Purpose of circuit in HP 3465B DMM power supply - Page 1
Forum section: electronics / repair

author: Boet
author_rank: Newbie
timestamp: October 13, 2019, 07:37:15 pm
content: I'm currently working on repairing an HP 3465B DMM - it works fine on battery power, but on mains power the numbers keep jumping around the proper measured value.Now, I found the service manual and have already identified a couple of bad caps, which will probably keep me busy for a while.But piece of the power supply leaves me puzzled about its purpose.I've attached both the circuit and the complete service manual (the power supply is on page 68).The scan isn't too clear, but both transistors are NPN.Q1 is a power darlington (MJE2100), Q2 is a CP4071 (a general purpose NPN according tohttps://www.eevblog.com/forum/repair/finding-a-suitable-replacement-for-transistor-from-hp-agilent-keysight-schematic/).My thoughts...R5 will put Q2 into conduction, pulling the right side of R4 close to Q2 its Uce (which will be < Ube Q1), thereby bypassing Q1, R1, R2, R3. All current will flow through Q2, which not makes sense since Q1 is a power transistor.But also, R1, R2 and R3 will form a voltage divider together with R5, reducing the base voltage on Q2 to a level < Ube, such that Q2 won't go into conduction (and the above story doesn't make sense anymore). R4 will then put Q1 into conduction, also slightly raising the voltage on Q2 its base because of the voltage drop across R3 (only 2.7 Ohms) however not enough to make it conduct.The network of R1, R2 and R3 doesn't make much sense to me either.Neither I do see a capacitance multiplier. A battery charging circuit could make sense, but I don't see how (and the whole meter is fed from the circuit as well).Simulating the circuit didn't help, that simply resulted in the output being exactly the same as the pulsed input (minus a couple of 100 mV loss, depending on load current).As you can guess, I'm quite puzzled.Anyone who has a clue about this circuit?

author: duak
author_rank: Super Contributor
timestamp: October 13, 2019, 08:07:37 pm
content: I believe it's a current limiter. Q1 is the pass transistor. R3 is the load current sense resistor. Q2 reduces base current to Q1 from R4 when the voltage across R3 is much above 1.5 V. R5 reduces the load current slightly when under high line or a short circuit load condition.

author: SeanB
author_rank: Super Contributor
timestamp: October 13, 2019, 08:11:46 pm
content: That is a simple constant current circuit, designed to charge the battery with a reasonably constant current, so long as the mains input is applied, and with a current that is constant, though there will be a varying voltage across the transistor Q1. The voltage drop across R3 is divided down, and used to turn on Q2 when the current rises, stealing base current from R4 from Q1, reducing current flow through it. R5 and C2 provide filtering, to prevent oscillation as the voltage varies during the mains cycle, along with reducing current in the charge circuit as voltage rises, to reduce power dissipation in Q1 as the voltage increases, or with a very flat battery connected, or if the battery is short circuit, to limit the power in a fault condition.. You probably want to replace C2 with a film capacitor, it will last longer there.There will be a drop of around 4V across the circuit before it regulates fully, but it is a nice simple rugged circuit provided you have adequate heatsinking on Q1 in operation, especially for short circuit operation.

author: Boet
author_rank: Newbie
timestamp: October 13, 2019, 08:34:23 pm
content: Thanks for your insight, duak and SeanB!A constant current supply was my first thought as well, but I wasn't familiar with this specific topology...With your explanations I now understand better how it operates. :-)I noticed an oscillation on the output voltage as well, I'll replace C2 and see if it helps.C2 is next to the non-heatsinked Q1, and judging from the colour of the PCB the cap is probably toast.