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Old 7th August 2008, 02:23 PM   #1
akis is offline akis  United Kingdom
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Default Independent (floating) supplies

Hello

I have a working circuit on which, as an addition, I need to sense the voltage across a resistor and then translate that voltage.

However I cannot think of an easya way to do it. For example I have made a small differential amplifier where there is no voltage difference between the bases of the two transistors (by biasing them very carefully) but in order for that arrangement to work the power supply to that amplifier must be isolated from the main circuit. Similar to how we attack the multimeter terminals across a resistor and we measure the voltage drop across it, without influencing the circuit.

So I have made all resistors around my noddy differential amplifier to be 220K (perhaps I should even increase them) so that I present a high input impedance. But how do I go about providing a separate power supply to it?

The only way I can think of is to attach another winding on the transformer, and take it all the way from back there.

But of course there may be a much easier way of doing this, I cannot be the first person hitting upon this problem ?
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Old 7th August 2008, 02:44 PM   #2
Elvee is offline Elvee  Belgium
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Default Re: Independent (floating) supplies

Quote:
Originally posted by akis


But of course there may be a much easier way of doing this, I cannot be the first person hitting upon this problem ?
No you aren't:
http://www.google.be/search?hl=en&as...i=&safe=images
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Old 7th August 2008, 03:00 PM   #3
akis is offline akis  United Kingdom
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wow, thanks, I am reading...
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Old 7th August 2008, 06:37 PM   #4
akis is offline akis  United Kingdom
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Ok after reading I see that what I am trying to do is not easy. There is talk of opto-isolators, hall-effect devices as well as high side and low side sense resistors and circuits. All of which either take their supply from what is being measured or require their own power supply. Unless I am missing something, and this would not be very unlikely as I am not formally trained in electronics, I still need a means of an independent power supply. Do I have to resort back to the mains transformer then, or is there a neater solution?
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Old 8th August 2008, 06:36 AM   #5
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I have also wondered about this a few times. I took a quick gander and seen this figure. It looks like this IC scenes voltage across a shunt and allows current to flow from Vin through Rout to ground thus raising Vout above ground.

Or its just 2:30 in the morning and i don't know what i am talking about...hehe
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Old 8th August 2008, 06:51 AM   #6
akis is offline akis  United Kingdom
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This is fine because the sense resistor is in line with the + supply rail. The diagram also does not show if the IC requires its own, individual power supply. Basically the two electrodes connecting to that sense resistor must be 100% neutral, as if they are not even there, they must not affect the voltages on the sense resistor and they must not get affected by the voltages present on that sense resistor.

PS: 7:50am here :-)
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Old 8th August 2008, 12:57 PM   #7
Bakmeel is offline Bakmeel  Netherlands
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Could you post a small schematic of what you intend to do?

Other than a current sense circuit I'm not sure what you want to achieve... If it is a current sense circuit, I may have something for you...
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Old 8th August 2008, 01:03 PM   #8
akis is offline akis  United Kingdom
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I am building a simple hFE transistor checker where the current going into the base of the transistor under test is sensed by a resistor. I can measure the voltage using my voltmeter and then do the calculations on a piece of paper, but I want an automatic way of doing that.

Fist step is to measure the voltage drop. There are more steps after that , but that would be my next question. :-)
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Old 8th August 2008, 02:05 PM   #9
Elvee is offline Elvee  Belgium
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Quote:
Originally posted by akis
I am building a simple hFE transistor checker where the current going into the base of the transistor under test is sensed by a resistor. I can measure the voltage using my voltmeter and then do the calculations on a piece of paper, but I want an automatic way of doing that.

Fist step is to measure the voltage drop. There are more steps after that , but that would be my next question. :-)
This can be done in a very simple manner: you need to insert a current mirror between the source for your base-bias (potentiometer, power supply..?) and the base resistor: common to the bias source, one leg to the base resistor, and the second leg is available to you and will source a "copy" of the base current into any ground (or anything) -referenced resistor.
The base-bias source will just have to supply twice the original current.
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Old 8th August 2008, 02:38 PM   #10
Bakmeel is offline Bakmeel  Netherlands
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Indeed.

I have attached a small schematic of a current-mirror based current sensor. It was originally designed for higher currents, but if you play around a bit with the sense resistor you can achieve more sensor gain.

The block in the centre is a dual monolythic PNP transistor. This will ensure that the current mirror operates more accurately. If you can't get one, try with well matched transistors.

The two other transistors are basically just followers to supply proper output impedance.

Hope this helps.
Bakmeel
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