DMM Chipset Power Supply Question - Page 1
Forum section: electronics / beginners

author: Unknown
author_rank: Unknown
timestamp: May 18, 2013, 08:22:02 pm
content: Hi, I'm working on a hobby project, which is a DMM using this Cyrustek chipsethttp://www.cyrustek.com.tw/spec/ES51990F.pdfI'm having problems understanding the power supply circuit on page 26 of the PDF, which is labelled "Regulator DC 3.0v".My questions are:1. Why does R4 have a value of 0 ?2. What's the circle with +1 in it ?3. If chip pins 91 and 90 (which are referenced as V-) are connected to V1-, why in the circuit does it connect to V- and V1-. Surely if V- and V1- are both connected then it should use either be V- or V1- for consistency?Thanks in advance!

author: codeboy2k
author_rank: Super Contributor
timestamp: May 19, 2013, 12:22:21 am
content: I took a look. it appears to me as follows (it makes sense to answer your questions in reverse order too)3. pins 90 and 91 are the power supply voltage into the chip, it is labeled V- on the chip. It is referenced to AGND. pins 93 and 93 are labeled V+ outputs from an on-chip DC-DC converter, to supply a positive rail to the rest of your circuitry, I presume. I can't see that it's used anywhere else.2. The circle with the +1 in it is actually the + and - outputs of a negative 3V regulator that you must supply. Normally this would bea battery powered DC-DC buck/boost converter. For example, use 3 AA cells, 3x1.5= 4.5V, and you buck/boost it to 3.0V, or a single 9V battery, etc. It's your choice.1. R4 connects the output of the regulator (the -3V, net labeled V-) to the power input of the chip, net labeled V1- . It is done like this because it is desirable to put a 0-ohm resistor in the power rails on the schematic, especially during prototyping, so that you can remove it and place a current sense resistor to measure the voltage across the sense resistor, or simply remove it completely and put an ammeter in series with the power supply. This way you can check the current draw of the device and estimate its useful battery life. Sometimes a 2 terminal jumper is used for this purpose too, especially on prototype boards, I do this all the time. If you don't need this, you can simply connect V- to V1- and continue without R4.

author: Unknown
author_rank: Unknown
timestamp: May 19, 2013, 12:55:19 pm
content: Thank you so much for explaining that; it was really helpful! :-)Another quick question is the transistor arrangement in this pic (Q5-Q8) I guess is it's some sort of over-voltage protection?, but is there a special name given to this arrangement?The other picture shows a suggested frequency input circuit with a different configuration (Q1-Q4); would I be right in thinking that this arrangement clamps any damaging input voltages approx over the input +- 600mV, and if badly overloaded, would give enough time for the PTC to react?

author: Bored@Work
author_rank: Super Contributor
timestamp: May 19, 2013, 01:06:42 pm
content: It is called "using a transistor as a diode".

author: codeboy2k
author_rank: Super Contributor
timestamp: May 19, 2013, 08:53:47 pm
content: Quote from: Bored@Work on May 19, 2013, 01:06:42 pmBAD BAW...You didn't say whyWhy use a transistor as a diode instead of a real diode? The reason here is because a typical PN juntion of a BJT has far far lower reverse leakage current than a diode, even a low leakage diode, which can be 5nA or so. The PN junction of a typical BJT has a leakage current spec of 20-30pA. Check the data sheet, you will see it there.This is important when used on the input of this DMM's integrator, in order to maintain accuracy and not have any leakage current that can affect the accuracy of the integration. So instead of real diodes for input protection, they used diode connected BJTs

author: Unknown
author_rank: Unknown
timestamp: May 19, 2013, 10:30:19 pm
content: thanks for your superb explanation! ;-) I now have all I need to finish off the DMM layout; and then it's time to write the code to talk to the chipset!