# current to voltage converter ( high current )

A

#### alex_tan

Dear all,

i want to know how to convert a high current(about 10A )to voltage. according to my research, the most current to voltage converter is less than 100 nano ampere. is it possible to convert high current to voltage?

can it use opamp to do it?

thanks & regards.
alex tan

B

#### Barrington

Pass the current through a resistor and you will get a proportional voltage drop across the resistor e.g. 4-20mA through 250 Ohm give 1 - 5 Volts.

What voltage range are you looking for?

K

#### Ken Irving

Alex,

If the current is AC (or perhaps otherwise varying) a current transformer might be used, along with a transducer to output a DC signal proportional to the source current. If it's DC you'll likely need to use an inline resistance (aka shunt) and measure the voltage drop according to Ohm's Law.

--
Ken Irving <[email protected]>

M

#### Michael Griffin

alex tan wrote:
<clip>
>i want to know how to convert a high current(about 10A )to voltage.
>according to my research, the most current to voltage converter is less
>than 100 nano ampere. is it possible to convert high current to voltage?
>can it use opamp to do it?
<clip>
I am not sure what you are asking. I think you are asking how to measure current, producing a voltage output which is proportional to the current flow. If this is the case, then consider the following.

First you must determine what your accuracy requirements are. Without doing this first, any additional effort is useless. You must also discover whether you wish to measure DC or AC current.

10 amps is not a high current at all. You can measure this with a current shunt, a current transformer, or a hall effect transducer. The best
means depends upon the accuracy you require, and whether you are dealing with DC or AC current.

**********************
Michael Griffin
**********************

O

#### Oly

If you are trying to measure or monitor large
current, the typical way this is done is to
use "current shunts." These come in current
rating increments e.g. 5A, 50mV [or 50 milliohms].

The shunt is connected in series with the current source or load and the voltage drop is measured and scaled to the right reading. You can use
a digital panel meter and scale it appropriately to read out in Amps.

Hopes this help.

Oly Jeon-Chapman

T

#### Thomas B. Bullock

Use a low ohmic value precision resistor in the 10 amp line. The precision of the resistor depends on the precision needed in your measurement. Then put the input of a differential operational amplifier across the resistor. The operational amplifier input and feedback resistors can be adjusted to give the desirable voltage swing on the output.
I used this before and it worked great.
Tom Bullock

M

#### MIG

Do you want to measure something that varies from 0 to 10A?
If so, you can build a simple current/voltage converter with op amp. The current must pass through a shunt resistor (ex.: 0.1 ohm or less). Pay attention to the wattage that you need for your shunt. Let's say for 0.1 ohm * 10 A * 10 A = 10 W dissipated. Pick something that can dissipate 5 times more. The rule is to use a very low resistance in order to reduce the voltage drop and of course the power dissipated.

Then you have converted 0_10A to 0_1V (if you had picked the 0.1 ohm. Now you can use a differential input amplifier (made of one op amp and some resistors)to get the voltage output needed.

C

#### Curt Wuollet

Use a current shunt to do this if your circuit can stand a few mv of voltage drop. For example, a 1mV/A shunt will give you a 0-10mv voltage for 0-10 A. Do this in the low side of the circuit so you can stay within the CMV rating of the input card. For AC you can use a current transformer
and a resistor.

Regards

cww

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J

#### Jim Cai

Just some hall-effect current transformer. It can convert the 10 A(ac) into 0-5 Vac

D

#### dooley

DC currents sensors are based on a heavy shunt resistor in series in the circuit. The shunt is sized to produce an IR DC voltage suitable for measurement. AC current sensors employ a current transformer having a small number of turns of
heavy gauge wire in the primary and a very large of turns of small gauge wire in the secondary from which the voltage is taken. The voltage is rectified and measured. In many cases the conductor is the primary. This type encloses the
conductor and can be split so that it is clamped on.

Important safety warning. Alex. Electricity is dangerous. This is not an area to play about with unless you know what you are doing .If this is a practical exercise it needs to be done by a suitably qualified person.

Vince

H

#### Hakan Ozevin

If you want to measure AC current, you can use a current transformer, or if you are a student, you can make your own. Wind a low resistance cable (we call secondary circuit) several times around the other cable that current will be measured (we call primary circuit). In the wound cable you will get a current proportional to the primary current and inversely proportional to the number of winds. Thus you can get a low current that can be measured. This second circuit is isolated from the primary one and can be measured with an op amp easily, without any problem.
Alternatively, for both AC and DC current, you can use a serial resistor (shunt) with low resistance but with high watt capacity. The voltage drop on the resistor is proportional to the current and we normally select it to get 10's of mV's. You can use an op amp to amplify this low voltage. In this case, your measurement circuit will not be isolated from the primary circuit, thus, be careful about the voltage difference to the ground (or power supply negative). Otherwise you can destroy your measurement electronics. Make your connection as: Power supply positive, load, shunt, power supply negative. Not: PS+, shunt, load, PS-.

I hope this helps.

Good luck

B

#### Burke

Alex:

Probably the cheapest way would be to use a meter shunt or resistors, as described above. There are current transformers available for both A.C.
and D.C. (Hall-effect devices), although the A.C. are the most common.

An extra word of caution about the current transformers, although I doubt you'll be using one for a ten amp experiment. Most of the industrial CTs are rated with a secondary output of 5A, xxx:5. They also have a burden (VA) value. If you run primary current through these, with the secondary open-circuited, the secondary terminal voltage will increase as necessary,
as it tries to dissipate the power. A person could be electrocuted easily.

I saw one system where they ran over the weekend with the CTs open-circuited and the voltage on the terminal board rose high enough to arc through the bakelite terminal board to the sheetmetal. It caught the CT terminal board on fire, which caught the nylon one above it on fire,
etc.

Made quite a mess. Be careful, and always short the secondary terminals together if you're not going to connect a device to them.

Paul Baker

J

#### Johan Bengtsson

Well, as I interpret the text below it get's wrong, I don't (of course) know how the rest of you interpret it but anyway:

Winding a cable *around* the primary will not make much of a transformer, you should either make a loop (or perhaps some more turns) on the primary and put the secondary winding side to side with that one. Or make the secondary winding appear beside the primary (not around it) so one edge of the circle is close to the wire:
|
|
| ---------------
| / \
| / \
|| |
|| |
| \ /
| \ /
| ---------------
|
|

if you want you can make the secondary formed as a spiral around the primary wire (ie a toroid)

The whole idea is that they should have a (at least partly) common magnetic field.

Either way, 10A isn't that much, it is probably easier to use a low ohm resistor and an opamp regardless of if it is AC or DC (for AC you might want to add some diodes, but do so after the
opamp, not before because the voltage drop would be too significant, well you need some other components too of course)

/Johan Bengtsson

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