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How would I make a Zener diode current source? Something like this? From Wikipedia

I don't know how to figure out what resistors/transistors/diodes I need.

I guess what I'm asking is, how do I calculate what resistors I need, and the rating for the transistor and the diode (I don't know how to tell if the circuit will burn them out).

Edit: I want the load to receive 1 Amp, the resistance of the load is 0.97232Ω, and I am supplying the circuit with either 8.4V, or 9.6V probably 9.6V.

What I'm looking for, is a set of equations to find values for the resistors. I(believe) I can assume that nothing in the circuit will experience current more than my target current of 1Amp? If that's true, I can get parts rated 1.5-2 times my target current.

Schematic of the circuit I want.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I know that if to much current goes through the diode/transistors, they will become burnt out. I just don't know how to determine how much current they can take. \$\endgroup\$ – CoilKid Aug 17 '14 at 21:23
  • \$\begingroup\$ Start with the current you want through the load. What size will this be? This value will determine the selection of transistor and all the other values of components. \$\endgroup\$ – JIm Dearden Aug 17 '14 at 21:30
  • \$\begingroup\$ I will need 1 Amp, but I don't know what to do from there :/ \$\endgroup\$ – CoilKid Aug 17 '14 at 22:23
  • \$\begingroup\$ What supply voltage do you need (or what is the load resistance range you want)? \$\endgroup\$ – Andy aka Aug 17 '14 at 22:26
  • \$\begingroup\$ CoilKid, instead of searching for solutions to a problem, like the current mirror you also posted, you should post a question like "How should I power this device?". \$\endgroup\$ – Vladimir Cravero Aug 17 '14 at 22:27
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With an 8.4 volt supply producing 1 amp to supply your load of 0.97232Ω you need to have a total resistance of 8.4 ohms (ohms law is still cool).

As you have only 0.97232Ω you need to put 7.42768 ohms in series.

That will produce 1 amp thru your load. It will also dissipate 7.42768 watts so choose a resistor that is rated for the power.

If, for whatever reason it must be a transistor current source and you need the accuracy of exactly 1 amp I wouldn't use the circuit in your question, I'd use an op-amp controlling the emitter voltage of a transistor like (7) in this below: -

enter image description here

The LM317 circuit I feel would also service your needs BUT, at the end of the day, with an 8.4 volt supply, something (usually the transistor) is going to be burning 7.42768 watts so get big components or a heatsink. It can be a BJT so don't worry too much about (7) showing a MOSFET.

Alternatively, use a switch mode current converter powered from the 8.4 volts and let the inductor and capacitor do what god made them for - taking energy in at one end (at higher voltage) and spitting it out at the other (but at lower voltage).

Smallprint - L and C can do the reverse too and I don't really believe in god, but it sounded cool to say that he invented L and C!

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  • \$\begingroup\$ Well, huh... I was actually looking at this circuit because it was suggested to have better efficiency than the idea I had... (Loss of 3.4W I think...) sigh. Thank you for the awesome answer. \$\endgroup\$ – CoilKid Aug 17 '14 at 23:11
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The resistor values required depend on the Zener diode's voltage and the current gain of the transistor. The voltage across R2 is equal to the Zener voltage minus the transistor's Base-Emitter voltage (typically 0.6V). Applying Ohm's Law, R2 = (Vzener-0.6V)/1A. For example, if the Zener is 5.6V then R2 must be 5 Ohms. Since 1A * 5 Ohms = 5W, R2 needs to be rated for at least 5W (10W would be safe).

R1 must supply enough current to bias both the Zener and the transistor. If the transistor has a current gain of 50 then it needs a Base current of 1A/50 = 0.02A. For good voltage regulation the Zener should be passing a similar current, so R1 needs to pass about 0.04A. The voltage across R1 is equal to the battery voltage minus the Zener voltage. Applying Ohm's Law again, (9.6V-5.6V)/0.04A = 100 Ohms.

The transistor has to dissipate about 4W of power, so it will need a good heat sink.

This circuit has very poor efficiency, but so would any linear circuit that needs to drop 9.6V down to less than 1V. To get better efficiency you either need to use a switching regulator, rewind your coil with more turns and/or thinner wire, or use a lower battery voltage.

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  • \$\begingroup\$ But how would I use a switching regulator? I'm gonna Google it now, but if its like the other electronics stuff, I won't find a lot. \$\endgroup\$ – CoilKid Aug 18 '14 at 1:04
  • \$\begingroup\$ Just buy a "CVCC" (Constant Voltage, Constant Current) switching regulator, set it to its lowest voltage and adjust the current to 1A. Here's one on eBay for $4.56 including shipping:- ebay.com/itm/… \$\endgroup\$ – Bruce Abbott Aug 18 '14 at 1:38
  • \$\begingroup\$ Eh! You're a lifesaver! I was thinking I would have to build one myself! :D Thanks so much! \$\endgroup\$ – CoilKid Aug 18 '14 at 2:50

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