuit schematic of this Automatic controller' PCB ??

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Dear Danishdeshmuk,
As Dear FVM said that is better try to repair damaged card but if you precis and want to build a replacement so please see some component that maybe help you in this way.
I introduce an IC to you that able to supply 0-32V and 10A :
http://cds.linear.com/docs/Datasheet/1038fa.pdf
Please see page 6 so you can build it Untitled1.png
Regard.

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If you can't find this component you can use this IC :IP138A 0-35V 5A
http://pdf1.alldatasheet.com/datasheet-pdf/view/42026/SEME-LAB/IP138A.html

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If you can't find this component you can use this IC :IP138A 0-35V 5A
http://pdf1.alldatasheet.com/datasheet-pdf/view/42026/SEME-LAB/IP138A.html
 

The card is burnt so badly that i can't repair it ...

In datasheet its written that its the circuit schematic of voltage regulator but i need current regulator circuit schematic .....
would it work the same way ??

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This IC : IP138A 0-35V 5A is for regulating (output) of about 5 Amps only .... ?
 

Okay, you should know that we don't have any feedback for control supply thus we should use constant current voltage regulator that mentioned on the datasheet.

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And you can control machine by some potentiometer as exist on your schematic.
 

but how with pot ? what would pot do & where to connect the pot ?
 

According to datasheet of LT1038 Page 6 you can connect your pot circuit instead of "20K OUTPUT ADJUST" potentiometer .
Also you should change your pot circuit for connecting to regulator.
I reviewed your pot circuit and I think you can use this scheme instead of your pot circuit :
1) Relay K2 use for ON/OFF clutch
2) Relay K1 use for change control by P1 or P3
Thus I think we can use this scheme instead of your circuit and connect instead of LT1038 potentiometer
scheme1.png
 

In this pot circuit where should be ground , where should be the supply connected ?

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Why do you want to change the pot circuit ?
 

Because your circuit need to vcc and ground and output terminal but if you want to use other circuit you should match a circuit to regulator board.
If you see regulators circuit usually you need only to use a pot. for tuning output voltage thus we can replace this pot by matching a circuit like above!
 

where should be vcc connected ? where should be ground connected ? where should be output terminal connected ?
 

As I said if you want to build a circuit you should change your circuit to achieve good circuit.
For example for connecting potentiometer to proposal circuit of LT1038 you should change potntiometer circuit to the circuit that mentioned last post:
connect same pins 1,2 at the both pictures
Untitled2.pngscheme1.png
 

ok

i try then tell you later

what would be its cost ?

How would be its performance ?
 

If you make it yourself so you can complete it around 120-150$
 

The pots that you referred in your given schematic is of what wattage (what power capacity ?) & what type of them single turn or multi turn ??

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asking about P1 , P2 & P3

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Also 20 K1 & 20 K2 relays are of what power capacity and of which type of coil ?
 

but as you are explaining that LT1038 is obsolete so i would have to use this IC : IP138A 0-35V 5A

And as the power handling capability of both ICs whether LT1038 & whether IP138A 0-35V 5A is different , then would the remaining and would the rest of the circuit remain the same ????
 

Just a brief comment. If you look at the original PCB and functional description in the datasheet, the LT1038 solution looks somehow overdone. The original device is apparently a simple PWM controller using the brake coil inductance as energy storage, the usual way to control proportional solenoid valves and other electromagnets in industry. The current range is said to be 0 - 2A.
 

so simple & short , don't try to use LT1038 solution ???
 

I don't say LT1038 can't serve the purpose. Personally I won't use it for this application. I would basically refer to the original design.
 

I don't say LT1038 can't serve the purpose. Personally I won't use it for this application. I would basically refer to the original design.

Would you tell why you won't use that design for this application ?

any damage to the machinery or its other accessories or any malfunctioning ?
 

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