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4 - 20 mA current source

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Aug 23, 2010
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hi to all ... i need a sink current source with big load output like 0-10k ohm ... any body have a schematic ..! thanks

Have you calculated the voltage you need to provide 20mA to a 10K resistor... only 200v

you right ...! i forgot this point .... load output is not important ...

The basic schematic of a voltage to current converter looks like this

The values are nor correct for what you want, they are for very low currents, also a rail to rail opamp would work better and you will need the R2 to be about 100 ohm.

The main idea of the operation is that the opamp constantly compares the voltage in the positive input with the voltage in the negative input which is coming from the voltage drop on R2.
Depending on the output current you get more or less voltage drop on that resistor (it works like a shunt resistor) but when this voltage is different from the voltage of the positive input the opamp drives the transistor with higher or lower current to equalize these two voltages.

An example, suppose that R2 is 100 ohm, for 10mA the voltage drop on the resistor will be 100*0.01=1v, if you set the positive input to 1v then the circuit will automatically balance so that the output current becomes 10mA so that both plus and minus opamp inputs have the same voltage (1v).

the above circuit uses a pot to set the positive input voltage, if you want to feed it a voltage coming from another source then it is not needed.

Sourcing 4-20mA can be a little more complicated, especially if you are using loop power. I suggest you read my article in Circuit Cellar Issue 241, of August 2010 entitled "The 4-to-20-mA Current Loop". It covers current sources made up of individual parts as well as dedicated ICs.

**broken link removed**

Circuit Cellar does charge a nominal amount for the article.

If you want an IC for the job the only one I know of is the AD694


thanks very much guys ... and i know ad694 before that but i should not use IC ... alex .. i'm going to use your suggestion schematic ..! can you explain about zener diode application in circuit ..???
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It's to provide a "stable" supply for the voltage reference.

I have seen your post but then you added the question as an edit so I never received an email notification.

amintlk is right, the schematic I have posted was intended to give an output current depending on the position of the pot.
If the pot is in a simple voltage divider feeding from the power supply then when the power supply voltage changes the current will change too because the voltage to the positive opamp input will change.
The purpose of the zener is to keep the voltage on the input of the voltage divider constant so that the output current stays constant too.
1N4728 is a 3.3v zener, the voltage going to the voltage divider is always Vcc to Vcc-3.3 so for 30v it is 30v-26.7v , for 24v it is 24v-20.7v etc.


i guess no ... for voltage reference is better use another way ..!

---------- Post added at 21:44 ---------- Previous post was at 21:42 ----------

It's to provide a "stable" supply for the voltage reference.
i guess no ... for voltage reference is better use another way ..!

I don't understand what you mean, what would you suggest?


Yes, you said
i guess no ... for voltage reference is better use another way ..!

guess no to what?
And what is the different way you suggest?
This circuit with the zener was made so that when you place the pot to a position for a specific output current then even if you change the power supply (for example from 12v<->24) the setting will still give (almost) the same current.


oh ...sorry .. it's just Misunderstanding ... i didnt understand about neddie's words ..

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