Hi Guys,

I have a question regarding output drive current of an opamp. I have LM324 hooked up with R1=100k and R2=10k non-inverting configuration. I am using LM324 and running simulations on Multisim with Vcc=5V and input to the opamp being digital 5V or 0V, I am trying to obtain a current of 1mA. But, the simulation shows that the output current is around 41mA.

I am attaching images of circuits I had designed on Multisim. please let me know If I have them right. Any suggestions or improvements most welcome. I am trying to achieve the specification to drive a manufactured PCB which says "Signal can be supplied (to the PCB I mentioned) directly from a PLC (6mA if the output from the PLC is 24V or 1mA if the output is 5v)"

On an other note, can someone help me with a voltage regulator circuit to step down 24V to 5V (0.5A) using 7805? I am trying to replace a blackberry adaptor (which supplies 5V 0.5A) with the voltage regulator that supplies the same rating.

Thanks a bunch!!

Best,
V

 

The output current will depend on the output voltage and the load. In your setup then you measure current the output is shorted to ground. And the setup is pointless. Remember that a ampere meter act as short. This is true also for the meter in your simulator. Then you measure output current from a source. You first apply a normal load for the circuit. Then you insert the meter in the loop. If you try your setup with a powerful power source like a car battery, or even worse the mains connector. You will fry your instrument very badly.

 

thank you for your reply. The load is actually a PCB manufactured by a company and the specification says that it needs 1mA current at 5V to drive the PCB to work. I really appreciate it. Would the third image setup work? I cannot directly connect the output of the opamp to the PCB inputs because it exceeds the current requirement. Hence, I placed a resistor to draw only the required current from the output of the opamp (which is the source for the PCB). Am I right here?

Thanks!

V

 

Also, On the PCB, I measured the impedance at the input terminals using a voltmeter. It shows 10k impedance at the inputs. So, would that impedance take care of the current rating? Or would it be safe to still put in a resistor before I connect to the input terminals of the PCB.

V

 

On another note, can someone help me with a voltage regulator circuit to step down 24V to 5V (0.5A) using 7805? I am trying to replace a blackberry adaptor (which supplies 5V 0.5A) with the voltage regulator that supplies the same rating.

If you try to use a 7805 regulator for your project, it will act as a small room heater.

Power dissipation in the regulator:
P=EI
P=(24v-5v)*0.5A
P=19*.5
P=9.5 Watts

Power dissipation in the load:
P=EI
P=5v*0.5A
P=2.5W

Total power dissipation: 9.5W+2.5W = 12W
Efficiency: 2.5W/12W = 20.83% - not good.

Look at something like an LM2675-5.0 simple switcher. You can get up to 96% efficiency. That IC, three capacitors, a diode and an inductor are all you need.

 

Thanks sgtwookie.... I found a similar switcher and I also happen to have it with me right now. I have a LM25574QMT by National Instruments. I am assuming that would work.

V

 

Thanks sgtwookie.... I found a similar switcher and I also happen to have it with me right now. I have a LM25574QMT by National Instruments. I am assuming that would work.

That's a National Semiconductor product. National Instruments makes software.

The LM25574 has an integrated N-ch MOSFET that has a 750m Ohm Rds(on), which means that you will have 187.5mW power dissipation in the IC itself.

Very generally, you are better off to use semiconductor products that are rated for 2x or better current/power than you require, which is why I suggested the LM2675-5.0.

 

Oh I see now. Thanks! Also would you mind sharing your thoughts on my previous question as well? I would really appreciate it. It has been a long time since I have done stuff on electronics and circuitry and I am having simple (or stupid) and basic doubts.

V

 

In your first posting you said you needed 1mA 5 volt. I think this is the current the input will draw from the source. The current will regulate it self. You can use a much simpler arrangement.

 

Oh I see now. Thanks! Also would you mind sharing your thoughts on my previous question as well? I would really appreciate it. It has been a long time since I have done stuff on electronics and circuitry and I am having simple (or stupid) and basic doubts.

t06afre's basically covered it.

If you wanted to, I suppose you could include a 1k resistor between the output of your opamp and the input to the PCB; that would limit the max current output of the opamp to 5mA. However, if the board needs 1mA current at 5v, the opamp would only supply 4v due to the drop across the 1k resistor.

You could somewhat compensate for that by moving R2's connection from the opamp output to the actual output; the current limit will then be determined by R3 and Vcc.

Note that real LM324's can't output higher than about Vcc-1.5v even at minimal current output. if you need 5v out, you'll need at least 6.5v for Vcc.