Buffered voltage reference questions, driving pass transistors - Page 1
Forum section: electronics / projects

author: PioB
author_rank: Regular Contributor
timestamp: February 22, 2016, 08:00:03 pm
content: Hi,I am trying to prototype a buffered voltage reference for a datalogger. It will need to drive some 6-8 350 Ohm Wheatstone bridge pressure sensors. As I can read them ratiometrically, the exact value of the voltage does not matter that much, just that it is as ripple-free as possible. So I thought that using an "LDO-type" topology might be the way to go. For prototyping I went with the single-PNP-configuration as found in e.g.http://www.analog.com/library/analogDialogue/archives/41-05/ldo.pdf(I would like to use a PMOS later, but didn't find one in my parts drawer; what will be the relevant values to look out for that?)I then assembled the circuit as per the attached schematic, but I only get 440 mV at the collector of the PNP output transistor for Uin=10 V. at Uin=8V I get 0 Vout. The final output transistors seem to work, if I drive the PNP to GND, I get 0 V out, if I drive it to the rail, I get full voltage (limited by the power supply current limiter)If anyone has an idea as to what I am overlooking I would be grateful for a hint. At the moment I am just using 1kOhm from Vout to GND as a base-load.

author: Kleinstein
author_rank: Super Contributor
timestamp: February 22, 2016, 08:15:48 pm
content: In the circuit the two transistors are both inverting, so the transistor stage give a large positive amplification. So the feedback is positive making the circuit bistable.The large gain of the two transistors is not easy to make them stable. So why not stay with the single PNP or a ready made LDO regulator.

author: PioB
author_rank: Regular Contributor
timestamp: February 22, 2016, 08:22:33 pm
content: Thank you very much!So with a single PNP, emitter would go to the rail and collector would be Voutput?

author: Kleinstein
author_rank: Super Contributor
timestamp: February 22, 2016, 08:56:18 pm
content: Quote from: PioB on February 22, 2016, 08:22:33 pmThat is right. To make regulation easier, a small resistor (e.g. 5 Ohms range) in the emitter line can help. Also a capacitor at the output might be a good idea. Keep in mind that that the LM358 will not go all the way to the positive rail.If true LDO performance is not needed, I would prefer an Emitter-follower (NPN) with resistor at the emitter. So just the classical circuit to drive capacitive loads.

author: Alex Trofimov
author_rank: Contributor
timestamp: February 26, 2016, 10:23:46 am
content: It seems like many people have seen this post, so it's not of much possibility that the reason is so dumb, but... Didn't you just swap inverting and noninverting imputs of the op amp?

author: Andreas
author_rank: Super Contributor
timestamp: February 26, 2016, 08:28:06 pm
content: Hmm,I would simply use a precision low drop regulator like the LP2950A/LP2951A (perhaps in 3.3V Version) with around 2V more regulated input instead of reference, Op-Amp and powerstage.Or if it has to be low noise the LT1763. (also available in 2.5V ).with best regardsAndreas

author: PioB
author_rank: Regular Contributor
timestamp: February 29, 2016, 03:50:40 pm
content: Hi,thank you all very much for your help; I didn't forget about my thread, but was swamped at work the past week.I think I will go the LDO route on the board. On the breadboard with the components at hand, I couldn't get a really stable voltage, at best I managed 5V + some mV/V of input voltage. As I am looking at a ratiometric measurement the LDO will be the way to go.Thank you very much!

author: Kleinstein
author_rank: Super Contributor
timestamp: February 29, 2016, 03:59:16 pm
content: A true ratio-metric measurement does not care about a stable supply. Its mainly 2nd order effects (e.g. self heating of resistors), that migh still need a reasonable stable voltage. So the LDO is a good solution, because it's simple, not so much be cause it's very accurate.