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I am trying to design a current amplifier circuit to amplify a signal current to 1A.

Application :

I am generating a signal using microcontroller (for ex : Pic16f877A). This signal is 5v having 10mA current (approx not calculated). I want to amplify this current upto 1A - 1.5A to drive a device.

Can anyone suggest me better way to solve this issue ??

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  • \$\begingroup\$ Is the signal a constant 5V that you wish to switch on and off to a device that draws ~1A when 5V is applied? \$\endgroup\$ – Andy aka Apr 13 '16 at 11:35
  • \$\begingroup\$ @Andyaka, actually the signal would be analog upto 5v. & I need this analog signal to drive a device which requires about 1A current. \$\endgroup\$ – skg Apr 13 '16 at 11:37
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    \$\begingroup\$ Is it a true analog signal or a PWM signal? What kind of frequency is the signal? \$\endgroup\$ – pjc50 Apr 13 '16 at 11:39
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    \$\begingroup\$ Strictly speaking, what you're describing isn't a current amplifier. You are describing a high current driver. \$\endgroup\$ – Scott Seidman Apr 13 '16 at 14:11
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    \$\begingroup\$ It doesn't make any sense to say "This signal is 5v having 10mA current." The reason is, if the source determines the voltage, then the load is what determines the current and vice versa. In this case, the load is the amplifier/driver that you are asking for help designing, which means that you haven't designed it yet, which means that if you know your source puts out 5V, then you don't yet know how much current the amp/driver will pull from the 5V source. OTOH, it does make sense to say, "The signal is 5V, and able to supply a maximum of 10mA..." \$\endgroup\$ – Solomon Slow Jun 13 '17 at 20:37
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Here is a basic voltage to current amplifier:

Due to the feedback thru the opamp, the voltage across R1 will be Vin. That means the current thru R1 is proportional to Vin. The only place that current can come from is thru the drain of the N channel FET. The net result is that this circuit will sink current at Iout proportional to Vin. The overall gain is 1/R1.

That's the basic concept, but in your case there are some details to consider. Your Vin varies from 0 to 5 volts. That means at full current, there will be 5 V across R1. Not only does that significantly eat into whatever supply voltage range you have for the load, but it will cause a lot of power to be dissipated by R1 at the full current of 1 A. (5 V)(1 A) = 5 W. Even if you're OK with using that amount of power, you will have to carefully consider how to deal with the heat.

The solution is to put a voltage divider between your 0-5 V signal and Vin. The lower the Vin range, the lower the power dissipation. However, you also decrease overall accuracy since the offset voltage and any other errors will be a larger fraction of the whole. You also have to consider how well the opamp does with both inputs near ground. You may want to give the opamp a small negative supply to keep it well within its normal active range.

I'd probably start with attenuating the 0-5 V signal by 5, so 0-1 V. That means the max dissipation in R1 is 1 W, so a 2 W resistor just soldered to the board without any special heat sinking should be fine.

Since you have a microcontroller already, making a small negative supply is easy. Many micros have a clock output, which can then drive a charge pump. I usually have the micro pin drive a back to back emitter follower pair, which then drives the charge pump. After all the voltage losses due to B-E junctions and diodes, you get about -2.2 V. However, that's plenty for many opamps.

You also need to consider the dissipation in the FET. It will see whatever the output current is thru it, but the voltage across it is not so clear. The worst case when the load is a short. You then have the supply voltage minus the R1 drop across the FET. If you have a 12 V supply, for example, and R1 drops 1 V at the full current of 1 A, then the FET can dissipate up to 11 W. That's doable for a power package, like a TO-220, but definitely requires a heat sink.

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  • \$\begingroup\$ ..Thanks alot for descriptive answer. Actually I am driving a 24v BLDC motor(electronic fitted). Motor needs "Enabe Input Signal (24v)" to start its drive. I have prepared a Signal switching circuit using Optocoupler to switch HIGH(5v) signal from microcontroller to 24v. Second signal required by the Motor is Ref.Voltage which is used to control its Speed. It needs 0-10v with 1A current. I will use DAC (0~10v). But its output current is approx 10mA. I need to amplify this current to 1A to drive the motor. So, I m looking for Current amplifier to get 1A current. \$\endgroup\$ – skg Apr 14 '16 at 4:36
  • \$\begingroup\$ @skg: That is totally different from what you asked about! You're not looking for a current amplifier at all, and the input signal isn't 0-5 V either. \$\endgroup\$ – Olin Lathrop Apr 14 '16 at 10:30
  • \$\begingroup\$ thanks for your comment. Actually its related to Current only as I have tested it practically. Applying 5v 1A drives the Motor 'ON' while 5v 10mA keeps the Motor in 'OFF' state. Reason behind using DAC 10v is just to drive the Motor at its High Speed. Motor can be drived directly via Microcontroller if 5v 1A is provide. Also please excuse me for Non-Technical terms as I m coming back to electronics after long time. \$\endgroup\$ – skg Apr 15 '16 at 4:32
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I'd consider this as a good starting point: -

enter image description here

I'd also use a Vsupply voltage of 6V and a rail-to-rail op-amp. With 6V at the collector and 5V at the emitter, the power dissipated in the transistor is about 1 watt and so a small heatsink will be required - I'd use a T0-220 packaged transistor.

If you expect to be able to deliver 1A to a 3V output then power dissipation will be 3 watts so you need to think about this but, if the current draw is proportional to output voltage then 3V will require 0.6 A and a max power dissipation in the transistor of 1.8 watts.

If you require to go exactly down to 0V then a negative 0.2 volt (or more negative) power rail on the op-amp is a good idea.

Also, I prefer using a BJT rather than a MOSFET for this because to adequately turn on a MOSFET to output 1 A at 5V might need a top supply voltage that is around 8V to 10V and power dissipation will be significantly higher when 1A is flowing to the load.

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You can use high current OP AMP, for example, LT1210. High current OP AMPS typically have current feedback, they are a little tricky to use. You shall need to design PCB carefully for such ICs.

In case linearity requirements are not high - you can use BJT with sufficient current ability. Use Common Collector circuit.

In any case, the power supply voltage must be higher than 5 V.

The design with common OPA and BJT (from the next answer) is good!

However: if your load is capacitive, this circuit easily becomes unstable. You can add several passive components to make it stable:

schematic

simulate this circuit – Schematic created using CircuitLab

Make sure to chose the suitable OP AMP and BJT according to current requirements.

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    \$\begingroup\$ Ouch. LT1210= $13 in small quantities. \$\endgroup\$ – JRE Apr 13 '16 at 12:11
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What I have done is to just use a Nmosfet and connect it to the load as low side switch. You can use the same idea, but with Pmos if you need a high side.

I think you can save an op-amp like that and probably additional supply circuits. Currently I'm using this circuit for a LED driver and it's been already tested according to industrial standards. You can use almost any NMOS in the market just double check the maximum drain current to be higher than your 1.5 A.

R3 is for discharching the NMOS capacitance whenever your micro is restarted, R2 is to limit the NMOS gate capacitor charging current, so make sure to have them.

schematic

simulate this circuit – Schematic created using CircuitLab

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