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I would like to inject 10 μA into my load, and to be able to switch the current on/off at up to 10 kHz. My first attempt is just based on the transistor current source in Horowitz/Hill.

schematic

simulate this circuit – Schematic created using CircuitLab

The relay switches the base between ground and a regulated 5V supply. The load current would then be (Vb-0.6)/R1, which is ~10 uA.

Two questions:

(1) First, since I have little experience in this area, is this approach even reasonable? (2) If so, how can I predict what switching speeds I could achieve?

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  • \$\begingroup\$ What's the relay for? That will drastically limit switching speed. 10ua is also rather small, what actually is the load? \$\endgroup\$
    – pjc50
    Mar 8, 2016 at 19:00
  • \$\begingroup\$ My thinking was that since the load current is proportional to the base voltage, I would want a better regulated base voltage than I would get from directly putting in the logic signal (e.g. coming from a microcontroller). The thinking behind the relay is that it would let me quickly switch this regulated supply on and off. The load is a human body (this is for an "impedance analyzer" essentially). \$\endgroup\$
    – Max
    Mar 8, 2016 at 19:08
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    \$\begingroup\$ 'Quickly' and 'relay' usually can't fit in the same sentence. How precise should the 10uA be? I doubt a single transistor will be able to output something more precise than 100uA, taking into account temperature. \$\endgroup\$
    – Vladimir Cravero
    Mar 8, 2016 at 19:23
  • \$\begingroup\$ Good to know, thank you. <10% would be ideal, though I'm not sure if that is asking too much. If this simple approach won't work, any suggestions on alternative techniques/areas to look into instead? \$\endgroup\$
    – Max
    Mar 8, 2016 at 19:26
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    \$\begingroup\$ Also, if your load model is any good you have a problem. 10 uA through a 100 pF capacitor for 50 us means you're trying to charge your capacitor to 20 V with a 10 V supply. \$\endgroup\$
    – The Photon
    Mar 8, 2016 at 21:10

1 Answer 1

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I'll start by answering your questions directly, then giving some thoughts you can use depending on your actual goal

(1) No, this is not a reasonable approach. Let's look at the period of your frequency at 10kHz -> 1/10,000Hz = .0001s, or .1ms. This means that your circuit needs to go from off, to on, to back off in .1ms.

Looking at a typical small-signal relay, such as the one below, the rated "on" time is 7ms, and the "off" time is 3ms. Meaning that if the circuit were being switched by this relay, the fastest it could reasonably go is about 100Hz (10ms = .010s, 1/.010 = 100) http://www.mouser.com/ds/2/307/en-g5v_2-536510.pdf

At the end of the day, a relay is something that takes an input and switches something else, and so is a transistor. However, a transistor is much faster and more efficient - it just can't do some of the things a relay can. You're better off just using a transistor.

Neglecting the relay, just start off thinking of your transistor as a perfect switch, then look up the specs (on-voltage, etc.) and transistor formulas that make it not perfect, and get a better idea of how your circuit will behave. What's the time constant of your RC-based load? Can it respond in .05ms? Look up the datasheet on the 2N3904, if you substitute in your formulas for a BJT, will it work? What current will flow through it, and thus your load?

After you look at how much current is going through/ into your BJT, is it reasonable? Will the transistor need a resistor on the base to protect it?

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  • \$\begingroup\$ Your last sentence is not true. The 440 K-ohm emitter resistor eliminates the need for a base resistor. \$\endgroup\$
    – Tut
    Mar 8, 2016 at 21:44
  • \$\begingroup\$ Right you are! Overall I meant that he should make sure he wasn't going to fry his transistor with base current, but neglected the size of that little guy. Post edited :) \$\endgroup\$
    – Drewster
    Mar 8, 2016 at 21:52

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