I'm a new poster here. I've posted a couple of times to SED, but always felt that the general level of SED was _way_ above me. SE_Basics_ is more my level.
My questions are about this battery tester (an "electronic load"):
Thanks, Bob
If you aren't designing, it doesn't belong there. You are trying to interpret an existing schematic.
This is all about loading the battery so you can see what the voltage is when under load. A battery may show decent voltage when unloaded, but then drop significantly when under load. This sort of thing means when you measure the voltage, it's more like when the battery is in the circuit.
You could just load the battery with a big resistor, but that has problems, so the fancier circuit.
U1 and Q1, the fet, form a constant current sink, meaning no matter what the voltage the FET sees, it keeps drawing the same current (a resistor across the battery would draw current varying with the voltage on the battery).
Since the circuit turns voltage into a current draw, if there was no voltage regulator, the drain would vary according to whatever is powering this circuit, the 9v battery. I'm not sure that the voltage regulator isn't overkill, but it doesn't add much in cost or size to the circuit, that sort of voltage regulator can be found in plenty of computer switching supplies, though maybe not that specific device.
U1 and the FET form a constant voltage sink.
The resistor is in the feedback circuit to the op-amp. No, it probably isn't needed, it's there for isolation (note that c1, the .01uf capacitor, is probably there to limit frequency response, and without R5, it wouldn't work as well).
U2 is the TL531 in the upper left corner, a precision voltage regulator (kind of like a fancy zener, where you can actually control the point at which the regulator regulates). It supplies a stable voltage to U1.
The other half of the 358, is simply labelled "U" and is completely out of the circuit (it's not used, so likely a dual op-amp was specified because the 358 has certain characteristics that means it works better in a single voltage power supply)
Michael
But technically, if not for the capacitor from the output of the opamp to the inverting input of the op-amp, the resistor is not needed.
It's no different from a voltage follower (or indeed, the other half of the op-amp that lies unused at the lower left, the output connected to the inverting input), though in this case, the FET is in that feedback loop.
But the opamp wants to see voltage, which it's already seeing on the non-inverting input. Since no voltage amplification is done in that stage, there's no absolute reason for the resistor, the voltage at the FET is the same voltage as at the inverting input of the op-amp.
But of course, it isolates the capacitor that's going from the op-amp output to the inverting input of that op-amp.
Michael
No, you need R5 there. It is the feed back sense required to keep U1 operating as a voltage comparator. The + input of U1 is the reference voltage required and the (-)input would be the comparing point, in this case, it is comparing the SOURCE (S) side of the mosfet transistor and will make what ever needed adjustment output on U1 to bias Q1 to get there. If the voltage exceeds at (S) of Q1, compared to what is sitting at (+) input of U1, U1 output will then drop in bias voltage on the gate (G) of Q1. This of course, will cause the Q1 to not conduct as much and lower the voltage at (S) of Q1 to satisfy the voltage comparator circuit of U1.
The output of U1 will not be following the + input reference voltage, it will be higher than the 2.5v max you would get with that circuit.
Looking up the transistor, the sheet tells me you'll need ~3 Volts Plus what is sitting at the Source of Q1 to get it to come on.
That device is like a programmable zener diode. Internally it has a
2.5v fixed reference (Band Gap), which is very stable. You can think of that as a zener diode. That is used as a internal voltage comparator against the control pin voltage.Since the internal is set for 2.5 volts, the component stops clamping the load when it hits 2.5 volts, because the internal comparator as hit the balance point of the internal fixed voltage reference.
Now, if you were to apply a scaled reference of what is appearing at the top side to the control pin, you can then force it to elevate its clamping voltage at a higher point. A scale reference would be like a voltage divider network where it derives its source from the top side of the programmable zener here.
Btw, U2 is the 3 terminal voltage reference, not the OP-AMP.
The 358 is a dual unit, it looks like they are simply terminating the leads due to lack of use so it won't damage the chip.
Hope that shed some light on the subject.
Jamie
Ha? I think you better look at that again, it is very needed..unless you want a run away system.
They terminate un-used op-amps in a package like that to insure no damage comes to them, which would most likely propagate over to the used side. Yes, it is a dual op-amp, to be exact.
Jamie
No, the R5 is needed. The cap there serves to pad the feed back down so it won't oscillate. Just think of adding more miller effect on top of what is already there inside of the op-amp.
In order to have a constant current you need the circuit to monitor the actual current and make calibrations at the gate drive. You can only do this if you have a way to monitor the current and R5 is that. It simply is monitoring the voltage which would be a function of current on those high power R's there.
Jamie
R5/R4/C1 keep both Q1 and U1 from oscillating. There are probably three oscillation modes: Q1 RF oscillating on its own, U1 oscillating from being loaded by the Q1 gate capacitance, and an overall loop oscillation.
