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Black Box Electronics An Introduction to Applied Electronics for Physicists 2. Analog Electronics: BJTs to opamps University of Toronto Quantum Optics.

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Presentation on theme: "Black Box Electronics An Introduction to Applied Electronics for Physicists 2. Analog Electronics: BJTs to opamps University of Toronto Quantum Optics."— Presentation transcript:

1 Black Box Electronics An Introduction to Applied Electronics for Physicists 2. Analog Electronics: BJTs to opamps University of Toronto Quantum Optics Group Alan Stummer, Research Lab Technologist

2 Active BJTs Class A with four resistors Class A with negative feedback

3 Drain to source resistance and/or current ≈ 1 / gate to source voltage. Zero gate current! Operate in active region (analog), or saturation or cutoff (digital). Most are N-channel (source grounded, positive drain and gate). Limited in current I D, voltage V DS and power P D. Very good saturation, R DS(on). Limited frequency range, capacitance dominates at high frequency. Select for current, voltage, speed and R DS(on) if saturating. Typical voltage V DS range 10V to 200V, extreme to 1.5KV. Typical current I D range 100mA to 10A, extreme 10mA to 100A. MOSFETs (Metal Oxide Semiconductor Field Effect Transistor)

4 Sample FET Transfer Curves A general purpose 48A, 60V N-FET from Fairchild (Digikey.ca). http://www.fairchildsemi.com/ds/ND/NDP6060L.pdf

5 NDP6060L Data Sheet – Parametric Section

6 Basic Q Switching NPN BJT Good speed. Poor saturation. Two components. Rule of thumb: I b ≈ I c /10 N-FET Good speed. Very good saturation. One component. (Note that Id is 0.1% of limit yet it works well)

7 Sample Linear FET Circuit Opamp uses a N-FET to increase current driving capacity. The opamp can supply only ten’s of mA but the laser needs ten times more. 1.The opamp can drive the FET from cutoff through to saturation. 2.All of the laser current goes through R9. 3.The opamp monitors the R9 voltage. 4.The opamp adjusts the FET gate voltage to control the FET conductance.

8 IGBT (Insulated Gate Bipolar Transistor) Current allowed through collector to emitter is proportional to voltage between gate and emitter. Voltage controlled like a FET, switches like a BJT. Used for power control such as motors. Pros: High voltage and current. Cons: slow, poor saturation. Thyristers: SCR (Silicon Controlled Rectifier) & Triac Anode/MT1 connects to cathode/MT2 once gate current exceeds threshold, stays on until anode/MT1 current drops below threshold. SCR is unipolar, Triac is bipolar. Used for motor and lighting controls (Φ control). Pros: High voltage and current. Cons: slow, poor saturation. Less Common Active Parts

9 Five terminal linear device (2 power, 2 inputs, one output). Hard Rule #1: The output goes positive if the input is positive (where input is defined as +ve input relative to –ve input). Soft Rule #2: The inputs and output can only range between the “rails” (the positive and negative supply pins). All else is imperfections: OPAMPs (Operational Amplifiers) Finite open loop gain (GBW) Finite bandwidth (GBW) Limited output voltage range Limited output current Finite slew rate (output dV/dt) Unstable with capacitive load Noise generation Limited input voltage range Finite input current Finite input offsets Phase shifting (lower gain, stability) Finite quiescent supply current Temperature sensitivity and aging of all parameters

10 Some OpAmp Errors I b : Input bias current, ranges fA to µA. I o : Input offset current, ~10-50% of I b. V io : Input offset voltage, ranges 10’s µV to mV. V n : Equivalent input noise, ranges 1-1000nV/√Hz A v : Open loop gain, usually 10 5 -10 6. GBW, f: Gain bandwidth product, ranges 1KHz to 1GHz. Slew Rate, dV/Dt: Output slew rate, ranges 0.1V/µS to 10KV/µS. I out : Output current limit, ranges 1mA to 10A. Plus temperatures, supply, load, R in, drifts, aging, interactions,…

11 Basic OpAmp Amplifiers A V = -R 1 / R 2 f -3dB = 1 / 2π R 1 C 1 In this example: A v = -100K / 10K = -10 f -3dB = 1 / 2π * 100K * 1nF = 1.59KHz 0V < V out < +5V V in range depends on Vref V out = 1 + R 3 / R 4 f -3dB = 1 / 2π R 3 C 2 In this example: A v = 1 + 100K / 10K = +11 f -3dB = 1 / 2π * 100K * 1nF = 1.59KHz 0V < V out < +5V 0V < V in < +5V

12 Ω The End Ω Next:3. Digital Electronics 4. Sample Circuits 5. Spice simulations Then: More in depth on anything? Suggestions?


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