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#21




now, the 2nd chart. Rather than one transistor of hFE=200ish, we are going to use a "super transistor" in the form of a darlington. its effective beta would be 40k + (200x+ bigger than in the first chart), and will drop two Vbe, right?
again, the green trace is the output current and the red trace is the effective beta of the output device. question: 1) do they look identical to the chart where no transistor or just one transistor is used? 2) why? once you work out the math  which is extremely simple, it will be clear to you. and it will also help you understand negative feedback and how it works here. 
#22




Thanks everyone! I notice that these circuits have low CMRR. I thought of using an in~amp instead, but I intend to learn about op amps first and how to cope with these constraints.

#23




Op~Amps are completely new to me and so I don't know how to apply the math here to get want I would like. Also, I noticed a couple schematics that had high value resistors at the output of the op~amp; why do they do this? I think It might have been for an oscillator circuit.
I went ahead and picked up a book from the Library, "Op amps for Everyone," by Ron Mancini. It seems to be a good book and could help me get the fundamentals down and also the math. Thanks! 
#24




there are quite a few such books on the net. Analog / Walter Jung has an excellent book on opamp, and full of historical perspective as well.
TI also has a great book on opamps. the book you have is excellent as well. 
#25




Thanks Millwood, you've been a big help. Could you explain how you came up with the resistor values of your last circuit? I'd appreciate it!

#26




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two principles when it comes to (ideal) opamps's DC performance: 1) they have infinite open loop gain. That means that once feedback is applied, the voltage differential between an ideal opamp's inverting input pin and noninverting input pin is always zero. 2) they have infinite input impedance. so no current is going into or output the input pins. once you establish those two principles, the rest is a breeze. 1) by principle #1, you can that the voltage on the inverting pin is also Vin. 2) take a look at node A. you know that all current going into that pin has to sum up to all current going out of that pin. so I3=I2+I4, thanks to principle #2. 3) but, since the voltage at node A is Vin, per Ohm's law, the following must be true: I3=Vin/R3 I4=(VrefVin)/R4, and I2=(VoVin)/R2 so we have Vin/R3=(VoVin)/R2+(VrefVin)/R4 solve for Vo: Vo/R2=Vin/R3+Vin/R2(VrefVin)/R4, or Vo=Vin(1+R2/R3)  (Vref  Vin) *R2/R4 the 1st term is the noninverting gain for Vin, and the 2nd term is the inverting gain for Vref. thus the output current on R1 would be Iout=Vo/R1. 4) take a look at node B. and you have the current going through the load R5 I5=Iout+I2. if R2 >> R1, you have I2<<Iout, so I5=Iout+I2 = ~Iout = Vo/R1. as you can see, I5 has nothing to do with R5's value. so when you play with the numbers, you want to make sure that the current going through the feedback network is sufficiently small. R1 is selected so that it doesn't overload the opamp: it needs to be sufficiently large so the opamp works below its maximum output current specification. R2/R3/R4/Vref are selected to give you the right Vin  Iout relationship. well, that's just in a nutshell. 
#27




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I said "It seems to me that the output current will be strongly influenced by the hfe of the transistor, rather than being determined by resistor ratios." You said: I wasn't talking about high beta transistors like you mention in another post. I was supposing that ordinary transistors with a wide production spread of betas would be installed in your circuit, and with fixed Vin and Vref, determine the load current. A 10 to 1 spread is not atypical; let the spread be 30 to 300, for example. Then the output current will have variations in the several percent range, enough to disqualify the circuit from being called a precision current source. If we similarly change the active device (the opamp) in the Howland circuit over the production spread of devices of the same type, the variation of output current will be orders of magnitude less. A good way around this problem is to use a mosfet, such as the BS170. Since there's no base current, the output current won't show the variations of output current as we plug in various BS170s from the production spread. I said "The original circuit can have an output impedance approaching infinity, and the current is determined by resistor ratios, and is therefore not much affected by temperature, or by parameter variations of a single transistor. " You said: Quote:
Good opamps are readily available, so you don't have to go much out of your way to get very good temperature invariance from the Howland circuit. You could get a multiple of matched, extremely low tempco resistors if you wanted ultimate performance. Doing that in your circuit would be a waste as long as that single bjt is there; it will dominate the temperature variations, and you won't be able to get low ppm performance. Using a mosfet would be much better. I said "Do you have an expression for the output impedance for your circuit? Won't it be essentially just the ro of the transistor? That may not be very high." Quote:
Can your circuit be trimmed to improve the output impedance? In another post you attempt to determine Zout of your circuit, but you've done it incorrectly. I'll address that in a response to that post. 
#28




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But THAT has a few papers on that so do check them out. Quote:

#29




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You've changed the circuit. In the first circuit you posted (in post #9), there was no R5. I had to guess which resistor was intended to be the load and I guessed R1. I will grant you that in this changed circuit, the drive R5 receives will be high impedance, an approximate current source. But, you've measured the output impedance incorrectly. In the first place, you should not connect the + input of the opamp to the  input. You should ground the + input because under normal operation of the circuit, it's driven from a zero impedance voltage source. Furthermore, the output impedance is not the impedance at the node between R5 and R1 with respect to ground. The output impedance is the impedance between the opamp output and the node between R5 and R1; that's the impedance seen by R5. What you should do is put an AC voltage source in series with R5 and determine the AC current out of that source. Then the ratio of that voltage to the current out of that source will be the output impedance. 
#30




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and those textbooks too, . 
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