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dlgoff
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How about a transistor to drive the LED? It'll cost a little board space however.
dlgoff said:How about a transistor to drive the LED? It'll cost a little board space however.
berkeman said:How about a couple opamp circuits? These show a couple of the most common errors that EE students (and new EEs!) make when working with simple opamp circuits. These are from Chapter 3 of Horowitz and Hill on opamps.
Same as before -- if you know the answers easily, please hang back to let others work on them. And for those posting possible answers, please do not guess! Check out some datasheets for typical representative components, and see if you can see some specifications that are being violated...
ranger said:[Circuit A]
Berkeman, I'm so used to working with ideal views of the op-amp, that when I need to take device limitations into account, I get stalled...
But looking at the circuit, I see no resistor, so when the diode is conducting, current will flow without limitation and will thus damage the very delicate 741?
Can give some more hints with regards to circuit A?-
berkeman said:There's one more problem with (a), and it has to do with the potentiometer.
ranger said:They could have used a rheostat to handle the higher power dissipations? And also, when pot is "maxed out", we'll be putting about 30V into the op-amp. According to the datasheet, we're pushing the limits of the 741 here. No?
antonantal said:So, in time it is possible that an oxide film will form at the contact of the wiper in the potentiometer. And because of the very low input current of the opamp, there will not be enough punch through voltage at the wiper contact to break down the oxide and the input of the opamp will remain isolated.
Would a resistor from the +15V or -15V rail to the wiper solve the problem?
antonantal said:So, in time it is possible that an oxide film will form at the contact of the wiper in the potentiometer. And because of the very low input current of the opamp, there will not be enough punch through voltage at the wiper contact to break down the oxide and the input of the opamp will remain isolated.
Would a resistor from the +15V or -15V rail to the wiper solve the problem?
berkeman said:Correct! It is a common mistake to ignore the minimum wiper current specification for a trimpot. The Bourns application note suggests 10uA to 100uA, but the potentiometer datasheet should list the number for whichever pot you are using. The LM741 input bias current is well below these numbers, so you can't just connect the wiper to the opamp input alone. And yes, you would do something more like what you are suggesting, where you connect the CW side of the pot to +15V, connect the wiper to the CCW side of the pot (Quiz Question -- why the CCW side?), and use another resistor down to -15V in a voltage divider arrangement to vary the input voltage to the opamp, while meeting the wiper current and power rating for the trim pot.
Good job. And now all of that should provide the hint necessary to solve circuit (b)...
berkeman said:Quiz Question -- why the CCW side?
ranger said:For circuit b, I think we need resistor from the non-inverting input to ground as a return path for the very small input current. Also if one wishes to work with small signals of varying frequency, the resistor will also help to set the cut-off frequency for ac signals (f3dB).
antonantal said:Because we want maximum input voltage to the opamp when the potentiometer is turned all the way in the CW direction.
If we connect the wiper to the CCW end, when we turn the pot all the way in the CW direction the wiper shorts the resistance of the pot and all the voltage will be droped on the resistor connected from the CCW end to the -15V rail, and we have maximum voltage at the input of the opamp.
If we connect the wiper to the CW end of the pot, when we turn it all the way in the CW direction the whole resistance of the pot enters in the voltage divider and we have minimum voltage at the input of the opamp.
ranger said:[Circuit I. - +15V regulator]
So far I see the following problem with it - the missing base resistor shows an improperly biased BJT.
ranger said:Berkeman, does this circuit have a problem with varying VE due to varying VB as the output of the op-amp varies depending on the voltage differential?
berkeman said:Why can't the opamp supply the needed base current?
ranger said:I don't understand this. Why not have a resistor to bias the BJT? The BJT has a relatively low input impedance. Shouldn't we want to avoid large output currents to prevent excessive power dissipation on the output stage of the op-amp
AlephZero said:Powering the op-amp from its own stabilised output won't work. To bias the transistor, the op-amp output would have to be higher than the supply voltage, and that can't happen.
Powering the opamp from the unstabilized 20 to 30V supply should be OK.
Also, for better regulation of the output voltage I would supply the zener through a resistor from the stabilized 15V supply, not from the unstabilized supply.
antonantal said:We can't have 15V at the output of the first opamp if it is powered from +5V.
berkeman said:AlephZero's comment about how to supply the zener is also a good idea.
The order of the figures must have changed in the 2nd edition of H&H, but I think these are what you're asking for. They're getting a bit harder now...ranger said:Hey berkeman,
I don't have a scanner right now. Could you post circuit B (x100 op-amp output stage for audio amp) and circuit G (x100 audio amp. (single supply)) so we can get some thoughts on it? You should of course post yours first.
Good point about the cap, but they are probably assuming a pretty small signal (no more than 15V/100 one would hope), centered around ground as the audio input.antonantal said:For the first circuit, I don't know what is the range of the input voltage but if it can reach a higher value than the potential on the base of the BJT, that electrolytic capacitor is not a good idea.
antonantal said:In the second circuit, there is no dc path to the ground for the inverting input.
ranger said:Circuit G:
It seems a coupling capacitor is missing from the output of the bjt amplifier to the input of op-amp. There should also be resistor to ground to cater for the bias current on the non-inverting input. Also the resistor will set the f3dB frequency; I guess we don't want to amplify all frequencies.
Shouldn't we also want to add a cap in series with the 1K to have unity gain at DC? We'd of course choose are to have the appropriate f3dB frequency.
ranger said:Unfortunately, I cannot quite put my finger on the last problem, berkeman
But does it have to do with trimming the off-set voltage?
ranger said:Unfortunately, I cannot quite put my finger on the last problem, berkeman
But does it have to do with trimming the off-set voltage?
berkeman said:Oh wait, you said "last problem", so I thought you meant the 2nd circuit. But if you're referring to my last comment for the first circuit, then yes, the problem is with the input offset voltage being amplified by x100. What do you typically do to minimize the input offset voltage component generated by the input offset current for the opamp? And assuming that you do that here, what are you left with as an input offset voltage for, say, an LM741 opamp? So what do you get for an output error after the x100 amplification?