LT1634CCZ Resistor value? - Page 1
Forum section: electronics / projects

author: Jon86
author_rank: Frequent Contributor
timestamp: September 19, 2013, 10:33:01 pm
content: I'm using an LT1634CCZ 5v reference to create a simple voltage regulator. I've got it working but the output from the reference is a bit on the low side, 4.8v rather than 5, it's 0.05% so it's well out of spec... I'm guessing I'm using the wrong resistor value? I'm using 3.3k, what am I doing wrong?

author: c4757p
author_rank: Super Contributor
timestamp: September 19, 2013, 10:40:28 pm
content: What is the voltage on the other side of that 3.3k resistor?I'd suspect a faulty chip, though. It looks incredibly happy and versatile, I can't think of many ways to get the wrong voltage out of it. With a 3k3 resistor you could use any input voltage down to 5.033V...Watch out - the LT1634CCZis only guaranteed to 0.2%, not 0.05%, but that's still way out of spec.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: September 19, 2013, 11:16:22 pm
content: Hi,To the original poster. What supply voltage are you using?Jay_Diddy_B

author: Jon86
author_rank: Frequent Contributor
timestamp: September 20, 2013, 05:35:00 am
content: My input voltage is around 11-12 volts. Maybe I've got a faulty DMM? That wouldn't surprise me...

author: Jon86
author_rank: Frequent Contributor
timestamp: September 20, 2013, 06:31:18 am
content: I've got a 1.25v one unopened, I could try that too? That'd rule out it being a chip problem I guess...Can someone give me a hand with the resistor calculations?

author: Clear as mud
author_rank: Regular Contributor
timestamp: September 20, 2013, 03:14:53 pm
content: Sure, those calculations are pretty simple.The data sheet (http://cds.linear.com/docs/en/datasheet/1634fe.pdf) suggests using the reference at a current between 10 uA and 20 mA. If the reference voltage is 5 volts and your input voltage is 11 to 12 volts, then your resistor should drop (11.5 - 5) = 6.5 volts. We don't really have to worry about the + or - half a volt, because this part of the calculation doesn't have to be too exact (as long as there is no load, or a very small load). Your LT1634 does the job of taking an inexact current and converting that to an accurate voltage.So, with 6.5 volts at 10 uA, Ohm's law gives a resistance of 650k. At 20 mA, you get 325 ohms. So, any resistance between those two values should work, provided that you have no other load on the circuit.If you have much of a load, you have to consider the load impedance. What is the lowest resistance to ground of the load, or equivalently (through Ohm's law) what is the highest level of current it draws? In terms of the current, you will want your load to draw less current than you used in your input resistor calculation. As long as that remains true, part of the current can go through the load, and part through the LT1634, which can then regulate the voltage as intended. If your load current goes up, even the tiniest amount above the quiescent (no load) current, then it will start dropping more than the proper amount of voltage across the input resistor, and the output voltage will go under 5 volts and become unregulated.Also, the data sheet suggests that the device regulates best at 10 uA (I see that all the other specifications are given under the condition of 10 uA current), which only gives you a very small amount of load current to play with. So, you can get an op-amp and set it up as a buffer as shown in the first part of the circuit on the product description pagehttp://www.linear.com/product/LT1634. The op-amp input impedance should be such that it's input current is much less than 10 uA. That way, the actual amount of current across the LT1634 doesn't change much, and you get the best regulation. Also, the op-amp's output current should be enough to drive your load, or otherwise you also need a pass transistor after the op-amp.However, if you know your load is always smaller than the current given by the input resistor you chose, you can do without the op-amp and/or transistor. In your case, you chose a 3.3k resistor. We need to calculate theminimumload current available through that resistor, and make sure the load isalwaysless than that amount. I = V/R, so we take the minimum voltage and the maximum resistance to do the worst-case calculation. Your stated minimum input was 11 volts. The maximum resistance should be 3.3k + tolerance. With a 5% resistor it will be 3.3k + 165 ohms = 3.465 k. So your available load current is 11V / 3.465kOhm = 3.17 mA. As long as your load draws less current than that, you don't have to use any other components to amplify the current coming out of the resistor/reference circuit.

author: Jon86
author_rank: Frequent Contributor
timestamp: September 20, 2013, 04:46:12 pm
content: Thanks so much for the detailed explanation, it all makes perfect sense nowI'll get breadboarding and see if that 1.25v one does the same thing.

author: Jibby
author_rank: Contributor
timestamp: October 03, 2013, 05:31:00 am
content: Yes. Reduce the res value.