differential amp

If you had, say, a current sensing resistor in the top rail, could you use a circuit like this: (view in monospace font)

I-->-+------Rs------+---->

| | | | 100k 100k | | | | +--, ,--+ | | | | | | | | | 100k 100k| | | | | | | | | | | 10k | | | '-/\\/\\/\\-' | | ^ | | | | ,-------, | | | | U1A | ,-10k--, | gnd | '-|-\\ | | | | | | >--+--1k--+-|-\\ | '----------|+\\ '---|+/ | >-+--out U1B | >--+-----1k----+----|+/ ,-|-/ | | U1C | | 10k '-------' | gnd U1: TLV274

The resistors connected to the 10k pot bring the signal down in the range where the TLV274 can measure it, and the pot nulls out error. Workable?

Reply to
kell
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It is workable, but requires that the 4 divider resistors drift the same way. If you use a pair of transistors in common base configuration (emitters to the current sense resistor), you can cut down on this requirement and get some voltage gain out of the voltage translation.

Reply to
John Popelish

kell ha scritto:

Search: High Side Current Sense in Google You'll found integrated circuit with all and more....

Emanuele

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Reply to
Emanuele

I'm trying to picture the circuit from your description. Does it look like

I-->-+------Rs------+---->

| | \\e e/ pnp |----------| pnp /c c\\ | | | | 100k 100k | | | | +--, ,--+ | | | | | | | | | 100k 100k| | | | | | | | | | | 10k | | | '-/\\/\\/\\-' | | ^ | | | | ,-------, | | | | U1A | ,-10k--, | gnd | '-|-\\ | | | | | | >--+--1k--+-|-\\ | '----------|+\\ '---|+/ | >-+--out U1B | >--+-----1k----+----|+/ ,-|-/ | | U1C | | 10k '-------' | gnd

Reply to
kell

You are getting close. You need to put the offset correction at the emitters. You also don't need the upper half of the voltage dividers, since the two collectors act as variable current sources. Oh, and you need a mechanism that applies a voltage to the 2 bases that produces an average voltage at the collectors that you need for the opamp inputs. You can linearize the emitter voltage to collector voltage gain by adding a pair of series emitter resistors.

Reply to
John Popelish

Next try

I-->-+------Rs------+---->

| | 1k 1k | 10k | +--/\\/\\/\\/\\/\\--+ | ^ | | | | | | | \\e | e/ pnp |----+-----| pnp /c | c\\ | | | | | | | | | | 10k | +--, | ,---+ | | | | | | | | | | | 100k | 100k | | | | | | | | | | | | | | | | | '---+---' | | | | | | | ,-------, | | | | U1A | ,-10k--, | gnd | '-|-\\ | | | | | | >--+--1k--+-|-\\ | '----------|+\\ '---|+/ | >-+--out U1B | >--+-----1k----+----|+/ ,-|-/ | | U1C | | 10k '-------' | gnd

Reply to
kell

U1: TLV274 Or instead of having the pot wiper connected to the transistor base node, I could connect the pot wiper to ground and use a fixed base-emitter resistor on each transistor.

Reply to
kell

Something like this might be made to work. Do you see how it gives you a lot more voltage gain than the original divider method? think about what you would have if you replaced the two collector resistors with a current mirror (to subtract one collector current from the other). Then you wouldn't need a subtractor output, but only a current to voltage converter.

There is also a much simpler way to produce the ground referenced single ended output with a single opamp that has a common mode range that includes its positive rail.

formatting link

Figure 8 shows the general concept.

Sorry if you just wanted a schematic of a finished high side current sense amplifier. I am more interested in exploring possibilities and circuit concepts with you.

Reply to
John Popelish

I'm totally on the same page with respect to intellectual curiosity.

I already knew about jfet op amps like the TL082 or LF353 where the common mode goes all the way to the top rail. In fact I've seen the very same circuit you pointed out in Figure 8 of the maxim app note in regard to high side current sensing, but kind of forgot. And well, to tell you the truth, this thread isn't really about high side current sensing. I do want to amplify a differential voltage at or very near the top rail, but there will be more impedance behind it. As a matter of fact, this is about the self-regulating heater we discussed in another thread on s.e.d. I'm still working on it! And here's the latest concept:

14v | ,----+---+----, | | | | .01 | | | | | | _ ,------+---------1k--------+ 15k / ^ | | | | \\ | | = 1000uF | | / | | | ,-----+---10k--|----+-->\\200k| | gnd | | | / | +---------, | =100uF load \\ | | | | | 0R5 | | \\c /+|-' | gnd 25uH | | |--< | | PTC heating | | /e \\-|---' coil | | | U1 | | | | '--------+----+ | | | ,----------------+-------100k--------+ | | | | | | = 100n | | | | | | | gnd | | '-|-\\ |\\U3 |-' | U2 | >---| >o---------------------|| +-----|+/ |/ |-, | | '------------------Rg--------------------+ | gnd

U1: TL082 U2: LM393 U3: inverting driver

Still using the concept of the resistance bridge to detect the temperature of the heating element, but in a different way. Before, I tried to do it going cycle by cycle, using the comparator to turn the heater off when it reached the setpoint and the big problem was, when you turn the heater off, everything stops. And even if there's a way to get around that issue, the heating coil has inductance, and when the comparator tries to turn the heating element off, it produces a huge spike of negative feedback.

In the latest version of the circuit the comparator keeps running as a voltage controlled oscillator that regulates the power through the heating element. The RC network connected to the heater resitance bridge filters out ac, smoothes out the voltage across the resistance bridge and the diff amp sees this mean dc voltage that follows the actual temperature of the heating element. The resistance of the heating element changes about 2 or 3 tenths of a percent for each degree change of centigrade. To keep the heater within a ten degree window, we want the diff amp to go to the rails when the voltage across the sense resistor changes about 2 or 3 percent. Now we have 25 amps going through 0.01 ohms for .25 volts, and 2 percent of that is 5 mV. Amplifying that to 15 volts requires a gain of 3000. So I guess Rg would need to be... 3 megs?

Reply to
kell

speaking roughly of course. And the circuit should look more like this. Maybe it doesn't have too many mistakes this time. I'll have to stick a pot in somewhere to null op amp error. Gain = Rg/Rd 14v U1A | /+|-----+--------, ,----+---+----, ,--+-< | | | | | | | | | \\-|--, | 100k .01 | | | | | | = 10uF | | | | _ | '-------' | '---+ 15k / ^ Rd gnd | | \\ | | | | / | | ,-----+--100k--|----+-->\\200k| | | | | / | +---------, | =10uF load \\ | | | | | 0R5 | | '-| /+|-' | gnd 25uH | | ||-< | | PTC heating | | ,-| \\-|---' coil | | | U1B | | | | '--------+----+ | | | ,----------------+-------100k--------+ | | | | | | = 100n | | | | | | | gnd | | '-|-\\ |\\U3 |-' | U2 | >---| >o---------------------|| +-----|+/ |/ |-, | | '------------------Rg--------------------+ | gnd

U1: TL082 U2: LM393 U3: inverting driver

Here's wishing you and yours a very Merry Christmas, John, and many more to come.

Kell

Reply to
kell

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