# Current-to-Voltage Converter

• posted

Hi all,

Does anyone know of a circuit that converts a (linear) current into an exponential voltage of the form V = e^(x) -1 (-1 term necessary to start at 0V).

I've looked at a Sharp application note (vorlon.case.edu/~flm/eecs245/ Datasheets/Sharp%20photodevices.pdf) where on page 6 they have a simple current - to - voltage conversion using photo transistor and op amp. The graph (fig 13B) indicates a somewhat exponential output Voltage - has anyone used this circuit before?

I appreciate the input is light intensity and not current, but was hoping to replace photo transistor with BJT.

Cheers, Ozzy

• posted

to

The following article shows a couple of ways of doing it with diodes and BJT's, but exponentiation is inherently difficult to achieve while maintaining stability against temperature changes and component variability.

-- Joe

• posted

to

I don't think that will be exponential... it looks linear down near the origin.

The Ebers-Moll equation gives you an exponetial relation between collector current and base-emitter voltage. Maybe you could get Vbe from some I*R drop and then turn Ic into a voltage with another resistor.

Perhaps there is a clever way to do that.

George H.

• posted

The current in a diode is exponential on voltage. So a current-voltage converter could be followed by a diode-based exponentiator.

These used to popular for doing analog math, and there were lots of ADI and Burr-Brown app notes, including temperature compensation and such.

What's the application?

John

• posted

to

John,

An Antilog circuit was described in a national app note. The problem is it requires an input voltage, hence an extra stage and would then require further amplification for my use... seems long winded. I don't have an application in mind at the moment, but just wanted to know if it was possible.

Ozzy

• posted

The simple method is to use a monolithic matched pair with their emitters connected together, drive a constant current through one, and use the collector of the other to make an exponential current source. Then you just shove that into a TIA and you're done:

*----R2R2----* | | | |\ U1 | | | \ | *--------------*----|- \ | | | >---*---0 Out | *---|+ / | +Vref | | / | | GND |/ I_C1 | | R V | 1 I_C1=exp(I_in*R3*e/kT)*(Vref/R1) | R | 1 V_out=R2*I_C1 | | |\ | | | \ U2 I_in | *----------*--------|- \ -> Q1 | | Q2 | | >---* |/ \| | *---|+ / | 0------*---| |----* --- | | / | | |>
• posted

If you want a circuit that just works, use a PIC. Unless you are after really crude approximation, an analog solution won't be neither simple nor accurate.

Vladimir Vassilevsky DSP and Mixed Signal Design Consultant

• posted

ry to

2R2----*

=A0 =A0 =A0 =A0 =A0|

=A0|\ =A0U1 =A0|

=A0| \ =A0 =A0 |

| =A0 >---*---0 Out

/

R3*e/kT)*(Vref/R1)

1

=A0 =A0 =A0 =A0 |\

=A0 =A0 =A0 =A0 | \ =A0U2

=A0 =A0| =A0 >---*

=A0*---|+ / =A0 =A0|

=A0 |

=A0 =A0 =A0 =A0 =A0 =A0 =A0|

=A0 =A0 =A0 =A0 =A0|

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|

*
t

Phil, thanks for the circuit diagram. Just a few questions though:

Would this work for currents in the micro amp range - i'm talking 2uA?

What does TIA stand for?

What font is best to view your circuit?

My thoughts were to get an exponential curve which could then be amplified via an op amp, to get up to 9V. Would a micro power op amp with it's low slew rate seriously distort the output? I guess I could just simulate.

Cheers, Ozzy

• posted

Might be a bit of an issue if you expect it to operate at 100MHz though..

• posted

View in any fixed-pitch font such as Courier.

A microamp is in the sweet spot, if you don't expect too much speed. The extrinsic emitter resistance becomes important above a few hundred uA.

You haven't said what you want it for, but I'm guessing some audio effect. You'll definitely want to use a FET input op amp, but I wouldn't do micropower unless you have to.

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal Consultant```
• posted

If you're converting a current to a voltage with R3 to drive the differential pair, wouldn't you want to use the opposite input? The input to Q1 is low impedance.

• posted

I_C1 is within +-2 orders of magnitude of I_C2, but you can do a wider

If you're using R3 to convert an input current to a voltage to drive the differential pair, wouldn't you want to use the opposite input? The input to Q1 is low impedance.

