low-droput regulator questions

i am a newbie at electronics and am trying to build this:

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it isn't working. i may be installing the LM2936-z5 incorrectly. i am working on the assumtion that the middle post is the ground. also, the guy at the electronics store said that the lm2936cz-5 was the same as the lm2936-5 just made by a different company. are either of these correct?

thanks.

Reply to
spencer
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The TO92 version (Like a transistor) has input - gnd - output when viewed holding the pins facing towards you and the flat on the bottom. Download the data sheet from the National site.

-- Regards ........... Rheilly Phoull

Reply to
Rheilly Phoull

Reading from left to right that is !!

Reply to
Rheilly Phoull

You might have fried the regulator. You can easily test it with a few batteries, a resistor, and a multimeter.

4 AA cells, connected up in series, along with a 1k resistor, hooked up as follows:

LM2936-5 .-------. |in out| .-----| |---. | | gnd | | | '-------' |

  • | | | --- | | - | o------------ DMM + | | | ^
  • | | | | --- | | | - | | | | / 1k | 5V
  • | | | --- | / | - | | | | | | v
  • | | o------------- DMM - --- | | - | | | | | | | | '---------o-------'

Hold the LM2936 case with the markings (flat side) towards you, and the pins down. Then the left pin is Vout, the middle pin is GND, and the right pin is Vin.

Use 1.5V AA batteries to test. 4 in series = 6V. The minimum input voltage is 5.5 for the LM2936

created by Andy´s ASCII-Circuit v1.25.250804

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Regards,
   Robert Monsen
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Reply to
Robert Monsen

The CTS-line will put out voltage before the supply voltage is generated. This will blow the temperature sensors. They will then have a huge current draw and the regulator cannot supply that and stays at a voltage around 1V. The whole design shows your digital abilities, but adaequate skills in analog seem to be missing.

--
ciao Ban
Bordighera, Italy
Reply to
Ban

CTS is an input to the PC. RTS or DTR might come up initially, but the inputs are protected using those 4.7k resistors and 5.1V zeners. I'm not sure how much protection that really gives, though. A diode from the SCL and SDA lines to the 5V rail might help if this is really a problem. That would certainly prevent the inputs from getting much higher than Vcc.

This would probably be a better parallel port project, actually, although I think the programming would be harder. Using the parallel port, he could dispense with the regulator altogether, since it's all 5V.

Also, I don't think the OP designed it.

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Regards,
   Robert Monsen
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Reply to
Robert Monsen

thanks for all the responses. i do/did have the regulator installed correctly. at the Vin side i am getting 9V, the Vout side nothing. this is a new regulator(it appears the old one was still good, but i used a new one to be thorough). is 9V to much?

thanks again to everyone for their suggestions and help.

Reply to
spencer

We do not have an RS232 protocoll here, but the designer tries to simulate a two wire bus with SCL (system clock) and SDA (bidirectional data) the sensor only functions if SDA receives the right address, so the SDA will always send from the PC to poll data from the sensors. even if the line cannot go above 6V with the 50mA output current of an RS232 driver, this is enough to drive the sensors into a latch up when the supply voltage is not yet at +5V, generating high supply current and eventually destroying the sensors. I do not think this circuit will work as drawn on the website, even if the guy claims to have won "design" prices, well maybe from Elektor. :-) The only reasonable solution would be to use a dedicated RS232 driver IC here.

--
ciao Ban
Bordighera, Italy
Reply to
Ban

The max input voltage is 40V. Take a look at the datasheet here:

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I'm not sure what to tell you. Did you test the regulator in a separate battery powered circuit, like I indicated in another branch of this thread?

Note that if Ban is correct, there will be a short through the temperature probe, and it'll pull the output voltage down. Thus, if you are testing the regulator in circuit, that may be your problem.

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Regards,
   Robert Monsen
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Reply to
Robert Monsen

sorry it has taken so long to reply, been really busy with work stuffs.

i took a new regulator and tested it with a new 9V battery. to do this i put the regulator on a breadboard, hooked the batterie's - side to the ground on the regulator. the the + side to the Vin. i set my volt meter to "20". at Vin it reads 9.66V at Vout it reads 7.29V. something interesting, according to the pdf datasheet below using the flat part of th to-92 the left pin is Vin. if i hook it up like that the voltage reads the same at both sides(9.33 i think).

i think that i must be missing something.

Reply to
spencer

actually nevermind. i got it to work. using batteries instead of trying to power it off the serial port. thanks for the help.

Reply to
spencer

No, you need to look at the datasheet again. That view is from the bottom. Thus, with the flat part facing you, and the pins down, the pins are

1 2 3 VOUT GND VIN

If you put a reverse voltage on most regulators, they will be destroyed in short order. However, this one has 'reverse battery protection'. I am not sure if that means you can reverse VOUT and VIN, or if it means you can reverse VIN and GND. It actually looks like it might withstand this from the schematic in the datasheet, so who knows?

You may also need a load on the regulator. Use a 1k from VOUT to ground.

Also, the datasheet calls for capacitors on input and output. However, since you are driven by a battery, that doesn't seem that important. If you have a 0.1uF cap and a 10uF cap, put the 0.1uF cap on the input, and the 10uF cap on the output.

Other than that, I don't have any suggestions.

Good luck.

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Regards,
   Robert Monsen
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Reply to
Robert Monsen

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