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What value resistor should I use for the base resistor for a transistor in front of a shutdown pin that uses virtually no current? If I do the math myself I get a crazy high number.

I am trying to control a shutdown pin that's pulled high to 5v via a 10k pullup resistor from a 3.3v signal. The shutdown pin is probably implemented as something like a mosfet that uses virtually no current.

Update:

The 5v 10k pullup resistor is tiny smd on a breakout board for the device I'm trying to interface with and I can't change it easily. And that pullup is why I can't drive the pin directly as I have a 3.3v MCU, it would presumably damage the 3.3v-only GPIO pin.

I just need a logic level shift to drive a small 5V signal low. I only have BJT transistors laying around and don't have any small signal mosfets. As I understand it the base resistor is (I think?) inversely proportional to the amount of current you expect to consume, and the shutdown pin will use virtually no current.

Update #2:

This is how I'm simulating what I'm trying to do (below). I assume that the shutdown pin is like a mosfet gate.

I found that making the base resistor very low (~100Ω) made driving the BJT consume about 30ma of current, which seems like a bit much. I found that 1kΩ-10kΩ base seemed to function exactly the same, and consumed about 1ma. Which I think is exactly what you would expect through the 10kΩ pullup. 100kΩ and it was not able to drive the mosfet fully. The logic is inverted of course but I don't care about that.

So I think this is what I'm going to go with unless I can find something else wrong with it.

enter image description here

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  • \$\begingroup\$ Then all you care about is the current that the pullup resistor requires when you control the pin, right? You can work out the base resistor for that much. \$\endgroup\$ – jonk Oct 27 '17 at 23:38
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    \$\begingroup\$ The only (?) reason to use a transistor is because you need to change voltage levels and / or you need to invert the control signal. Are you sure you can't drive it directly? \$\endgroup\$ – Transistor Oct 27 '17 at 23:46
  • \$\begingroup\$ the 10k 5v pullup is a tiny smd one breakout board and I can't change it easily. I can't drive the pin directly because of that 5v pullup, it would fry the 3.3v MCU pin. & I only have NPN BJTs, I don't happen to have any small signal mosfets laying around. \$\endgroup\$ – J Halcres Oct 28 '17 at 1:38
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    \$\begingroup\$ 2.6V*Hfe*10,000R/5V \$\endgroup\$ – Trevor_G Oct 28 '17 at 2:34
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    \$\begingroup\$ If you add a pulldown resistor, of value 20k ohm, to the input, would that mean you could connect it directly to your 3.3V logic? Some inputs for 5V logic don't need the full 5V to be reliably HIGH. \$\endgroup\$ – Whit3rd Oct 28 '17 at 3:59
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Your illustration is a wiring diagram and, unfortunately, doesn't show us the scheme of the circuit. For example, which pin is the base on the transistors? Why are there two? We need something more like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Simple level converter.

... and consumed about 1ma. Which I think is exactly what you would expect through the 10kΩ pullup.

  • The 10k pull-up will pass \$ \frac {V}{R} = \frac {5}{10k} = 0.5 \ \mathrm{mA} \$ when connected to GND.
  • If we make a conservative calculation that the transistor has a gain of 100 then we need 5 uA into the base.
  • The voltage across R1 will be 3.3 - 0.7 = 2.6 V due to the forward voltage drop across the base-emitter junction. Let's use 2.5 V to make the calculation easy.
  • \$ R1 = \frac {V}{I} = \frac {2.5}{5u} = 500k \$.

Most of us, at this stage, would use a 100k resistor to apply a large safety factor and ensure that the transistor is definitely driven into saturation to ensure that the input is pulled low.

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