# current amp help

Discussion in 'The Projects Forum' started by minkey01, Sep 2, 2014.

1. ### minkey01 Thread Starter Member

Jul 23, 2014
185
0
hi. could someone help me with a circuit design? this should be pretty easy, but i'm still learning.

i'm trying to amplify my current from 1 mA to around 20mA.

the control signal is 5V when on and can give about 1 mA. coming from this :

http://www.expert-sleepers.co.uk/es5.html

but needs to trigger this :

http://www.sainsmart.com/4-channel-5v-relay-module-for-pic-arm-avr-dsp-arduino-msp430-ttl-logic.html

my power supply is around 5.3 V. so probably just a simple transistor circuit with a resistor before the transistor limiting the current to the 1mA? and then the 5.3V power into the transistor. could someone sketch up the circuit and list part numbers? so i don't foul this thing up.

thanks a bunch!

2. ### #12 Expert

Nov 30, 2010
17,818
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This looks like a job for the 2N7000.

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3. ### MikeML AAC Fanatic!

Oct 2, 2009
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Here you go. Note that 900uA makes much more than 10mA at the collector.

R2 is just there for an illustration. It can be almost anything greater than 47Ohms....

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4. ### minkey01 Thread Starter Member

Jul 23, 2014
185
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thanks guys. a few questions though.

#12 : do you know what the out current would be for your recommended circuit? also i see a difference in that the out is before the resistor compared to MikeML's.

MikeML : isn't that output current way too high for my application? i'm looking for around 20 mA out. same question as to why the out is after the resistor compared to #12s before the resistor.

thanks so much!

5. ### minkey01 Thread Starter Member

Jul 23, 2014
185
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also what does that 56k resistor to ground do?

6. ### #12 Expert

Nov 30, 2010
17,818
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The 56k resistor keeps the gate from floating if you don't have a ground path through the thing that was supplying 5 volts. The output current is whatever the 150 ohm resistor limits it to after you consider how much voltage the opto-coupler is going to use up and exactly what your supply voltage is. You might need 180 ohms or 220 ohms. Measure the circuit and look at the opto spec sheet.

Last edited: Sep 2, 2014
7. ### wayneh Expert

Sep 9, 2010
13,415
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The output current depends on your load. Once you supply "enough", the load won't care whether you could supply more.

Another way to say what #12 just wrote, the pull-down resistor makes sure the transistor fails to "off" if the control signal fails to open. Doo-doo happens.

#12 likes this.
8. ### #12 Expert

Nov 30, 2010
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Thanks, wayneh.
I seem to be having difficulty connecting with this one.

9. ### minkey01 Thread Starter Member

Jul 23, 2014
185
0
ok thank you. i understand now about stopping the floating gate. great idea.

also now understand that as long as the opto circuit gets enough current it will trigger and not need to draw more. thanks again.

two last issues :

supply voltage is 5.3V and I believe the opto circuit is needing 5V, that is what it is listed at. i dont think there is a spec sheet available. could you give more detail as to what i need to measure exactly to figure out this resistor value (the 180-220 ohm one)?

also why did mike's circuit have the opto connection at this red junction, but yours is on the other side of the resistor? see attached pic.

thank you for the help!

10. ### #12 Expert

Nov 30, 2010
17,818
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The advertisement does not specifically say that it already equipped with a resistor on the opto-isolator, but I think it is. Look on the circuit board inputs and see if there are resistors already installed in the range of 150 to 220 ohms.

If this is true, my circuit will not allow enough current with that resistor in it. You can measure the current the way I drew the circuit. If it is obviously low, take out the 150 ohm resistor that I drew and connect the opto from 5V to the drain of the transistor.
Do not add anything else where you drew the red line.

In Mikes drawing, the 47 ohm resistor represents where the opto-isolator terminals are connected. He is showing the concept. I am showing the circuit. If you can't derive the details required to finish Mikes circuit, try mine.

Last edited: Sep 3, 2014
11. ### #12 Expert

Nov 30, 2010
17,818
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Maybe this is what you need...

I think we've been assuming you are more versatile than you are, talking in short-hand, dealing in symbols. It's hard to think on your level after 40+ years of doing this.

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12. ### minkey01 Thread Starter Member

Jul 23, 2014
185
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ok thank you.

i did find this schematic of one of the relays if that helps.

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13. ### wayneh Expert

Sep 9, 2010
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What more do you need? You need to control the current thru U7 with your ~1mA input?

14. ### #12 Expert

Nov 30, 2010
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Pick one.

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• ###### MikeMl circuit.png
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Last edited: Sep 3, 2014
15. ### minkey01 Thread Starter Member

Jul 23, 2014
185
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slight change in the info here. sorry about this. seems the seller listed the driving amps wrong. the relay only needs about 1.5 mA per channel. it was a misprint. i measured and this is indeed true. so i will not need to amplify anything.

the only thing needed would be to invert the logic of my control signal. and maybe incorporate that pullup to stop the gate from floating.

new circuit from you possible?

thank you. i appreciate the understanding of my noobness.

16. ### minkey01 Thread Starter Member

Jul 23, 2014
185
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found this on the web.

do you approve?

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17. ### minkey01 Thread Starter Member

Jul 23, 2014
185
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ok i'm testing that circuit i just posted.

but using a M 2N3904 i had laying around. i like how the inverter part works, but the current draw is around 2.87 mA. anyway we can get it down to the 1.5 mA?

is it possible for you to edit that circuit i just posted to get 5V and 1.5mA coming out of the collector end (RELAY BOARD INPUT)?

thanks, you guys are the best!

18. ### #12 Expert

Nov 30, 2010
17,818
9,143
First, I just deleted a mistaken entry. All of the circuits will produce, "drive on = relay on".
Use the MikeML circuit and add some resistance in the collector circuit. Probably less than 100 ohms.
However, I don't know why you want to mess about trying to take out 1.35 milliamps as if the relay board designer wanted a 5 volt circuit minus a tiny error caused by the value of R14.

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19. ### wayneh Expert

Sep 9, 2010
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You could eliminate the 10K and increase the 2.2K resistors. Both would reduce current consumption.

20. ### #12 Expert

Nov 30, 2010
17,818
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Difference of opinion here.
I don't like trying to adjust the input resistance to achieve a certain gain because it is temperature dependent. This should be a hard switching action with the current limited in the collector circuit by R14. MikeML demonstrated that a 2N3904 would do the job nicely. If you want to trim a milliamp off, add some resistance in the collector circuit.

My opinion is, "Quit messing with it". You have the 5 volts applied to the relay board. That's what it wants.