• posted

ry to

2R2----*

=A0 =A0 =A0 =A0 =A0|

=A0|\ =A0U1 =A0|

=A0| \ =A0 =A0 |

| =A0 >---*---0 Out

/

R3*e/kT)*(Vref/R1)

1

=A0 =A0 =A0 =A0 |\

=A0 =A0 =A0 =A0 | \ =A0U2

=A0 =A0| =A0 >---*

=A0*---|+ / =A0 =A0|

=A0 |

=A0 =A0 =A0 =A0 =A0 =A0 =A0|

=A0 =A0 =A0 =A0 =A0|

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|

*

t- Hide quoted text -

That's nice! Cute idea for temperature control of the differential pair.

George H.

• posted

an

sary to

5/

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-R2R2----*

=A0 =A0 =A0 =A0 =A0 =A0|

=A0 =A0|\ =A0U1 =A0|

=A0 =A0| \ =A0 =A0 |

=A0 | =A0 >---*---0 Out

• /
/

n*R3*e/kT)*(Vref/R1)

_C1

=A0 =A0 =A0 =A0 |\

=A0 =A0 =A0 =A0 | \ =A0U2

=A0 =A0| =A0 >---*

=A0*---|

• /
=A0 =A0|

=A0 =A0 |

=A0 =A0 =A0 =A0 =A0 =A0 =A0|

=A0 =A0 =A0 =A0 =A0 =A0|

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|

```--
> > =A0 =A0 =A0 =A0 =A0 =A0 |
> > =A0 =A0 =A0 =A0 =A0 =A0 |```
• posted

Not sure I understand. R3 will drop several tens of millivolts at most, which is usually within the meaning of 'low impedance' at a given current level. You want a low input impedance when the input is a current.

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal Consultant```
• posted

I snitched it from an old PMI app note, and have been using it for 20 years or so. It's in every noise canceller I build.

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal Consultant```
• posted

an

sary to

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---R2R2----*

=A0 =A0 =A0 =A0 =A0 =A0|

=A0 =A0|\ =A0U1 =A0|

=A0 =A0| \ =A0 =A0 |

=A0|

=A0 =A0 |>---*---0 Out

---|+ /

| /

=A0|/

(I_in*R3*e/kT)*(Vref/R1)

R2*I_C1

=A0 =A0 =A0 =A0 =A0 |\

=A0 =A0 =A0 =A0 =A0 | \ =A0U2

\

=A0 =A0 =A0 =A0|>---*

=A0 =A0*---|+ / =A0 =A0|

=A0 =A0 |

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|

C1 =A0 =A0 =A0 =A0 =A0 =A0|

=A0| =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|

------*

R4

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t- Hide quoted text -

Simpler to just drive the gain input of an AGC amp- they have gone to great lengths to do all the temperature and feedback compensation- a single package solution and cheap. "Easy enough for a caveman" type of thing...

• posted

If the equation for the differential input voltage is given by deltaVbe = Vt*log (Iref/IC_1), then since the base of Q2 is at ground it looks like from your equation the deltaVbe is meant to be created by the input current at I_in against Q1's base resistor. But as current through Q1 increases, and since the current through Q2 is fixed, the output voltage of the op amp across R4 will decrease to "suck away" the excess current.

It seems like an increasing voltage at Q1's base while the output voltage of the op amp is decreasing to compensate for the increased current would make Q1's input low impedance, and I'm wondering if that needs to be taken into account when sizing R3 in your equation I_C1=exp(I_in*R3*e/kT)*(Vref/R1).

Since the current through Q2 is fixed by feedback, if the voltage at Q2's base increases or decreases the output voltage of the op amp will follow right along, making that input high impedance.

• posted

Q1's emitter potential is fixed by feedback to be the bias required for Q2 to sink Vref/R1, no more and no less, completely independent of anything Q1 does (within reason). Its base is shunted by a low resistance to ground. How is that going to be high impedance????

Try spicing it if you don't believe me.

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal Consultant```
• posted

I'm sorry if I'm not explaining myself well - it's Q1 that I'm suggesting has a lower input impedance than Q2, if we can take R3 out of the picture for a moment. I will try it in Spice and see what I get.

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