Please help me understand this HP voltage regulator circuit - Page 1
Forum section: electronics / repair

author: davefr
author_rank: Contributor
timestamp: January 15, 2023, 03:34:35 pm
content: This is a "simple" battery regulator circuit in a old HP 427A multimeter that I'm in the process of debugging. I'm trying to understand the theory of operation but I'm stumped. Input varies from about + and - 8-15V (battery or DC output from upstream PSU). The circuit produces/regulates + and - 6.7VDC at the outputs.I understand how the -6.7V rail is generated. Vzener + Q3 C-E + Q4 B-E = -6.7V. HP says the +6.7 rail is produced using the -6.7 VDC as a reference. Can anyone please explain this? I don't see how Q2 produces the +6.7 VDC at it's emitter? Q4 and Q5 obviously provide some feedback path based on the load. Attached is the schematic and I've noted the typical voltages from the HP manual.TIA

author: wasedadoc
author_rank: Super Contributor
timestamp: January 15, 2023, 04:32:57 pm
content: For the moment, mentally short circuit CR3, 4 and 5. R8 and R9 form a potential divider. Negative feedback of Q5 and Q2 act to make the junction of R8 and R9 (which is the base of Q5) be at one Vbe above zero volts. As R8 and R9 are almost the same value it means that the positive and negative output rails are almost the same voltage in magnitude but opposite polarity. Now mentally remove the short across one of the 3 diodes. That balances Q5's Vbe. Not completely clear to me what the other 2 diodes are for. R8 is slightly lower resistance than R9 because it also has to supply Q5 base current

author: PartialDischarge
author_rank: Super Contributor
timestamp: January 15, 2023, 04:35:19 pm
content: The negative output is determined by the zener and Vbe of Q3The positive is determined by the base voltage of 0,6V at Q5. r8, r9, cr3..5 (which are 0.23V drop germanium diodes) fix this

author: PartialDischarge
author_rank: Super Contributor
timestamp: January 15, 2023, 04:37:07 pm
content: Quote from: wasedadoc on January 15, 2023, 04:32:57 pmThe trick is that those are Ge diodes

author: davefr
author_rank: Contributor
timestamp: January 15, 2023, 05:27:59 pm
content: But what generate the +7.3V going to the base of Q2 and thus 6.7V at Q2's emitter? (one Vbe drop). Is it the combination of CR1, R2 and Q1 that create this 7.3V reference voltage for Q2's base? CR2 and Q3 also generate a 6.7V reference voltage (negative) but I don't see how this influences the base of Q2 and the positive rail of this supply. I understand that the junction between R8 and R9 will be +.5v.It seems like there are two reference sources. CR1, R2 and Q1 for the positive rail and CR2 and Q3 for the negative rail?Thanks for taking the time to reply.

author: PartialDischarge
author_rank: Super Contributor
timestamp: January 15, 2023, 05:35:29 pm
content: The negative feedback between the output and the collector voltage of q5, it self regulates to give the desired base voltage of q5

author: wasedadoc
author_rank: Super Contributor
timestamp: January 15, 2023, 05:46:57 pm
content: R3 and CR1 bias Q1 to provide base current to Q2. Q5 steals the appropriate amount of that to maintain Q2 emitter at the voltage which makes the base of Q5 be at one Vbe above zero.

author: davefr
author_rank: Contributor
timestamp: January 15, 2023, 05:47:23 pm
content: Quote from: PartialDischarge on January 15, 2023, 04:35:19 pmSo the +.5V at the junction of R8 and CR3 AND the -6.7V reference voltage generated by CR2/Q3 determine the current flow thru the divider network. (ignoring Q5's base current). So the voltage at the top end of R8 becomes positive by approx. the same magnitude as the bottom half? And the rest of that circuitry regulates based on loading? I think it's becoming more clear.

author: wasedadoc
author_rank: Super Contributor
timestamp: January 15, 2023, 06:13:33 pm
content: Quote from: davefr on January 15, 2023, 05:47:23 pmNot quite. The voltage at the top of R8 affects the voltage at Q5 base. If Q5 base is lower than the nominal 0.6 Volts then Q5 conducts less leaving more base current for Q2 which raises the voltage at top of R8. If Q5 base goes higher than the nominal 0.6 Volts then Q5 conducts more and Q2 gets less base current so voltage at the top of R8 falls. Equilibrium is when the voltage at the middle of the R8, R9, CR3,4,5 potential divider is close to 0.6 Volts. The resistor values of R8 and R9 are designed so that the positive and negative voltages are nominally equal.

author: floobydust
author_rank: Super Contributor
timestamp: January 15, 2023, 06:17:23 pm
content: Quote from: PartialDischarge on January 15, 2023, 04:37:07 pmWho's germanium? Must be at least some of the transistors. There's no way for Q1 to turn on unless its VBEis much less than CR1 VF.

author: wasedadoc
author_rank: Super Contributor
timestamp: January 15, 2023, 06:35:08 pm
content: Section 4.29 ofhttp://hparchive.com/Manuals/HP-427A-Manual.pdfprovides a brief description of the battery regulator.The parts list in the same document shows that Q1 and Q4 are Germanium transistors. CR3,4,5 also listed as Germanium while CR1 is Silicon.

author: davefr
author_rank: Contributor
timestamp: January 15, 2023, 06:39:10 pm
content: All voltages are within spec except the ones noted in boxes. Output is +1.8V and -1.1V (versus + and - 6.7VDC). Voltage across the CR2 Zener diode is +.5V (vs. 6.2V). Base of Q2 is 2.4V (vs. 7.3v)I suspect CR2 is bad but will probably replace Q2 at the same time. Q1/Q4 and CR2,4,5 are all Germanium components.

author: floobydust
author_rank: Super Contributor
timestamp: January 15, 2023, 07:02:50 pm
content: To me it sorta looks like a lazy Q1, it's not on much if the voltage drop across R2 is so small. Wild guess its design is around 0.7mA CCThese circuits are fun to figure out and get humbled by. HP or Tek did a lot with discretes, the pre-IC era and PNP Ge you had to be clever.I think the diode string is temperature compensation for the battery pack as well.

author: wasedadoc
author_rank: Super Contributor
timestamp: January 15, 2023, 07:19:18 pm
content: If there is only half a volt across the zener then either it or Q3 has a problem. Even if Q2 were not conducting at all, there would be enough current through emitter to base of Q4 and then through Q3 to measure a significant voltage across the zener.

author: floobydust
author_rank: Super Contributor
timestamp: January 16, 2023, 08:03:46 pm
content: I see this regulator also generates mid-point common, making a virtual ground from the floating battery pack.There's no current-limiting resistor for Q3 E-B/CR2 so any monkey business causing the -6.7V rail to go more negative i.e. battery (+) short to COM (holder) would damage them.I would just lift the legs on a few parts and diode-test check them, full analysis is fun but a repair can go quicker just doing quick tests.OP do you have an R22? The ECN Change 3 is confusing, I think they added it:"Remove resistor A1R22 (390Ω) from collector of A1Q4." {should be add it?}"NOTE: A Printed Circuit Board 00427-66501 with revision c or beyond has A1R22 added in collector circuit of A1Q4. Resistor A1R22 was added to reduce power dissipation in A1Q4 during overload."