# voltage to current converter

Discussion in 'Electronic Design' started by john, Nov 16, 2007.

1. ### johnGuest

Hi,

I am trying to simulate a DC block circuit in PS PICE. I am using the
following practical circuit

http://profile.imageshack.us/user/ai466/images/detail/#69/circuitdiag...

I replaced the voltage to current section with the current source
symbol given in PS pice. Please find the circuit diagram in the
http://img253.imageshack.us/my.php?image=circuitdiagram2gl3.png

I have following questions

1. How can I see the capacitor charging and discharging in PSPICE? I
found the capacitor always charged whenever I run the simulations.

2. What is the difference between voltage to current converter and
constant current source? The INA circuit that I am using could be a
constant current source ( plz. refer link1) and can be replaced with
the current source given in PSPICE. I tried to use the PSpice models
of INA and opa131 but did not work .

3. I am trying to get rid of the DC leakage current produced by the
INA 133 circuit using capacitors. I have discussed this issue in my
previous post but need more advice! There problem is

a. At the power up the DAC outputs DC -2.5 volts. The inverting input
of the INA133 is grounded and the 470uF capacitor is reduing the -2.5
volts to -0.5 volts accross the 384ohm resistor. There is a volage
difference between the inputs which reflects at the output. I solved
this problem partly by using the waveforms that ends at zero. So,
after sending the waveform out of DAC, the DAC holds the zero and no
more leakage. But at power up its an issue.

b. The leakage current is constant and I came up with the circuit
that
I mentioned in the link2 to get rid of it. Any comments!!

Regards
John

2. ### Jan PanteltjeGuest

None basically.
I would like to understand what exactly you mean by 'DC leakage'.
In a previous post I showed you need no capacitor,

So then for a 0 to +ma output current you need -2.5 V on the other input
of the INA.

No output cap either, but your load impedance must be low enough
so the requested current can flow.
If yo uhave a supply of plus and minus 10V, and a 5 k series resistor,
then whatever yo udo,m the output current cannot be more thne +or- 10 / 5000.
If your load is say 1K, then the output current can never be more then
+or- 10 / (5000 + 1000).
So this limits the input voltage range, as I out is U1-U2 / 5000 and
can never be greater then Iout max.

Using the 1M makes no sense as for 1mA you need 1000V.

So, practially the calc goes a bit like this:

U1-U2 = 2.5V, R=5000, Iout = 500uA, for 10V supply Rload total should be
10 / .0005 = 20000, you already have 5000 for R, so Rload must be < 15000 Ohm.

If I did not make any errors, but thisis the system.

3. ### johnGuest

Hi,

The DC leakage current will go into the load without the capacitors.
The -2.5 volts offset is only there at the power up not after the DAC
outputs the waveform. After the DAC outputs the waveform, the voltage
at the non inverting input is zero so leakage gets low but still its
there. I think that DAC zero and the ground zero has some offset and
need trimming but can not do trimming because of PCB issues.

John

4. ### Jan PanteltjeGuest

At power up the caps have charge zero, so the ywill not help,
but perhaps even increase startup surges / currents in teh load.

OK I give up.

5. ### johnGuest

Hi,

The laod is in the range of 10kohm to 200kohm.

John

6. ### johnGuest

Hi,

The resistor in parallel with the current source will not let the
capacitor charged upto +/- 15 volts. so when the power is turned off ,
the capacitor will have less charge to dump in the load. What do you
think?

Regards
John

7. ### Jan PanteltjeGuest

OK, now think for a minute.

The output was (U1-U2) / R, where R was 5000.
Your DAC does 2.5 V or more ful lsignal.
the output current is then 2.5 / 5000 = .5 mA
..5 mA in 200.000 Ohm = 100 Volt.
Go back to start and pay 100\$

8. ### Jan PanteltjeGuest

I was thinking about the other capacitor, it is a short at power up.
But I was also thinking that none of this can work because see my other post.

10. ### johnGuest

Hello,

Would you please confirm two more things

1. What is the output impedance of this voltage to current converter?
Ideally current sources have infinite output impedance.

2. If the voltage across the load is 1000 volts and power supply is
10volts. What would likely to happen
a. the current will not flow
b. the current will flow

Regards
John

11. ### Jan PanteltjeGuest

Ideally, yes.
The output impedance is set by the open loop gain.
The current CANNOT flow, as you need 100V for it to flow.
As you only have a much lower supply vltage, only a lower current
will flow, resulting in the amplifier hanging one way against the supply.

12. ### Jan PanteltjeGuest

Ideally, yes.
The output impedance is set by the open loop gain.
The current CANNOT flow, as you need 100V for it to flow.
As you only have a much lower supply vltage, only a lower current
will flow, resulting in the amplifier hanging one way against the supply.

13. ### JosephKKGuest

Jan Panteltje posted to
sci.electronics.design:
Gosh, a fixed value source versus a controlled source but no
difference? That does not make sense to me. Spice has controlled
sources.

15. ### Jan PanteltjeGuest

look at this in context, he has a controlled source.
Any constant current source can be controlled in some way,
If you really wanted to be pedantic.

F*ck spice.
Reality rules.

16. ### JosephKKGuest

Jan Panteltje posted to
sci.electronics.design:
It seems unlike you to not follow the chain properly.

17. ### Jan PanteltjeGuest

Reality rules.