I am designing a constant current source(35mA) circuit using transistor.Thought of usingenter image description herethis below circuit but the problem with that is my resistance R is fixed( 2.3 ohm ) which I could not change due to the requirement.Could you please suggest some design ideas how to meet the requirement.

Provide constant current source of 35mA Resistance R is fixed with 2.3ohm

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    \$\begingroup\$ 35mA into 2.3\$\Omega\$ is only 80.5mV. You will need an op-amp or a differential amplifier made from discrete transistors. \$\endgroup\$ – Spehro Pefhany Nov 23 '15 at 14:57
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    \$\begingroup\$ Why not add a 22 ohm resistor on top of your constrained 2.3 ohm resistor \$\endgroup\$ – Marla Nov 23 '15 at 14:58
  • \$\begingroup\$ Due to certain requirements,my resistance across the circuit should always see 2.3ohm \$\endgroup\$ – ANONYMOUS Nov 23 '15 at 14:59
  • \$\begingroup\$ Resistance across What circuit? \$\endgroup\$ – Marla Nov 23 '15 at 15:01
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    \$\begingroup\$ Dead End here : Even if you could get 2.3 ohms across Vcc to ground, Vcc would determine the current. Sounds like a concept problem. Perhaps describing your project might help. \$\endgroup\$ – Marla Nov 23 '15 at 15:13

If you want a constant current and you want decent stability you ought to use an op-amp controlled BJT current source: -

enter image description here

Vset forces the op-amp to control the BJT so that Vset appears across Rset. With Vset constant and assuming Rset is also constant then the current through the load is also constant.

You can do a PNP version as well.

  • \$\begingroup\$ It would be kind enough if you tell me the constraints that can be taken into account while selecting the op amp and transistor for this circuit \$\endgroup\$ – ANONYMOUS Nov 23 '15 at 15:01
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    \$\begingroup\$ If you want to be a designer it would be an excellent exercise to find those constraints yourself ! \$\endgroup\$ – Bimpelrekkie Nov 23 '15 at 15:30
  • \$\begingroup\$ If R is constant and I should be constant, then set the "vset" as just a constant resistor divider so that the voltage across Rset when Rset = 2.3 ohms, is the same as the voltage divider \$\endgroup\$ – KyranF Nov 23 '15 at 18:00
  • \$\begingroup\$ Made a similar arrangement and wanted to measure the voltage across load resistor on microcontroller ADC input. However the correct voltage appear on multimeter, but no luck in getting ADC value. Any suggestions will be helpful... \$\endgroup\$ – Prasan Dutt Mar 3 '17 at 13:00
  • \$\begingroup\$ @PrasanDutt it's likely that you should ask a new question as this Q&A sound unrelated. \$\endgroup\$ – Andy aka Mar 3 '17 at 13:07

If your supply voltage is fixed, the squib's resistance is fixed, and the current through it is fixed at 35 milliamperes, then to get that current all you have to do is connect a resistor between the squib and the supply such that Rt = E/I and subtract the squib's resistance from Rt .

What you'll be left with will be the value of the series resistance which will give you the fixed current you want.

For instance, if you have a 12 volt supply and you want to take 35 milliamperes from it:

$$ Rt = \frac{E}{I} =\frac{12V}{0.035A} = 342.85\Omega $$

Then, since your squib's resistance is 2.3 ohms, the series resistor needs to be :

$$ Rs = Rt - Rsquib =342.85\Omega - 2.3\Omega = 340.56\Omega $$

However, from the data sheet you linked to yesterday, it appears the resistance isn't fixed at 2.3 ohms, but can vary from a low of 1.8 ohms to a high of 2.6 ohms. if such is the case and you want to measure the resistance by passing a constant current through the squib's detonating wire and measuring the voltage drop across it, you'll need a constant current source, like the one @Andy aka posted, which will work.

As an aside, do you know how to select the current sensing resistor and the reference?

  • \$\begingroup\$ @ EM Fields:Thank you very much for the detailed explanation.I understand now that the resistance is not fixed.So I need to use the constant current source.Shall I use this similar circuit to serve the purpose like this bristolwatch.com/ccs \$\endgroup\$ – ANONYMOUS Nov 23 '15 at 20:42
  • \$\begingroup\$ No. All you need to do is to follow our advice since most of us know what we're talking about and won't lead you astray and can help you get to where you need to be. \$\endgroup\$ – EM Fields Nov 23 '15 at 21:51

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