Path: utzoo!utgpu!news-server.csri.toronto.edu!mailrus!cs.utexas.edu!rice!uw-beaver!sumax!amc-gw!richm From: richm@amc-gw.amc.com (Rich Moran) Newsgroups: sci.electronics Subject: Re: JFET Confusion Keywords: devices fet op-amp Message-ID: <1165@amc-gw.amc.com> Date: 16 Mar 90 22:18:50 GMT References: <7310@rice-chex.ai.mit.edu> Reply-To: richm@the-end.amc.com (Rich Moran) Organization: Applied Microsystems, Redmond, WA Lines: 94 In article <7310@rice-chex.ai.mit.edu> mikec@ai.mit.edu (Mike E. Ciholas) writes: }I must not understand JFETs yet. } }Background... } }I am trying to build a circuit that measures current flow very accurately }without inducing a significant voltage drop. The following circuit is shown }on page 2-25 of the Motorola Linear and Interface Integrated Circuits }databook. ("+" indicates connection of crossing wires) } } +-----------------------------+ } | | } +------|------+ | } | | | | } | | |-----\ | } | | | V+ \ |------| } INPUT-----+--R1--+---|- \ | | } | | \ | LOAD | } R2 | LF355 )--+ | | } | | / | |------| } +----------|+ / | | } | | V- / | | } | |-----/ | GND } 2N4092 |--| | | } JFET |<---------|---------+ } |--| | }OUTPUT------+ GND } | } R3 } | } GND } }OUTPUT = R3*(R1/R2)*(current in load) } }Typically, R1 is quite small, about .1 ohms. R2 is about 100 ohms and }R3 is about 5K. This gives 5V per amp in load as output. } }Now the question... } }The way I understand JFETs, N channel types in particular, is that you must }supply a negative gate voltage on the order of -5V to turn them off }(the 2N4092 is a depletion mode JFET). } }Since the LF355 is supplied from the INPUT voltage to ground, its output }range must be within this voltage, therefore it cannot turn off the }JFET. } }Is this true? Does this circuit work? } It does work. This circuit is balanced whenever the voltages at the + and - inputs of the LF355 are equal. The voltage at the - input is fixed by the input voltage minus the drop across R1. The voltage at the + input is fixed by a voltage divider consisting of R2, R3 and the JFET. The circuit balances itself by feeding back on the JFET, thereby modifying the voltage divider. With the circuit balanced, the current through R2 will be .001 (.1/100) times the current through R1. Since the current through R2 also passes through R3, the voltage developed across R3 is a direct representation of the current through R2 (and R1). Assume a 1A current through R1. This yields a 1mA current through R2 and R3, and +5V at the output (and the source terminal of the JFET). Note that the JFET will never be completely turned off, it is used as a variable resistance to balance the circuit. Its gate to source junction can be biased as a result of voltage drop across R3. If the JFET IS ever turned completely off, or not biased at all (completely ON), then the circuit is operating outside of its range. There are two limiting factors to the operation of the circuit : First, the input voltage must be at least 5.1 times the LOAD current to allow the circuit to balance. This 5.1/1 ratio can be modified by changing the value of R3, and thereby changing the ratio of volts/amp on the output. If the LOAD current is too high to be measured by the circuit at the given input voltage, the JFET will be biased completely off. For higher currents at lower voltages, R3 should be made smaller. The second limiting factor is low-current operation. If a very small current is drawn by the LOAD, the drop across R3 will be too small to allow the LF355 to bias the JFET. In this case, the JFET will be unbiased, completely on. For low current operation, R3 should be made larger (again, changing the volts/amp ratio on the output). -- ============================================= Rich Moran richm@amc.com =============================================