lab power supply design - Page 1
Forum section: electronics / projects

author: xReM1x
author_rank: Regular Contributor
timestamp: October 05, 2015, 12:36:58 pm
content: Schematic:https://i.gyazo.com/5a083c121a1b25fbf66f0b0f7ac5e045.pngis there anything I should change?I'm going to use MAX472 instead of the differential amplifier.and OP275 op amp instead of the LT1013.

author: dom0
author_rank: Super Contributor
timestamp: October 05, 2015, 01:48:51 pm
content: There are several flaws in this circuit, but it might still just work. Transient response will be sluggish at best and current regulation will be very poor due to U4 (a simple op amp current divider will have no errors beyond basic resistor inaccuracy and the alpha error of the "downstream" pass transistor).If you want to know how a proper lab-grade PSU is done, look at the schematic of commercially available units (e.g. HP/Agisight).

author: c4757p
author_rank: Super Contributor
timestamp: October 05, 2015, 01:55:44 pm
content: Q9-Q8 has extremely high voltage gain and so will be hard to control - you may find it's unstable without a load on the output. A resistor in series with Q8's emitter could solve that.What is the purpose of M1?

author: xReM1x
author_rank: Regular Contributor
timestamp: October 05, 2015, 02:25:17 pm
content: Quote from: dom0 on October 05, 2015, 01:48:51 pmI'm going to use MAX472 and not U4..Quote from: c4757p on October 05, 2015, 01:55:44 pmcan you explain more? just a resistor in Q8 doesnt work..the purpose of M1 is that I added it because without it, the current regulation will go max to ~3A and not 5A.

author: c4757p
author_rank: Super Contributor
timestamp: October 05, 2015, 02:34:15 pm
content: How is M1 supposed to provide current regulation?Anyway - yes, I can explain more, but what do you mean by "just a resistor in Q8 doesnt work"? I promise you, it'll work for what I'm saying.The voltage at Q8's emitter is constant, so it can be considered to be a non-emitter-degenerated common-emitter amplifier. The gain of such an amplifier is - assuming low output current, say 10V output with no load except the 10mA drawn by R21:AV= -ROUTgm= -ROUTIC/VT= -1k (10mA / 26mV) = -385Where:AV= gain of the amplifierROUT= output impedance, here R21 = 1kgm= transconductance of Q8IC= collector current of Q8VT= thermal voltage = about 26mVThat is further multiplied by the gain of the amplifier formed by Q9 and its resistors, which is -330/14 = -23.6.So the total Q9/Q8 gain is -385 * -23.6 = 9066.Waaaaaay too much. U5 isn't meant to control something so fussy. It's like trying to drive a car that has a very, very sensitive accelerator while having a heavy weight tied to your foot...Adding an emitter resistor to Q8 will decrease its gain significantly from 385.

author: xReM1x
author_rank: Regular Contributor
timestamp: October 05, 2015, 03:04:06 pm
content: Quote from: c4757p on October 05, 2015, 02:34:15 pmM1 isnt providing current regulation, but without it the current is limited for unknown reason to 3A and not 5A.it does some magics that make this psu work correctly.what is the resistence value for the resistor R1?is it supposd to be connected like this?

author: c4757p
author_rank: Super Contributor
timestamp: October 05, 2015, 05:35:45 pm
content: Yes, it should be connected like that. Rule of thumb - choose the resistor to drop about a tenth of VBE(so 65mV) at the full load current. It can be chosen more 'properly' via loop stability analysis but I think that's a bit beyond our scope right now