Buy LMMP – TEXAS INSTRUMENTS – Fixed LDO Voltage Regulator, 15V in, V Dropout, V/mA out, SOT at Farnell element order. The ohm resistor is only required for the V version of the LD to maintain minimum regulation, with a 10 mA minimum load. This is the basic LDV33 voltage regulator, a low drop positive regulator with a V fixed output voltage. This fixed regulator provides a great amount.

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Dropout can be as low as 1V at 20 mA at C!!!

If this was a mistake, please flag the comments so a mod can delete both. Could someone tell that if there is any advantage of using the resistor here? SO the ohm resistor shown for R1 would not meet the LM minimum load requirement worst case.

No real 3.33v of note except that typical value given for R1 in about every example circuit I’ve seen violates data l,117 spec for minimum current at no load. With R1 set, R2 can now be dimensioned to achieve the desired output voltage.

R1 will always have 1.

As others have noted, R2 need to be small enough such that Iadj voltage drop in R2 can be ignored or it must be allowed for. When ‘designing” a circuit rather than just “making it work” it is essential that the worst case parameters are used. The datasheet has suggested to use two capacitors, 0. Post Your Answer Discard By clicking “Post Your Answer”, you acknowledge that you have read our updated terms of serviceprivacy policy and cookie policyand that your continued use of the website is subject to these policies.

Post Your Answer Discard By clicking “Post Your Answer”, you acknowledge that you have read our updated terms of serviceprivacy policy and cookie policyand that your continued use of the website is subject to these policies.

That saida ohm resistor is only a 27 mA load at 3.

Either a lower value of R1 must be used or a minimum external load suitable to bring the total up to at least 10 mA must always be present. A maximum minimum is a nice concept: This doesn’t have much of anything to do with 33.3v op-amp. Home Questions Tags Users Unanswered. When I remove the resistor and leave the capacitors, it does go back to 3.

By clicking “Post Your Answer”, you acknowledge that you have read our updated terms of serviceprivacy policy and cookie policyand that your continued use of the website is subject to these policies. When I remove the resistor, I am getting the expected Vout 3. Even a 20k here would cause a change of 0.

### voltage – Resistors values to use with LM – Electrical Engineering Stack Exchange

The adjust pin current has a maximum of uA. It’s given by the following equation: When I follow the circuit, I am getting Vout as 3.

The datasheet has a spec section for multiple output voltages. You also have to take into account Iadj, which is around uA.

## 3.3V Regulator Board Using LM1117-3.3V IC

You may need a larger input capacitor depending on how far away your power supply is. It’s unneeded for the 3. Worst case, with llm117 load, R1 provides a convenient way of providing the 10 mA while also providing a nicely “stiff” divider. My question is, what if I use 2K and 3. So efficiency must be lower than the maximum possible in most cases. If it’s not within range at a higher load up to mA it may be bad. Looked at data sheet – you were correct: So the higher you have R1, the more “error” Iadj will cause, as it starts to become a significant part of the overall current.

The output voltage is determined not by the lm17 of R1 to R2. IF the external load always draws 10 mA or more then all is well. Changed my answer to accomodate. What constitutes ‘worst” will vary with the parameter and, in some cases, you may have to use the mimimum value of a parameter for one design calculation and the maximum value of the same parameter for another calculation. JGord – Did your comment end up on the wrong post?

Make sure your capacitors are as close to the LDO as possible. Where Vin is more than about 2V above Vout it is the regulators job to drop the excess voltage. I know that ratio R1 to R2 determine the output voltage of LM