U2A appears to do nothing. It looks like the designer attempted to keep it running closed-loop to keep it from affecting the other section, but he didn't do it right.
It would be more sensible to use U2A to buffer the reference, increase R1, and save a bunch of 9V supply current.
-- John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom timing and laser controllers Photonics and fiberoptic TTL data links VME analog, thermocouple, LVDT, synchro, tachometer Multichannel arbitrary waveform generators
Both R5 _and_ R4 are to isolate the OpAmp from a capacitive load, to prevent instability.
The second OpAmp is not U2, it's the second section of U1... it's not used and is "dummy" connected" to prevent undefined currents.
U2 is the TL431, an adjustable shunt regulator, connected here simply as a 2.5V _zener_. ...Jim Thompson
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| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
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I love to cook with wine. Sometimes I even put it in the food.
But quite practically, if not for C1 and its associated resistors then the circuit will oscillate strongly. That circuit is a classic for driving a MOSFET gate without oscillations.
It's considerably different from a voltage follower into a resistive load, though, because the MOSFET gate is very capacitive, and the LM358 (like lots of op-amps) Really Doesn't like driving capacitive loads, and will oscillate like mad if you try.
Unless, of course, you want the circuit to actually work correctly.
-- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
First, you need _some_ connection between the MOSFET source and the negative feedback to the op-amp, because it's by servoing that source voltage to your set voltage that you're controlling the current.
The reason for R5 is because you need R4 and C1 in there for stability: the MOSFET gate is highly capacitive, which slows the response of U1 down considerably. If you just connected U1 straight to Q1, then there would be enough added lag in the loop formed by U1 and Q1 that the circuit would oscillate. R4 isolates U1's output from the MOSFET gate (but leaves in the lag). C1 lets U1's output 'talk' directly to U1's input (which introduces lead, which counteracts the lag). R5 gives C1 a chance to work, while still providing a DC path from the MOSFET source voltage (which is what you want to set) to U1's negative input.
The second op-amp is mislabeled (as is the first). The first one should be U1A, and the second one (which is in the same package) should be labeled U1B.
U2 is the thing that looks like a zener with a wire sticking out of it, hooked back to the cathode -- it's a shunt voltage reference: stick a reasonable current into it, and it'll hold a defined voltage.
-- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
Well, at least it gives people something to talk about. But I still can't see how he expects the circuit to be a constant current source for the test load if you don't bother to account for the sharp knee on the gate turn on voltage point and the load varying due to a battery discharging while under test.
I almost get the impression that maybe he thinks that is a jfet or non-enhanced, when in fact, it's not. Maybe using the correct foot print may have removed the mystery behind that.
Jamie
The op-amp accounts for that. If you put your thumb over C1, R4, and Q1, then what's left is a voltage follower that drives the voltage at the top of R13 to be equal to the voltage at pin 5 of U1. Since (at DC at least) the MOSFET takes no gate current, that voltage is proportional to the battery current, and hence it is the battery (or power supply) current that is being servoed.
The circuit should do a pretty darned good job of holding the load current steady, in fact.
-- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com
I fully understand how the circuit works, but my comment to the original statement is, that R5 is needed when it was thought it could've been removed because it looked like it wasn't needed. That is far from the truth. That is all I was trying to convey, R5 is absolutely needed here.
This is a 101 constant current circuit using a differential circuit to maintain it's current at R13 and R14.
Oh well, maybe I should pick a different brand beer. This Coors Light is getting to me, then again, it could be the fact that I just got done with my Taxes. And I hope all the free loaders enjoy getting my money that I have to additionally pay on this year.
Jamie
I'm sure they'll enjoy the three bucks that you didn't blow on cheap beer.
-- You can't have a sense of humor, if you have no sense.
Ah -- I thought you were commenting on the OP's post, not Mr. Black's post. My error.
Yes, and I was kind of surprised that you didn't realize it. But I've seen some pretty astute analog circuit designers who've seen this circuit (or similar ones) and not realized what was up.
Coors Light. Ick. Surely in this day and age you can get some decent craft beer with real taste and color!
-- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com
??? the circuit compares the volage across in r13 + r14 with the preset voltage in the non-inverting input and turns the mosfet up or down to match the voltages. the mosfet response doesn't need to be linear, only monotonic,
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nah, a capacitor from output to the inverting input reduces the AC gain producing even more lag, this stops the op-amp from overshooting and also ensure that it won't oscillate.
-- ?? 100% natural --- Posted via news://freenews.netfront.net/ - Complaints to news@netfront.net
Jason, Could you elaborate on that? Like write us an equation that shows how it reduces the AC gain? ...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
Reduces the AC gain of what? The loop from inverting input to inverting input?
Keep the difference between your overall circuit response and the loop gain straight, please.
-- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com
You mean it lowers the BW ?
Jamie
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