# How to calculate this output(load) current of this differ. amp ckt?

Discussion in 'Homework Help' started by SandiegoSD, Oct 12, 2012.

1. ### SandiegoSD Thread Starter New Member

Oct 12, 2012
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Hi guys, I'm trying to tackle a precision current source design. This is an application note i found from AD. The one shown above is a simplified version of this:
And the I(output) =
with Rg1 = Rg2 = Rf1 = Rf2 for the complicated one. Source here:

http://www.analog.com/library/analogDialogue/archives/43-09/current_source.html

I'm just trying to start with the simpler one I showed up top but can't even get past that. Could anyone enlighten me on that?

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Show as you work. Do yo use nodal analysis ?

Oct 12, 2012
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4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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To solve for IL current we need use a circuit theory.
The circuit diagram look like this

As you can see load current is equal to

VL/RL = I4 + I5 = (Vp - VL)/R4 + (Vout - VL)/R5

So we need to know Vp and Vout first.

So I write nodal equation for Vp.

(Vin - Vp)/R3 = (Vp - VL)/R4

and I solve for Vp.

Vp = Vin*R4/(R3+R4) + VL*R3/(R3+R4) = (R4*Vin + R3*VL)/(R3 + R4)

Next I write for node Vn

Vn/R1 = (Vout - Vn)/R2

Solve for Vout

Vout = Vn * (R1+R2)/R1

Notice that if we assume ideal op amp we can assume Vp = Vn

So we can write
Vout = Vp * (R1+R2)/R1 = ( Vin*R4/(R3+R4) + VL*R3/(R3+R4) )*(R1+R2)/R1

Now we can back to our first equation

VL/RL = (Vp - VL)/R4 + (Vout - VL)/R5

we substitute and we get this

$\frac{VL}{RL}=\frac{\frac{R4*Vin + R3*VL}{R3 + R4} - VL}{R4}+\frac{(\frac{R4*Vin + R3*VL}{R3 + R4}*\frac{R1+R2}{R1})-VL}{R5}$

And all you need to do is to solve this good luck

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5. ### SandiegoSD Thread Starter New Member

Oct 12, 2012
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Thank you Jony for the detailed illustration. Following your eqns I got for the current I= (R5+2R4)/(R5*(Rload + 2R4)) * Vin
assuming R1= R2=R3=R4= 40Kohm, as with that specific amp. This result turns out to be the same as what they(Analog Devices) have for this circuit

source:figure 2. http://www.analog.com/library/analogDialogue/archives/43-09/current_source.html
Now that's a bit confusing cuz this circuit has an extra transistor but somehow they get the same answer for the current as the circuit you used for illustration.
But thank you for the detailed answer. That's a great start as I haven't worked with analog for a while now.

Last edited: Oct 15, 2012
6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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You shouldn't be surprised by that. The BJT in this diagram work as a emitter follower (voltage follower), so the Vout voltage want change. Also negative feedback also helps here.
The only purpose of this BJT is to increase the op amp output current capabilities.

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7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Also I want to add that if we assume R1 = R3 = R4 = R5 = R and R2 = 2R then
Iout/Vin = transconductance (Gm) = 2/R

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8. ### SandiegoSD Thread Starter New Member

Oct 12, 2012
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Yes.. the good old emitter follower! but I have some doubts about the bjt i'm using
I'm trying to bump up the load current(output current) to 500mA range. I did some simulation in spice and it seems the Iβ is only 4.74mA (green circle)

So I'm looking at a gain (DC hfe) of about 500/4.7=~100 on the bjt. I'm putting TIP31A here only temporarily . but it seems like (from the TIP31A datasheet) at 25°C with Ic = 0.5A, the hfe is only around 40~50. (apparently the spice model used in the simulation software isn't that comprehensive)

My Vce is ~12V while in this graph it's 4V, but I don't think the gain is that much different (I'd need hfe 100).
So I'm wondering, should I look for a bjt with large(>100) hfe at 500mA ? Is it true that the larger hfe the merrier it would be for me?

Last edited: Oct 19, 2012
9. ### SandiegoSD Thread Starter New Member

Oct 12, 2012
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Maybe one with gain like this one below?

That hfe is definitely over 100.

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
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You should try using a Darlington stage for more current gain. Or maybe you shoudl use a MOSFET (BUZ11) instead of a BJT.
Also don't forget about power dissipation in the transistor and in R1 resistor.

Ptot = (Vcc - Vout) * Iout = (22V - 10V)*0.5A = 6W

P = I^2*R = 0.5^2 * 20Ω = 5W

You have a lot of power to dissipate, so as you can see you need a heatsink for your transistor and for resistor too.

Last edited: Oct 20, 2012
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11. ### SandiegoSD Thread Starter New Member

Oct 12, 2012
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Yes. I'd need heat sinks definitely. Also I guess by dialing down the voltage I could take some burden off the sinks

I replaced the bjt with a MOSFET
There's only BUZ21 in the simulator , though it has the same threshold voltage as BUZ11. Looks like the idea will work except for one comment I saw somewhere else regarding gate voltage on CMOS: that Vgate has to be relatively high? or maybe it was talking about Vgs on a MOSFET. I'm not sure. I thought I only need to exceed the threshold voltage

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
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If you worried about Vgs then use logic level mosfet, for example IRL530.

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