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Diode Simulation Question

 Hello,

 

I am having a problem understanding what is happening when I try to simulate a photodiode going into a transimpedance amplifier. 

Why does a simulation show a very high negative voltage when I try to use the interactive DC current source? See the yellow area in the schematic below. 

 

The diode I am try to simulate is a Perkin-Elmer VTB1113. It should output current in the range of 0 to 60 uV. In the simulation below, the current is approximately correct (source 21uV and the probe #1 shows approximately 19.6uV. But what is going on with the voltage that appears at Probe #1? A negative 9,780 V ??? It must be some side effect of the simulation math in MultiSim14, but what to do about it? I tried placing a 14V Zener diode on it and it did clamp the voltage, but the current then went whacko. 

 

Can I ignore this very high voltage as the current produced is approximately correct?

 

Any ideas or comments? Thanks in advance.

 

MS14-PhotoDiodeSimulation.jpg

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Hello dwen,

 

I have encoutered similar issues on the past with opamps, giving high voltages without reason. This is is usually due the spice model. Can you attach your files to review them.

 

Regards,

Randy @Rscd27@
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Hi Randy,

 

Thanks for your response. Attached is the net list file for you to look at.

 

Thanks,

Doug

 

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Hello guys

 

first of all - there are mistakes in schematic -

 

check out polarity of your negative battery and C7, C4.

 

second one - why do you need all these blocking caps for simulation?

 

third one - what is the purpose od r4 and c5??

 

4 - all your block caps are electrolitic - it is a nonsense.

 

5 - what is a purpose of R2 and why it is so high?

 

6 - value of R1 - looks too high to me.

my calculations give me that =

current = 2.1 *10pow-5    A

voltage  = 10 v(opamp max V out)

R1 = 10v / 2.1 *10pow-5  = around 500 KOhm;

 

7 - your Rl\L is too high too.



with hope it helps
good luck
thanks a lot
Michael

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The 0.021 mA current source driving 7 Gohms in parallel with 501 Mohms is 9800 V.

 

Lynn

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Thanks for your response Michael and Lynn,

 

Let explain some more about this circuit. This circuit is part of an instrument that has been on the market for 5 years. It appearently works. The designer of this circuit abruptly left the company about 4 years ago and either took or discarded all documentation about this circuit. I have been asked to try to restore information on how this circuit works. That is why I am trying to simulate it to see what it does under circumstances.

 

Michael, the opamp circuit is a transimpedance amplifier. I read that it is suppose to convert current into a voltage. C1 and R1 are the actual values used and I've checked them on the circuit board. 

 

The simulated diode was taken from some app notes from Vishay, TI, and Analog Devices similar to the VTB1113 device. R2 is the shunt resistance taken from the VTB1113 datasheet. Likewise C3 is the PN junction capacitance. The photodiode is a current source which generates current proportionally to the illumination. The polarity of V1 may be incorrect, but it is set to -27V on the positive looking terminal. The current range from 0 to 60uA. Attached is the actual circuit I am trying to understand.

 

RL makes no difference. I have remove it and the simulated outputs are the same with it or without it. AYou are right, I probably don't need the bypasses in a simulation.

 

Lynn, you're correct according to ohms law. You addressed my qquestion exactly, Lynn. The simulation math maybe correct, but the simulation of the pdiode is not true. If I measure the photodiode on the circuit board it never goes above approx -25V. I dare say that a circuit containing -27V, +12V, and -12V could never produce -9.78 kV. 

 

Thanks all for your input. I have emailed Randy at NI a copy of the SPICE net. I may try this same circuit in TI's TINA, but I don't like it as much as MultiSim.

 

Doug

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Hello guys, thanks for response

 

there are a few things -

I did not say polarity of V1 is wrong, I said polarities  of  N12V and C4 and C7 are wrong. - take a look at your schematic.

 

C5 is connected reversed if V1 is -27 v.. actually you do not need no R4 and no C5 for simulations.

 

the diode data sheet does mention shunt resistance of 7G but it has nothing to do with that case.

 

Lynn was right but just partially, because diode's current does not flow throught R2, it flows from OPAMP pin 2 toward R4 and then to V1

 

there is nothimg wrong with MS, problem is with your schematic -

there is why -

if OPAMP works properly, pin 2 behaves as virtual ground - we can surely assume it's voltage equal to zero. ALWAYS, what ever happens, it is zero.

 

in order to do that , current sum that income and currents sum which outcome should be equal by value and reversed by direction -

Kirchhoff's first law

 

now - in order to have it , current outcomes to diode should be equal to current  through resistor R1. (I assume the OpAmp input current is zero, which actually close to reality - it is 5*10pow-9 A)

 

I_R1 = Vopamp out \ R1.

 

from here  Vopamp_out = R1 * I_R1 = R1 * I_D.

 

that why you are getting 10KV,   - - because R1 is too high.

 

for 20uA  current from  diode and 10 V output from opamp , R1 should be equal =

10V  / 2*10 pow-5  = 5*10pow5 = 500 KOhm.

 

the R7  is useless here because the diode is connected to circuit.

 

francly I am wondering about these 500GOhm resistors.  it is very unusual to see suct in low voltage electronic.

 

I would definitely double check it . which is a  problem by it self too. most resistance meters have top limit as 50 MOhm, and to use megger in that circuit is out of questions because it will fry all components with high voltage.

 

How are sure about these resistors?

 

 


with hope it helps
good luck
thanks a lot
Michael

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The LF347 input bias current is 50 pA, typical. So no more current than that will flow through pin 2 of the op amp. If the op amp is not operating in its linear region, pin 2 may be at any voltage as determined by the external circuit. 

 

Current loops related to I3: 

1. I3, R2.  If this is the only path, R2 has 147 kV across it.

2. I3, R4, V1, RL, U1A output, R1. Note that this is really two or more loops with the U1A output returning to ground through the op amp power supply.

 

If linear operation of the op amp is assumed, then pin 2 is at virtual ground (although it is important to remember that the actual current flows through R1). For I3 = 21 uA the voltage at pin 1 of the op amp must be -I3*R1 = 10500 V. Since the op amp cannot produce voltages that large, the op amp will not be operating linearly and this analysis does not apply. However, it shows that either the photodiode current is much too large or R1 is too large.

 

Resistors of 500 megohms are not uncommon in circuits like this. That value is appropriate for photocurrents less than 20 nA.  If the photocurrents will be as large as 21 uA, reduce R1 to 470 kohms.

 

Another comment on simulations: Many op amp models do not accurately represent the non-linearities which occur when voltages or currents exceed the data sheet limits. The results of the simulation must be analyzed carefully when out of range values are encountered.

 

Lynn

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