LM336 to replace LM4040C25 - Page 1
Forum section: electronics / beginners

author: belzrebuth
author_rank: Frequent Contributor
timestamp: November 26, 2019, 10:04:37 am
content: I'd like to try a board while waiting for the correct part to arrive and thought since I've got a LM336 I could use that instead of the LM4040.To my understanding both parts are zeners but the 336 also has an adj pin.Problem is that 336 needs more current to operate and is less precise but of the circuit can handle the current am I going to be okay temporarily?Schematics to follow since I'm on the road..

author: newbrain
author_rank: Super Contributor
timestamp: November 26, 2019, 05:12:53 pm
content: I see no problems as long as the 3300resistor is brought down to ~1000.This will provide (3.3-2.5)/1000 - 150E-6 = 650 µA to the LM336, enough to have it working correctly.The ADJ pin can be left unconnected.

author: not1xor1
author_rank: Frequent Contributor
timestamp: November 26, 2019, 05:21:25 pm
content: Neither LM336 nor LM4040 are zener diodes. They are shunt regulator ICs with a bandgap reference.A proper design of a shunt regulator circuit must take into account:- maximum load current- minimum working current of the shunt regulator IC- power dissipation in the shunt regulator IC and resistorso the value of R8 in your circuit must be:(3.3V - 2.5V) / (max_load_current + minimum_regulation_current)for instance:if your 2.5V load never needs more than 10mA and you use an LM336 you have to use a:(.8 / (10e-3+400e-6)) resistor i.e a 76.9 ohm onetaking into account resistor and voltage reference tolerance (and current leakage through C17) you have to use a standard and lower value resistor i.e. 68 or even better 56 ohmBTW if your load just takes few µA you have to take into account the capacitor leakage that would probably be the same order of magnitudeand of course power dissipation in the resistor and shunt regulator would be:.8^2 / R8 and 2.5 * (.8 / R8)

author: newbrain
author_rank: Super Contributor
timestamp: November 26, 2019, 05:37:27 pm
content: Quote from: not1xor1 on November 26, 2019, 05:21:25 pmAll true, but the schematics indicates a 150µA load, dutifully taken into consideration.Plus I do not expect anything close to a µA o.o.m. leakage from a ceramic 100nF cap...Ma dove li compri condensatori così scamuffi?A 56resistor with no load (or a light one) would force 0.8/56 ~ 14 mA through the LM336 - exceeding recommended operating conditions.

author: not1xor1
author_rank: Frequent Contributor
timestamp: November 27, 2019, 05:54:39 am
content: Quote from: newbrain on November 26, 2019, 05:37:27 pmI was misled by the polarized capacitor symbol (with reverse polarity) and that led me to think it was a 100 µF capacitor...I would not trust much such schematic...You're right about the LM336 limit... I forgot to check the datasheetI think it is better to provide some more information to beginners so one may understand how circuits work and when such information is wrong (like the oversight on LM336 current of my example) somebody would correct that

author: newbrain
author_rank: Super Contributor
timestamp: November 27, 2019, 07:52:28 am
content: Quote from: not1xor1 on November 27, 2019, 05:54:39 amThat is Style 2 from the IEEE/IEC/ANSI standard. It might have been more common in days past, and it's not the preferred one, but still perfectly acceptable.See Chapter 2.2 at page 24 (44 in pdf).I don't understand what you mean with reverse polarity, even considering it polarized.Quote from: not1xor1 on November 27, 2019, 05:54:39 amProbably due to the above misreading; I see nothing wrong with it.Quote from: not1xor1 on November 27, 2019, 05:54:39 amThis is true, I should have made the calculation explicit.

author: not1xor1
author_rank: Frequent Contributor
timestamp: November 28, 2019, 06:04:45 am
content: Quote from: newbrain on November 27, 2019, 07:52:28 amI had not noticed the "+" on the top as it was partially hidden by "C17".That's probably due the fullHD resolution of my laptop and the poor state of my eyes.