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Modifying Voltage Regulator Design !!!

Please find the attached circuit.Its a linear voltage regulator providing an output of 1.2V (up to 1A) by the resistor divider network and feeds the 1.2V to POWER TEST POINT. I want to modify the above circuit . that provides 1V to the same POWER TEST POINT.What modification should i do to get 1.2V and 1 V from the same voltage Regulator !!!
 

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KrisBlueNZ

Sadly passed away in 2015
This is a straightforward linear regulator. The output voltage is determined by the ratio of R46 to R47.

The reference voltage in the MAX8869 is 0.8V. This is the voltage that will appear at the SET pin (pin 11) when the output voltage is correct.

With R47=30k, the R46 and R47 current will be I = V / R = 0.8 / 30,000 = 26.667 µA. For a 1V output voltage, the voltage across R46 will be 0.2V so R46 will be R = V / I = 0.2 / 0.000026667 = 7500 ohms.

Short answer: Change R46 from 15k to 7k5. This will change the regulator's output voltage from 1.2V to 1.0V.

You cannot get 1.2V and 1.0V simultaneously from a single regulator. It only has one output.
 
Sir. that i know. my question is whether by adding any other circuitry to this can we produce 1V , like any control to enable both 1.2V or 1V ? Or can you help me designing a circuit that provides both 1.2V and 1 V with single output, maximum 100mA-500mA
 
Or instead of intermittently breaking the feedback loop with a SPDT switch, use a SPST switch to add a resistor to the voltage divider in order to more cleanly switch between the two output voltages.
 
Sir I am designing a circuit that works with 1.2V and 1V,these are the input voltages to the system.1.2V is the normal operating voltage and the requirement of 1V is to check a Low Voltage operation of the system. That is when 1V is applied it should give a low voltage indication. So my requirement is to provide 1.2V and 1V respectively. So my concept is to control a circuit that gives 1.2V and 1V. and the problem is no human interaction should be there to control this circuit. ( you said about SPDT switch). it should be software controlled. I have to give these voltages to 4 PCB's. I have earlier designed a circuit using Quad OPAMP. But that was not a feasible solution.
 
i am attaching the circuit i used earlier. I used Quad OP-AMP AD 8534 (not LP324 as in picture) . The 4 circuits give 1.2V and 1V and 0V respectively by setting/resetting OE and F_VIN. The circuit that uses power (1.2V, 1V,0V ) requires 10mA. and AD 8534 gives 150mA output current. and the sometimes the circuit works well. but sometimes the output voltage of OP AMP drops to 500mV. This random behavior of the above circuit forced me to discard this design. Can you suggest any alternative for this.?
 

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KrisBlueNZ

Sadly passed away in 2015
So you want to modify your circuit of post #1 to provide a software-switchable output voltage of either 1.2V or 1.0V?

The simple way to do that is to add a second resistor in parallel with R47 which is enabled by a MOSFET. Unfortunately, turning that MOSFET ON will increase the output voltage, and you really want a circuit where the output voltage is normally 1.2V and decreases when you signal it.

Here's my suggestion. Put the new resistor normally in circuit by pulling the MOSFET's gate up to the +5V rail. The output voltage will be 1.2V by default, and the MOSFET's gate must be driven to ground to drop the output voltage to 1.0V.

suggestion 1.png

BTW when you post diagrams, you should use GIF or PNG format, not JPG. It wastes space and you get nasty compression artifacts in high contrast areas. Zoom in on your image in post #1 and you'll see the mottled background. Using JPG format also makes it hard for others (i.e. me) to edit without wasting space or making a mess.
 
Sir what about using a NPN transistor to enable the newly added resistor ? like the uploaded image ?
 

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KrisBlueNZ

Sadly passed away in 2015
No. A junction transistor has a saturation voltage that's not zero, and poorly defined. This would mean that when you turn the transistor ON, the collector does not pull fully to 0V - it may sit at 50 mV, 100 mV, even 200 mV. The regulator output voltage won't be correct, and it won't be reliable.

Also you need to ensure that the switching device is normally ON, to give you a 1.2V output in the normal state.

Stick with the design I posted. You can use a different MOSFET - just make sure it is guaranteed to have a relatively low ON-resistance (10Ω or less) at VGS=4.5V.
 
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