ICL7107CPL voltmeter configuration

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kenwn1
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ICL7107CPL voltmeter configuration

Post by kenwn1 »

Hi,
I just built a digital voltmeter kit with an ICL7107CPL chip. The meter works perfectly, however the circuit is configured so that the meter's inputs are isolated from the chip's 5V supply inputs. I would like to use this voltmeter in an automotive application to monitor the charging system. I powered it from the vehicle's 12V system through a 7805 regulator but when I try to monitor the vehicle's electrical system, the meter reads 2.2V less than the actual voltage with the IN+ terminal connected to the vehicle's wiring and IN- left unconnected; and 4.4V less than actual when the IN- terminal is grounded to the vehicle. How do I reconfigure the circuit so that it reads correctly?
russlk
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Re: ICL7107CPL voltmeter configuration

Post by russlk »

The inputs have to be at least 1 volt above ground. If you are measuring the voltage, just move the sense resistor up the divider by 1 volt.
Dimbulb
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Re: ICL7107CPL voltmeter configuration

Post by Dimbulb »

if the auto electric sample of sufficient uA is divided by 1000 the meter can be set to mV. This requires two resistors set to divide.
tonmwamuka
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Re: ICL7107CPL voltmeter configuration

Post by tonmwamuka »

Can someone send me icl7107 configurations to measure 0-50 volts. Please.
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Edd
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Re: ICL7107CPL voltmeter configuration

Post by Edd »

Tonmwamuka:<p>Use the contained info in Intersil’s companion application note:
http://www.intersil.com/data/an/an023.pdf <p>Then what you would be doing is reducing the specified 50VDC (or up to 200) down to a 200 MV level by using an
Input resistive voltage /divider bridge along with doing the fixed decimal point re positioning for the new voltage range/scale. <p>ET AL:
For all of you guys that have been in a repetitive common dilemma, that of wanting to enact the metering of a cars
battery voltage, while using that common auto battery as its powering source, thus not requiring an isolated separate
meter battery.
I find that in no way can I build up a circuit and its time/components to circumvent that problem for less cash
outlay that I can get with this panel meter unit, that will work in both manners.<p>MET1031:
http://www.bgmicro.com/pdf/page4.pdf <p>
73's de Edd
[email protected] ..….....(Interstellar~~~~Warp~~~Speed)
[email protected]….......(Firewalled-Spam*Cookies*Crumbs)
:) <p>Image<p>OH BOY ! I can see that it I am not the one stretching the format width, since your scanned info has already accounted for that.
Here is a break down of the sectionalizing of the PS unit schema.
Top left quadrant…….The raw DC power supply
Center top……………....The main power switch mode power components TR1-TR2, Sw power xstrs
and T3 power switching transformer with T2 isolation and driver transformer.
Far right half ……..…Isolated low voltage output sections
Center ………..……..Controller IC for Main power supply
Bottom left quadrant...Standby 5VDC power supply<p>The section that is of interest to you will be in the left center and bottom quadrants. At the very bottom left corner is located the +5 VDC standby, which you will see is fed by the far two lines dropping down along the left side of the schema from the raw DC supply at the very top. The 5 V supply is its own self contained power osc circuit that is powered up at the time its IEC power connector is inserted. The main Switch Mode supply is DC powered up also, but its operation is dependant upon receiving drive from the PWM controller IC….TL 494 in this unit.
Now if you will examine transistor Q10-Q1 combo above that supply, you can then see how that power turn on that you were querying is accomplished. The omnipresent 5 V SBY feeds thru R 23 to the base of Q10 which turns it on and its resultant closed C-E loop holds back the operation of Q1 by virtue of its collector loop to its supply resistor (R17) being negated. Therefore complete circuit to the controller IC's pin 5 can established and no pulse drives from its pins 8 and 11 are established to its power driver circuitry.
Go back to your PS-ON test point at Q-10 base control circuitry and bring that point low and that transistor will no longer be turned on and remove its shunting of Q1…. Then that circuit will activate TL494's drive output and then power up the main switch mode power supply.
Coincidentally, there are also three transistors in the center left quadrant that are coupled into that same power circuitry thru steering diode to that shared (R17) collector resistor, in this case it is Q5's collector that is also sharing it. Thereby, an over voltage fault sensing will shut down the power supply as a protective feature.
That's it…now isn't that ever so much better than a "black box"
I have a wee bit more , but will come back later as an addenda.<p>73's de Edd
[email protected] .........(Interstellar~~~~Warp~~~Speed)
[email protected]........(Firewalled*Spam*Cookies*Crumbs)<p>I'm as frustrated as an arsonist in a petrified forest.<p>[ September 30, 2005: Message edited by: Edd Whatley ]</p>
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