How do I make a current source using a zener diode and transistor?

Discussion in 'General Electronics Chat' started by skymin0202, Feb 12, 2019.

1. skymin0202 Thread Starter New Member

Feb 11, 2019
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I have a load that has varying resistance with external pressure. In order to measure the change, I am trying to measure the voltage on the load, when constant current is applied. And the circuit in the attachment is known to supply constant current using a transistor and a zener diode. But I have no idea what resistor/ transistor/ diode I need. I do not know if the circuit will burn them out.
The current on the load has to be approximately 6 mA, I think. And the supply voltage might be 5V or 3.3V. If impossible, the supply voltage can be changed. If needed, the transistor and zener diode I purchased are KTC3198, and 1N5231B-5.1V, respectively. The minimum current zener diode needs is 0.25 mA. Please comment, if any further information needed

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2. dl324 AAC Fanatic!

Mar 30, 2015
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Welcome to AAC!

You can't make a current source using a 5.1V zener with a 5V (or 3.3V) voltage source.

What is the resistance range for the load?

3. skymin0202 Thread Starter New Member

Feb 11, 2019
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i'm sorry. the supply voltage will be 10V

4. skymin0202 Thread Starter New Member

Feb 11, 2019
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0
the resistance will be in range of 6-400? (one digit ohm ~ three digits ohm)

5. skymin0202 Thread Starter New Member

Feb 11, 2019
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0

i'm sorry. the supply voltage will be 10V.
the resistance will be in range of 6-400? (one digit ohm ~ three digits ohm)

6. dl324 AAC Fanatic!

Mar 30, 2015
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You should operate the zener well beyond the knee. Closer to 20mA than 0.25mA.

Current is set by R2.

Use 750 ohms or use a parallel combination to get 733.

7. dl324 AAC Fanatic!

Mar 30, 2015
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750 ohms in parallel with 33k will give 733 ohms.

8. WBahn Moderator

Mar 31, 2012
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How precise do you need the current to be?

Notice that I'm not asking about accuracy -- since you said it needed to be approximately 6 mA, I'm assuming that anything in that general ballpark will be suitable.

I'm asking about precision or, in this context, stability. If it's, say, 6 mA at 400 Ω, how close to 6 mA does it need to still be at 6 Ω?

That circuit is not going to have a very high output resistance, but you are operating it at a sufficiently low current that it might be good enough -- but you do need to be able to articulate how good constitutes good enough.

There are other circuits that can significantly increase the output resistance of the source.

Mar 14, 2008
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10. Heyvargasmon New Member

Feb 13, 2019
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ZENER DIODE TESTING The testing of zener diodes requires a variable dc power supply. A typical test circuit can be constructed, as shown in figure 4-19. In this circuit, the variable power supply is used to adjust the input voltage to a suitable value for the zener diode being tested. Resistor R1 limits the current through the diode. With the zener diode connected as shown in figure 4-19, no current will flow until the voltage across the diode is equal to the zener voltage. If the diode is connected in the opposite direction, current will flow at a low voltage, usually less than 1 volt. Current flow at a low voltage in both directions indicates that the zener diode is defective.

11. crutschow Expert

Mar 14, 2008
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The TS is not testing Zeners.

12. MrAl AAC Fanatic!

Jun 17, 2014
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Hello,

Here is an approximation to the current in the collector assuming there is enough overhead voltage from Vcc:
Ic=(10*Vd-7)*(B-1)*(B+1)/(10*B^2*R1)

where:
Ic is the collector current,
Vd is the zener voltage,
R1 is as in post #9 the emitter resistor,
B is the transistor Beta.

Fooling around with that equation tells us the minimum Beta should be around 20 which should be easy to obtain. This is based on the values in post #9. Also if the Beta is 20 or above the collector current is relatively insensitive to a change in that Beta.

13. WBahn Moderator

Mar 31, 2012
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Even if the TS were testing zener diodes, what use is a post referring to figures that aren't included and component references in schematics that aren't included.

14. WBahn Moderator

Mar 31, 2012
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I think the equation would cause less head scratching if it weren't forced to have only integers. As it is, it's a head-scratcher for a while about why the zener diode voltage is multiplied by ten and then seven is subtracted from it.

Ic = (Vd - 0.7 V)/R1 * (B² - 1)/B²

15. kubeek Expert

Sep 20, 2005
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Andy why not simply Ic = (Vd - 0.7 V)/R1?
The (B² - 1)/B² term is basically one for any practical beta. At extremely low B=5 the error is 4% when you just use 1 instead of calculated 0.96. You will get grater error from the accuracy and temperature dependance of zener voltage and temperature dependence of Vbe.

16. WBahn Moderator

Mar 31, 2012
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Agreed (even the actual Vbe compared to the assumed 0.7 V will have a larger impact), but I think MrAl was specifically making a point about β and that's hard to do if you don't include those terms.

Mar 30, 2015
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18. Alec_t Expert

Sep 17, 2013
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Over what temperature range must your circuit operate?

19. MrAl AAC Fanatic!

Jun 17, 2014
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Hi,

Yes i dont always reduce the solution to the simplest possible form and using floats.
It's good to show it that way too so we can immediately spot the base emitter voltage approximation too.
So we end up with:
Ic = (Vd - Vbe)/R1 * (B² - 1)/B²

and the multiplying effect of Beta becomes more apparent too.

20. Ylli Active Member

Nov 13, 2015
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dl324 and kubeek like this.